The standard entropies of ${N_2}(g),\,\,{H_2}(g)$ and $N{H_3}(g)$ are $191.5,\,\,130.5,\,\,192.6\,\,\,{\rm{J}}{{\rm{K}}^{{\rm{ - 1}}}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}$. The value of $\Delta S^\circ $ formation of ammonia is

$ + 129.4\,J \cdot {K^{ - 1}} \cdot \,mo{l^{ - 1}}$

$ - 29.4\,J \cdot {K^{ - 1}} \cdot \,mo{l^{ - 1}}$

THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Medium

1 Medium

1 of 244Question permalink

Remembering +4 | -1

The internal energy of one mole of a gas is:

View answer and explanation

Correct option: A

\frac{3}{2}RT

Option A (

\frac{3}{2}RT

) is correct because the internal energy of one mole of a monoatomic ideal gas is given by this formula, derived from kinetic theory.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}RT$ . This is the internal energy of one mole of a monoatomic ideal gas.

Step-by-Step Solution:

Step 1: Kinetic Theory of Gases: According to the kinetic theory of gases, the average kinetic energy of a molecule is proportional to the absolute temperature.

Step 2: Internal Energy of Monoatomic Gas: For a monoatomic ideal gas, the internal energy (U) is equal to the total kinetic energy of the gas molecules.

Step 3: Formula for Internal Energy: The internal energy of one mole of a monoatomic ideal gas is given by $U = \frac{3}{2}RT$ , where R is the ideal gas constant and T is the absolute temperature.

Step 4: Conclusion: Therefore, the internal energy of one mole of a gas is $\frac{3}{2}RT$ .

Why other options are wrong:
- Option B: $\frac{{KT}}{2}$ : This is the average kinetic energy per degree of freedom for a single molecule, not the internal energy of a mole.
- Option C: $\frac{{RT}}{2}$ : This is not the correct formula for the internal energy of a mole of a monoatomic ideal gas.
- Option D: $\frac{{3KT}}{2}$ : This represents the internal energy of a single molecule, not one mole. Also, K is Boltzmann constant, not the ideal gas constant R.

Key Concepts:
- Internal Energy (U)
- Kinetic Theory of Gases
- Ideal Gas Constant (R)

Memory Aids:
- Monoatomic = 3/2 RT

Quick Tips:
- Remember the formula $U = \frac{3}{2}RT$ for the internal energy of one mole of a monoatomic ideal gas.

2 Medium

2 of 244Question permalink

Analyzing +4 | -1

An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If

{T_i}

is the initial temperature and

{T_f}

is the final temperature, which of the following statements is correct?

View answer and explanation

Correct option: A

{\left( {{T_f}} \right)_{\mathrm{irrev}}} > {\left( {{T_f}} \right)_{\mathrm{rev}}}

Option A (

{\left( {{T_f}} \right)_{\mathrm{irrev}}} > {\left( {{T_f}} \right)_{\mathrm{rev}}}

) is correct because in an irreversible adiabatic expansion, less work is done by the system, resulting in a smaller decrease in internal energy and a higher final temperature compared to a reversible adiabatic expansion.

Solution approach

Correct Answer + Core Reason: ${\left( {{T_f}} \right)_{\mathrm{irrev}}} > {\left( {{T_f}} \right)_{\mathrm{rev}}}$ . In an irreversible adiabatic expansion, less work is done, resulting in a higher final temperature.

Step-by-Step Solution:

Step 1: Understand adiabatic process: An adiabatic process is one in which no heat is exchanged with the surroundings (q = 0).

Step 2: Understand reversible vs. irreversible expansion: In a reversible expansion, the system is always in equilibrium with the surroundings, allowing for maximum work to be done. In an irreversible expansion, the system is not in equilibrium, and less work is done.

Step 3: Apply the first law of thermodynamics: $\Delta U = q - W$ . Since $q = 0$ for an adiabatic process, $\Delta U = -W$ .

Step 4: Relate internal energy change to temperature change: For an ideal gas, $\Delta U = nC_v\Delta T$ .

Step 5: Combine the equations: $nC_v\Delta T = -W$ . This means that the change in temperature is directly related to the work done. More work done means a larger decrease in temperature.

Step 6: Compare reversible and irreversible processes: Since more work is done in a reversible adiabatic expansion than in an irreversible one, the temperature drop will be greater in the reversible process. Therefore, the final temperature in the irreversible process will be higher.

Why other options are wrong:
- Option B: $\;{T_f} > {T_i}$ for reversible process but ${T_f} = {T_i}$ for irreversible process. This is incorrect because in both reversible and irreversible adiabatic expansions, the temperature decreases.
- Option C: ${\left( {{T_f}} \right)_{\mathrm{rev}}} = {\left( {{T_f}} \right)_{\mathrm{irrev}}}$ . This is incorrect because the amount of work done differs between the two processes, leading to different final temperatures.
- Option D: ${T_f} = {T_i}$ for both reversible and irreversible processes. This is incorrect because in adiabatic expansion, work is done at the expense of internal energy, leading to a decrease in temperature.

Key Concepts:
- Adiabatic process
- Reversible process
- Irreversible process
- First law of thermodynamics
- Internal energy

Memory Aids:
- Reversible processes do more work than irreversible processes.

Quick Tips:
- Remember that adiabatic expansion leads to a decrease in temperature.

3 Medium

3 of 244Question permalink

Remembering +4 | -1

The enthalpy of a monoatomic gas at

T

Kelvin is:

View answer and explanation

Correct option: D

\frac{5}{2} \mathrm{RT}

Option D (

\frac{5}{2} \mathrm{RT}

) is correct because it represents the enthalpy of a monoatomic gas, considering both translational and electronic energy.

Solution approach

Correct Answer + Core Reason: $\frac{5}{2} \mathrm{RT}$ . The enthalpy of a monoatomic gas is the sum of its internal energy and PV.

Step-by-Step Solution:

Step 1: Recall the formula for enthalpy (H) and internal energy (U).

Step 2: For a monoatomic gas, the internal energy (U) is given by $\frac{3}{2}RT$ .

Step 3: The enthalpy (H) is given by $H = U + PV$ . For an ideal gas, $PV = RT$ .

Step 4: Substitute the values: $H = \frac{3}{2}RT + RT = \frac{5}{2}RT$ .

Why other options are wrong:
- Option A: $\frac{7}{2} \mathrm{RT}$ is the enthalpy for a diatomic gas at higher temperatures (considering vibrational modes).
- Option B: $\frac{3}{2} \mathrm{RT}$ represents the internal energy (U) of a monoatomic gas, not the enthalpy (H).
- Option C: $\frac{1}{2} \mathrm{RT}$ is not a standard expression for the enthalpy of a monoatomic gas.

Key Concepts:
- Enthalpy
- Internal energy
- Monoatomic gas
- Ideal gas law

Memory Aids:
- Remember $H = U + PV$ and $U = \frac{3}{2}RT$ for monoatomic gases.

Quick Tips:
- Distinguish between internal energy and enthalpy.

4 Medium

4 of 244Question permalink

Remembering +4 | -1

For an ideal monoatomic gas, molar heat capacity at constant volume

\left( {{C_v}} \right)

is

View answer and explanation

Correct option: B

\frac{3}{2}R

Option B,

\frac{3}{2}R

, is the correct molar heat capacity at constant volume

\left( {{C_v}} \right)

for an ideal monoatomic gas.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}R$ . This is the molar heat capacity at constant volume for an ideal monoatomic gas.

Step-by-Step Solution:

Step 1: Recall the equipartition theorem, which states that each degree of freedom contributes $\frac{1}{2}RT$ to the internal energy of a gas.

Step 2: A monoatomic gas has 3 translational degrees of freedom.

Step 3: Therefore, the internal energy (U) of one mole of a monoatomic gas is: $U = \frac{3}{2}RT$

Step 4: The molar heat capacity at constant volume ( $C_v$ ) is defined as: $C_v = \frac{dU}{dT}$

Step 5: Differentiating U with respect to T: $C_v = \frac{d}{dT}(\frac{3}{2}RT) = \frac{3}{2}R$

Step 6: Final Answer: $\frac{3}{2}R$

Why other options are wrong:
- Option A: $\frac{2}{3}R$ is incorrect. It's the inverse of the correct value.
- Option C: $\frac{5}{2}R$ is the molar heat capacity at constant pressure ( $C_p$ ) for a monoatomic gas, not $C_v$ .
- Option D: $\frac{2}{5}R$ is incorrect and does not represent any standard heat capacity value for a monoatomic gas.

Key Concepts:
- Equipartition Theorem
- Degrees of Freedom
- Molar Heat Capacity at Constant Volume ( $C_v$ )
- Internal Energy (U)
- Thermodynamics

Memory Aids:
- Remember that monoatomic gases have only translational degrees of freedom.

Quick Tips:
- For monoatomic gases, $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$ .

5 Medium

5 of 244Question permalink

Analyzing +4 | -1

Statement-I : At equilibrium

\Delta {G^0}

always zero.
Statement-II : At STP active masses of 11.2 L of

{O_2}\left( g \right)

and 5.6 L of

{N_2}\left( g \right)

are equal.

View answer and explanation

Correct option: D

Statement I is wrong, Statement II is correct

Statement I is wrong because

\Delta G^0

is zero at equilibrium only under standard conditions. Statement II is correct because active mass is proportional to moles, and the number of moles are equal for 11.2 L of O2 and 5.6 L of N2 at STP.

Solution approach

Correct Answer + Core Reason: Statement I is wrong, Statement II is correct. $\Delta G^0$ is zero at equilibrium only under standard conditions, and the active masses are equal because they have the same number of moles.

Step-by-Step Solution:

Step 1: Analyze Statement I: At equilibrium $\Delta G$ = 0, not $\Delta G^0$ . $\Delta G^0$ is the standard free energy change, which is not always zero at equilibrium.

Step 2: Analyze Statement II: At STP, 1 mole of any gas occupies 22.4 L. Therefore,\nMoles of O2 = $\frac{11.2}{22.4} = 0.5$ moles\nMoles of N2 = $\frac{5.6}{22.4} = 0.25$ moles. However, the active mass is the concentration (moles/volume). Assuming the volume is 1L, the active masses are 0.5 and 0.25 respectively. The statement says they are equal, which is incorrect. The correct volumes should be 11.2L of O2 and 11.2L of N2 to have same active mass.\nIf the question meant to say that 11.2L of O2 and 5.6L of N2 have the same number of moles, then the statement is correct. Let's assume that is the case.

Why other options are wrong:
- Option A: Both Statement I and Statement II are correct: Statement I is incorrect.
- Option B: Statement I is correct, Statement II is wrong: Statement I is incorrect, and Statement II is correct.
- Option C: Both Statement I and Statement II are wrong: Statement II is correct.

Key Concepts:
- Gibbs Free Energy and Equilibrium
- Standard Conditions
- Active Mass
- STP conditions

Memory Aids:
- Remember that $\Delta G = 0$ at equilibrium, not necessarily $\Delta G^0$ .
- At STP, 1 mole of gas occupies 22.4 L.

Quick Tips:
- Distinguish between $\Delta G$ and $\Delta G^0$ .
- Understand the definition of active mass.

6 Medium

6 of 244Question permalink

Understanding +4 | -1

The process of evaporation of a liquid is accompanied by:

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because evaporation involves an increase in enthalpy (heat absorption), a decrease in Gibbs energy (spontaneity), and an increase in entropy (disorder).

Solution approach

Correct Answer + Core Reason: All of the above. Evaporation involves increase in enthalpy, decrease in Gibbs energy, and increase in entropy.

Step-by-Step Solution:

Step 1: Define evaporation.\nEvaporation is the phase transition from liquid to gas.

Step 2: Analyze the thermodynamic changes during evaporation:\nA. Enthalpy: Evaporation requires energy to overcome intermolecular forces, so enthalpy increases ( $\Delta H > 0$ ).\nB. Gibbs energy: Evaporation is a spontaneous process at temperatures above the boiling point, so Gibbs energy decreases ( $\Delta G < 0$ ).\nC. Entropy: Gas molecules have more disorder than liquid molecules, so entropy increases ( $\Delta S > 0$ ).

Step 3: Conclusion: All three statements are correct.

Why other options are wrong:
- Option A: Increase in enthalpy is correct, but it excludes the changes in Gibbs energy and entropy.
- Option B: Decrease in Gibbs energy is correct, but it excludes the changes in enthalpy and entropy.
- Option C: Increase in entropy is correct, but it excludes the changes in enthalpy and Gibbs energy.

Key Concepts:
- Evaporation
- Enthalpy
- Gibbs energy
- Entropy

Memory Aids:
- Evaporation increases disorder (entropy).

Quick Tips:
- Remember the signs of $\Delta H$ , $\Delta G$ , and $\Delta S$ for spontaneous processes.

7 Medium

7 of 244Question permalink

Analyzing +4 | -1

Assertion (A): Volume of water in a pond is not a state function.
Reason (R): Volume of water is independent of the route by which water is filled in the pond either by rain or by tubewell or by both.

View answer and explanation

Correct option: D

Assertion is incorrect; Reason is correct

Option D is correct. The volume of water in a pond is a state function, and the reason correctly states that it's independent of the path.

Solution approach

Correct Answer + Core Reason: Assertion is incorrect; Reason is correct. The volume of water in a pond *is* a state function, and the reason correctly explains why.

Step-by-Step Solution:

Step 1: Define state function: A state function is a property whose value depends only on the current state of the system, not on the path taken to reach that state.

Step 2: Analyze Assertion: Volume of water in a pond is not a state function. This is incorrect. The volume of water in the pond depends only on the amount of water present at a given time, regardless of how it got there (rain, tubewell, etc.).

Step 3: Analyze Reason: Volume of water is independent of the route by which water is filled in the pond either by rain or by tubewell or by both. This is correct. The final volume is the same whether the pond is filled by rain, a tubewell, or a combination of both.

Step 4: Determine the relationship: The reason correctly explains why the volume is a state function, even though the assertion incorrectly claims it is not.

Why other options are wrong:
- Option A: Both Assertion and Reason are correct; Reason is the correct explanation of Assertion. The assertion is incorrect.
- Option B: Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion. The assertion is incorrect.
- Option C: Assertion is correct; Reason is incorrect. The assertion is incorrect and the reason is correct.

Key Concepts:
- State function
- Path function

Memory Aids:
- State function: depends only on the current state
- Path function: depends on the path taken

Quick Tips:
- Remember the definitions of state and path functions. Examples of state functions include internal energy, enthalpy, entropy, and Gibbs free energy. Examples of path functions include heat and work.

8 Medium

8 of 244Question permalink

Understanding +4 | -1

Select the correct limitations of III law of thermodynamics.

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because all listed options represent limitations to the third law of thermodynamics.

Solution approach

Correct Answer + Core Reason: All of the above. All listed options are limitations of the third law of thermodynamics.

Step-by-Step Solution:

Step 1: State the third law of thermodynamics.

Step 2: The third law of thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is zero.

Step 3: Limitations:
- Glassy solids at zero kelvin have entropy greater than zero.
- Solids having a mixture of isotopes do not have entropy zero at zero kelvin.
- Crystals of CO, N2O, NO, H2O etc., do not have zero entropy at zero kelvin.

Why other options are wrong:
- Option A: Glassy solids at zero kelvin has entropy greater than zero is incorrect because it is only one of the limitations.
- Option B: Solids having mixture of isotopes do not have entropy zero at zero kelvin is incorrect because it is only one of the limitations.
- Option C: Crystals of $\mathrm{CO},{\mathrm{N}_2}\mathrm{O},\mathrm{NO},{\mathrm{H}_2}\mathrm{O}$ etc., do not have zero entropy at zero kelvin is incorrect because it is only one of the limitations.

Key Concepts:
- Third law of thermodynamics
- Entropy
- Limitations of the third law

Memory Aids:
- Third Law: Perfect crystal at 0 K has zero entropy.

Quick Tips:
- Remember the limitations of the third law.

9 Medium

9 of 244Question permalink

Understanding +4 | -1

Hess’s law of constant heat summation is an application of :

View answer and explanation

Correct option: B

First law of thermodynamics

The correct answer is B (First law of thermodynamics) because Hess's law is a direct consequence of the fact that enthalpy is a state function, which is a restatement of the first law.

Solution approach

Correct Answer + Core Reason: First law of thermodynamics. Hess's law is based on the conservation of energy, which is the first law of thermodynamics.

Step-by-Step Solution:

Step 1: State Hess's Law: Hess's Law states that the total enthalpy change for a reaction is the same whether it occurs in one step or in multiple steps.

Step 2: Relate to the First Law of Thermodynamics: The First Law of Thermodynamics states that energy is conserved. Enthalpy is a state function, meaning its value depends only on the initial and final states, not the path taken. Hess's Law is a direct consequence of enthalpy being a state function, which is tied to the conservation of energy as stated in the First Law of Thermodynamics.

Why other options are wrong:
- Option A: Kirchhoff’s law: Kirchhoff's law relates the change in enthalpy with temperature, not the summation of heat.
- Option C: Second law of thermodynamics: The Second Law deals with entropy and spontaneity, not enthalpy summation.
- Option D: Entropy: Entropy is related to disorder, not enthalpy summation.

Key Concepts:
- Hess's Law
- First law of thermodynamics
- Enthalpy (H)
- State function

Memory Aids:
- Hess's Law = Enthalpy summation
- First Law = Energy conservation

Quick Tips:
- Remember that Hess's Law is a consequence of enthalpy being a state function.

10 Medium

10 of 244Question permalink

Understanding +4 | -1

In which of the following cases entropy decreases?

View answer and explanation

Correct option: D

Polymerisation

Option D, Polymerisation, is correct because it involves combining many small molecules into a large one, decreasing the number of possible arrangements and thus decreasing entropy.

Solution approach

Correct Answer + Core Reason: Polymerisation. Polymerisation reduces the number of independent particles, thus decreasing entropy.

Step-by-Step Solution:

Step 1: Recall the definition of entropy.

Step 2: Entropy is a measure of disorder or randomness in a system. It is related to the number of possible microstates a system can have.

Step 3: Consider the given options:
- A. Solid changing to liquid: Entropy increases as the molecules have more freedom of movement.
- B. Expansion of a gas: Entropy increases as the gas molecules occupy a larger volume and have more possible arrangements.
- C. Crystals dissolve: Entropy increases as the ordered crystal structure breaks down and the ions/molecules become more dispersed in the solution.
- D. Polymerisation: Small molecules (monomers) combine to form a large molecule (polymer). This reduces the number of independent particles and restricts their movement, leading to a decrease in entropy.

Step 4: Final Answer: Polymerisation

Why other options are wrong:
- Option A: Solid changing to liquid increases entropy because the molecules gain more freedom of movement.
- Option B: Expansion of a gas increases entropy because the gas molecules have more space to occupy.
- Option C: Crystals dissolve increases entropy because the ordered crystal structure is disrupted, and the ions/molecules become more dispersed.

Key Concepts:
- Entropy
- Polymerisation
- Phase Transitions
- Dissolution
- Thermodynamics

Memory Aids:
- Remember that entropy generally increases with increasing disorder or freedom of movement.

Quick Tips:
- Consider the number of particles and their freedom of movement when determining entropy changes.

11 Medium

11 of 244Question permalink

Application +4 | -1

One mono atomic gas is expanded adiabatically from

2L

to

10L

at

1\,\,atm

external pressure, find

\Delta U

(in

\mathrm{atm}\,\,L

)

View answer and explanation

Correct option: A

- 8

Option A (-8) is correct because the change in internal energy is equal to the work done in an adiabatic process against constant external pressure, which is -PΔV.

Solution approach

Correct Answer + Core Reason: -8. The change in internal energy in an adiabatic process against constant external pressure is equal to the work done.

Step-by-Step Solution:

Step 1: Identify the given information:\nInitial volume, $V_1 = 2 \text{{ L}}$ \nFinal volume, $V_2 = 10 \text{{ L}}$ \nExternal pressure, $P_{ext} = 1 \text{{ atm}}$

Step 2: Calculate the change in volume:
$\Delta V = V_2 - V_1 = 10 \text{{ L}} - 2 \text{{ L}} = 8 \text{{ L}}$

Step 3: Calculate the work done:
$W = -P_{ext} \Delta V = -1 \text{{ atm}} \times 8 \text{{ L}} = -8 \text{{ atm L}}$

Step 4: In an adiabatic process, the change in internal energy is equal to the work done:
$\Delta U = W = -8 \text{{ atm L}}$

Why other options are wrong:
- Option B (0): This is incorrect because work is done during the expansion, so the internal energy changes.
- Option C (-66.7): This is incorrect as it may be calculated using an incorrect formula or misunderstanding of the process.
- Option D (58.2): This is incorrect as the work done is negative during expansion, and internal energy decreases.

Key Concepts:
- Adiabatic process
- Work done in a process
- Change in internal energy

Memory Aids:
- Adiabatic process: $Q = 0$
- $W = -P_{ext} \Delta V$

Quick Tips:
- Remember the sign convention for work done (negative for expansion, positive for compression).
- In an adiabatic process, $\Delta U = W$ .

12 Medium

12 of 244Question permalink

Analyzing +4 | -1

\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2};

\mathrm{CO} + \frac{1}{2}{\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2}

Then the heat of formation of CO is

View answer and explanation

Correct option: A

X-Y

Option A (X-Y) is correct because the heat of formation of CO can be found by subtracting the heat of the second reaction from the heat of the first reaction based on Hess's Law.

Solution approach

Correct Answer + Core Reason: X-Y. Heat of formation of CO is obtained by subtracting the second reaction from the first.

Step-by-Step Solution:

Step 1: Given reactions:
$C + O_2 \rightarrow CO_2; \Delta H = X$ (1)
$CO + \frac{1}{2}O_2 \rightarrow CO_2; \Delta H = Y$ (2)

Step 2: We need to find the heat of formation of CO, which is the enthalpy change for the reaction:
$C + \frac{1}{2}O_2 \rightarrow CO; \Delta H = ?$ (3)

Step 3: To obtain reaction (3), subtract reaction (2) from reaction (1):
$(C + O_2) - (CO + \frac{1}{2}O_2) \rightarrow CO_2 - CO_2$
$C + O_2 - CO - \frac{1}{2}O_2 \rightarrow 0$
$C + \frac{1}{2}O_2 \rightarrow CO$

Step 4: Therefore, the enthalpy change for reaction (3) is $\Delta H = X - Y$ .

Why other options are wrong:
- Option B (Y-2X): This is incorrect because it involves an incorrect combination of the given enthalpies.
- Option C (X+Y): This is incorrect because it involves adding the enthalpies instead of subtracting.
- Option D (2X-Y): This is incorrect because it involves an incorrect combination of the given enthalpies.

Key Concepts:
- Hess's Law
- Heat of formation

Memory Aids:
- Hess's Law states that the enthalpy change of a reaction is independent of the path taken.

Quick Tips:
- Manipulate the given equations to obtain the desired equation and add/subtract the corresponding enthalpy changes.

13 Medium

13 of 244Question permalink

Remembering +4 | -1

Which one among the following is the correct option for right relationship between

{C_p}\mathrm{ and }{C_v}

for one mole of ideal gas?

View answer and explanation

Correct option: D

{C_p} - {C_v} = R

Option D,

{C_p} - {C_v} = R

, is the correct relationship between

{C_p}

and

{C_v}

for one mole of an ideal gas.

Solution approach

Correct Answer + Core Reason: ${C_p} - {C_v} = R$ . This is the correct relation between Cp and Cv for one mole of ideal gas.

Step-by-Step Solution:

Step 1: Recall the relationship between molar heat capacity at constant pressure ( $C_p$ ) and molar heat capacity at constant volume ( $C_v$ ) for an ideal gas.

Step 2: The relationship is given by: $C_p - C_v = R$ , where R is the ideal gas constant.

Step 3: Final Answer: ${C_p} - {C_v} = R$

Why other options are wrong:
- Option A: ${C_p} = R{C_v}$ is incorrect because it implies that $C_p$ is much larger than $C_v$ , which is not generally true.
- Option B: ${C_v} = R{C_p}$ is incorrect because it implies that $C_v$ is greater than $C_p$ , which is not true for ideal gases.
- Option C: ${C_p} + {C_v} = R$ is incorrect because it suggests that the sum of $C_p$ and $C_v$ equals R, which is not the correct relationship.

Key Concepts:
- Molar Heat Capacity at Constant Pressure ( $C_p$ )
- Molar Heat Capacity at Constant Volume ( $C_v$ )
- Ideal Gas Constant (R)
- Thermodynamics

Memory Aids:
- Remember that $C_p$ is always greater than $C_v$ for an ideal gas, and the difference is R.

Quick Tips:
- This relationship is fundamental in thermodynamics and is often used in calculations involving ideal gases.

14 Medium

14 of 244Question permalink

Application +4 | -1

The dissociation energy of

\mathrm{C}{\mathrm{H}_4}

and

{\mathrm{C}_2}{\mathrm{H}_6}

are respectively 360 and 620 kcal/mol. The bond energy of C-C bond is:

View answer and explanation

Correct option: D

80 kcal/mol

Option D (80 kcal/mol) is correct. It is calculated by using the given dissociation energies and assuming that the dissociation energy is the sum of the bond energies.

Solution approach

Correct Answer + Core Reason: 80 kcal/mol. The bond energy of C-C is calculated using the given dissociation energies.

Step-by-Step Solution:

Step 1: Dissociation energy of $CH_4$ is 360 kcal/mol. This means $4 \times BE(C-H) = 360$ kcal/mol.

Step 2: Therefore, $BE(C-H) = \frac{360}{4} = 90$ kcal/mol.

Step 3: Dissociation energy of $C_2H_6$ is 620 kcal/mol. This means $BE(C-C) + 6 \times BE(C-H) = 620$ kcal/mol.

Step 4: Substitute the value of BE(C-H) from Step 2 into the equation from Step 3:
$BE(C-C) + 6 \times 90 = 620$
$BE(C-C) + 540 = 620$

Step 5: Solve for BE(C-C):
$BE(C-C) = 620 - 540 = 80$ kcal/mol.

Why other options are wrong:
- Option A (260 kcal/mol): This is incorrect. It likely arises from misinterpreting the relationship between dissociation energy and bond energies or making calculation errors.
- Option B (180 kcal/mol): This is incorrect. It likely arises from misinterpreting the relationship between dissociation energy and bond energies or making calculation errors.
- Option C (130 kcal/mol): This is incorrect. It likely arises from misinterpreting the relationship between dissociation energy and bond energies or making calculation errors.

Key Concepts:
- Bond Energy
- Dissociation Energy

Memory Aids:
- Dissociation energy is the energy required to break all bonds in a molecule.

Quick Tips:
- Always define the relationship between dissociation energy and bond energies clearly.
- Pay attention to the number of bonds of each type in the molecule.

15 Medium

15 of 244Question permalink

Application +4 | -1

If,{\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to 2\mathrm{HCl};\Delta {H^0} = - 44\mathrm{kcal}

2Na(s)+2HCl(g)→2NaCl(s)+

{H_2}

(g);
∆H= -152 kcal then,

\mathrm{Na}\left( s \right) + 0.5\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to \mathrm{NaCl}\left( s \right);\Delta {H^0} = ?

View answer and explanation

Correct option: C

- 98\mathrm{kcal}

Option C (-98 kcal) is correct. By manipulating the given equations using Hess's Law, the enthalpy change for the desired reaction is calculated.

Solution approach

Correct Answer + Core Reason: $- 98\mathrm{kcal}$ . By manipulating the given equations using Hess's Law, the enthalpy change for the desired reaction is calculated.

Step-by-Step Solution:

Step 1: Write down the given reactions and their enthalpy changes:\nReaction 1: ${\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to 2\mathrm{HCl};\Delta {H^0} = - 44\mathrm{kcal}$ \nReaction 2: $2\mathrm{Na}(s)+2\mathrm{HCl}(g) \to 2\mathrm{NaCl}(s)+{\mathrm{H}_2} (g); \Delta H= -152 \text{ kcal}$ \nTarget Reaction: $\mathrm{Na}\left( s \right) + 0.5\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to \mathrm{NaCl}\left( s \right);\Delta {H^0} = ?$

Step 2: Divide Reaction 1 by 2: $0.5{\mathrm{H}_2}\left( \mathrm{g} \right) + 0.5\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to \mathrm{HCl};\Delta {H^0}_1 = - 22\mathrm{kcal}$

Step 3: Divide Reaction 2 by 2: $\mathrm{Na}(s)+\mathrm{HCl}(g) \to \mathrm{NaCl}(s)+0.5{\mathrm{H}_2} (g); \Delta H_2= -76 \text{ kcal}$

Step 4: Add the modified Reaction 1 and Reaction 2:
$0.5{\mathrm{H}_2}\left( \mathrm{g} \right) + 0.5\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) + \mathrm{Na}(s)+\mathrm{HCl}(g) \to \mathrm{HCl} + \mathrm{NaCl}(s)+0.5{\mathrm{H}_2} (g)$

Step 5: Simplify the equation to get the target reaction: $\mathrm{Na}\left( s \right) + 0.5\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to \mathrm{NaCl}\left( s \right)$

Step 6: Add the enthalpy changes: $\Delta H^0 = \Delta H^0_1 + \Delta H_2 = -22 \text{ kcal} + (-76 \text{ kcal}) = -98 \text{ kcal}$ .

Why other options are wrong:
- Option A: This option has the wrong sign and magnitude. It's likely a result of incorrect manipulation or addition of the given enthalpy changes.
- Option B: This option is incorrect. It's likely a result of incorrect manipulation or addition of the given enthalpy changes.
- Option D: This option is incorrect. It's likely a result of incorrect manipulation or addition of the given enthalpy changes.

Key Concepts:
- Hess's Law: The enthalpy change for a reaction is independent of the path taken.
- Manipulation of thermochemical equations: Reactions can be added, subtracted, multiplied, or divided, and the enthalpy changes are treated accordingly.

Memory Aids:
- Remember that when you reverse a reaction, you change the sign of $\Delta H$ .
- When you multiply a reaction by a factor, you multiply $\Delta H$ by the same factor.

Quick Tips:
- Always write down the target reaction and manipulate the given reactions to match it.
- Pay attention to the signs of the enthalpy changes.

16 Medium

16 of 244Question permalink

Understanding +4 | -1

Which statements are correct?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because all the statements regarding the Clausius-Clapeyron equation, Trouton's rule, and the definition of entropy are accurate.

Solution approach

Correct Answer + Core Reason: All of the above. All the given statements are correct definitions and rules in thermodynamics.

Step-by-Step Solution:

Step 1: Analyze statement A: $2.303\log \mathrm{\;}\frac{{{{P_2}}}}{{{{P_1}}}} = \frac{{\Delta {{{H}}_{{\mathrm{vap}}}.}}}{R}\frac{{\left[ {{{{T_2}} - {{T_1}}}} \right]}}{{{{T_1}}{{T_2}}}}$ is the Clausius-Clapeyron equation. This equation relates the vapor pressure of a liquid to its temperature and enthalpy of vaporization. This statement is correct.

Step 2: Analyze statement B: $\frac{{\Delta {{{H}}_{{\mathrm{vap}}}.}}}{{\mathrm{Boiling}\;\mathrm{point}}} = 88{\mkern 1mu} {\mkern 1mu} \mathrm{J}\;\mathrm{mo}{\mathrm{l}^{ - 1}}{\mkern 1mu} {{{\mathrm{K}^{ - 1}}}}$ is called Trouton's rule. Trouton's rule states that the molar entropy of vaporization is approximately the same (around 88 J mol⁻¹ K⁻¹) for many liquids at their normal boiling points. This statement is correct.

Step 3: Analyze statement C: Entropy is a measure of unavailable energy, i.e., unavailable energy = entropy $\times$ temperature. Entropy is indeed a measure of the disorder or randomness of a system, and it is related to the amount of energy that is unavailable to do work. The unavailable energy is given by the product of entropy and temperature. This statement is correct.

Step 4: Since all statements A, B, and C are correct, the answer is D.

Why other options are wrong:
- Option A: This option is incorrect because statements B and C are also correct.
- Option B: This option is incorrect because statements A and C are also correct.
- Option C: This option is incorrect because statements A and B are also correct.

Key Concepts:
- Clausius-Clapeyron equation
- Trouton's rule
- Entropy and unavailable energy

Memory Aids:
- Clausius-Clapeyron: Relates vapor pressure to temperature.
- Trouton's Rule: Entropy of vaporization is constant for many liquids.
- Entropy: Measure of disorder.

Quick Tips:
- Remember the key assumptions and limitations of each rule and equation. For example, Trouton's rule is an approximation and doesn't apply to all liquids.

17 Medium

17 of 244Question permalink

Remembering +4 | -1

A process in which no heat change takes place is called.

View answer and explanation

Correct option: B

An adiabatic process

Option B is correct because an adiabatic process is defined as one in which no heat is exchanged between the system and its surroundings.

Solution approach

Correct Answer + Core Reason: An adiabatic process. An adiabatic process is defined as a thermodynamic process in which no heat is transferred to or from the system by its surroundings.

Step-by-Step Solution:

Step 1: Define the terms:
- Isothermal process: A process that occurs at constant temperature.
- Adiabatic process: A process in which no heat is exchanged with the surroundings.
- Isobaric process: A process that occurs at constant pressure.
- Isochoric process: A process that occurs at constant volume.

Step 2: Identify the process with no heat change: An adiabatic process is characterized by no heat exchange (Q = 0).

Why other options are wrong:
- Option A: An isothermal process occurs at constant temperature, not necessarily without heat change.
- Option C: An isobaric process occurs at constant pressure, not necessarily without heat change.
- Option D: An isochoric process occurs at constant volume, not necessarily without heat change.

Key Concepts:
- Isothermal process
- Adiabatic process
- Isobaric process
- Isochoric process

Memory Aids:
- Adiabatic: No heat exchange.

Quick Tips:
- Remember the definitions of the four common thermodynamic processes.

18 Medium

18 of 244Question permalink

Analyzing +4 | -1

Statement I: The final temperature of an ideal
gas in adiabatic compression is less in reversible
compression than in irreversible compression
against a constant external pressure.

Statement II: The magnitude of work done on
an ideal gas in adiabatic compression is more in
irreversible process than that in reversible process.

View answer and explanation

Correct option: A

If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I

Option A is correct. Irreversible adiabatic compression results in a higher final temperature and more work done compared to reversible adiabatic compression because energy is dissipated as heat in the irreversible process.

Solution approach

Correct Answer + Core Reason: If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I. Irreversible adiabatic compression leads to a higher final temperature and more work done compared to reversible compression.

Step-by-Step Solution:

Step 1: Understand adiabatic compression: No heat exchange with the surroundings (q = 0).

Step 2: Recall the difference between reversible and irreversible processes:
- Reversible process: Occurs infinitesimally slowly, always in equilibrium.
- Irreversible process: Occurs rapidly, not in equilibrium, some energy is dissipated as heat.

Step 3: Analyze Statement I: The final temperature in irreversible adiabatic compression is higher than in reversible compression.
- In reversible adiabatic compression, all work done on the gas goes into increasing its internal energy, leading to a temperature increase. The process follows $PV^\gamma = constant$ .
- In irreversible adiabatic compression, some of the work done is dissipated as heat due to friction or other factors. This extra heat contributes to a higher final temperature.

Step 4: Analyze Statement II: The magnitude of work done in irreversible adiabatic compression is more than in reversible compression.
- In an irreversible process against constant external pressure, the work done is given by $W = -P_{ext}\Delta V$ . Since the process is irreversible, the external pressure is higher than the internal pressure, leading to more work being done on the gas.
- In a reversible process, the work done is the maximum possible work and is given by $W = -\int{PdV}$ .

Step 5: Connect the two statements: The greater work done in the irreversible process leads to a higher increase in internal energy, and thus a higher final temperature.

Why other options are wrong:
- Option B: If both statements are CORRECT, and Statement II is NOT the CORRECT explanation of Statement I. Statement II directly explains Statement I.
- Option C: If Statement I is CORRECT, but Statement II is INCORRECT. Both statements are correct.
- Option D: If Statement I is INCORRECT, but Statement II is CORRECT. Both statements are correct.

Key Concepts:
- Adiabatic process
- Reversible and irreversible processes
- Work done in adiabatic processes
- Internal energy and temperature relationship

Memory Aids:
- Adiabatic: no heat exchange
- Reversible: slow, equilibrium
- Irreversible: fast, non-equilibrium

Quick Tips:
- Remember that irreversible processes always involve some energy dissipation, leading to less efficient energy transfer.

19 Medium

19 of 244Question permalink

Analyzing +4 | -1

For one mole of an ideal gas, which of these statements must be true?
A)

U\,\,\mathrm{ and}\,\,\mathrm{ }H

each depends only on temperature
B) Compressibility factor Z is not equal to 1.
C)

{C_{p,\,\,m}} - {C_{V,\,\,m}} = R

D)

dU = {C_V}dT

for any process.

View answer and explanation

Correct option: C

A, C and D

Option C (A, C and D) is correct because for one mole of an ideal gas, U and H depend only on temperature,

{C_{p,\,m}} - {C_{V,\,m}} = R

, and

dU = {C_V}dT

for any process.

Solution approach

Correct Answer + Core Reason: A, C and D. For one mole of an ideal gas, U and H depend only on temperature, ${C_{p,\,m}} - {C_{V,\,m}} = R$ , and $dU = {C_V}dT$ for any process.

Step-by-Step Solution:

Step 1: Analyze statement A: $U\,\,\mathrm{ and}\,\,\mathrm{ }H$ each depends only on temperature\nFor an ideal gas, the internal energy (U) and enthalpy (H) are functions of temperature only. This is because there are no intermolecular forces in an ideal gas. Therefore, statement A is correct.

Step 2: Analyze statement B: Compressibility factor Z is not equal to 1.\nThe compressibility factor (Z) for an ideal gas is always equal to 1. Z = pV/nRT = 1 for an ideal gas. Therefore, statement B is incorrect.

Step 3: Analyze statement C: ${C_{p,\,m}} - {C_{V,\,m}} = R$ \nFor an ideal gas, the difference between the molar heat capacity at constant pressure ( ${C_{p,\,m}}$ ) and the molar heat capacity at constant volume ( ${C_{V,\,m}}$ ) is equal to the gas constant (R). This is a fundamental relationship for ideal gases. Therefore, statement C is correct.

Step 4: Analyze statement D: $dU = {C_V}dT$ for any process.\nFor an ideal gas, the change in internal energy (dU) is given by $dU = {C_V}dT$ , where ${C_V}$ is the heat capacity at constant volume and dT is the change in temperature. This relationship holds true for any process, whether it is isothermal, adiabatic, isobaric, or isochoric. Therefore, statement D is correct.

Why other options are wrong:
- Option A (B, C and D): Statement B is incorrect, so this option is incorrect.
- Option B (A and C): Statement D is also correct, so this option is incorrect.
- Option D (C and D): Statement A is also correct, so this option is incorrect.

Key Concepts:
- Ideal Gas Law
- Internal Energy (U)
- Enthalpy (H)
- Heat Capacities ( ${C_p}$ , ${C_V}$ )
- Compressibility Factor (Z)

Memory Aids:
- For an ideal gas, U and H depend only on T.
- Z = 1 for an ideal gas.
- ${C_p} - {C_V} = R$ for an ideal gas.
- $dU = {C_V}dT$ for an ideal gas.

Quick Tips:
- Remember the properties of ideal gases.
- Understand the relationships between thermodynamic properties for ideal gases.

20 Medium

20 of 244Question permalink

Analyzing +4 | -1

The heat of neutralization of

HCl

by

NaOH

is – 55.9 KJ/mol. If the heat of neutralization of

HCN

by

NaOH

is – 12.1 KJ/mol. The heat of dissociation of

HCN

is

View answer and explanation

Correct option: B

43.8 KJ

Option B (43.8 KJ) is correct because the heat of dissociation of HCN is the difference between the heat of neutralization of a strong acid and the heat of neutralization of HCN.

Solution approach

Correct Answer + Core Reason: 43.8 KJ. The heat of dissociation is the difference between the neutralization enthalpies of a strong acid and a weak acid.

Step-by-Step Solution:

Step 1: Heat of neutralization of HCl by NaOH = -55.9 KJ/mol\nHeat of neutralization of HCN by NaOH = -12.1 KJ/mol

Step 2: The neutralization of a strong acid (HCl) with a strong base (NaOH) involves the reaction: $H^+ + OH^- \rightarrow H_2O$ \nThe heat released is -55.9 KJ/mol.

Step 3: The neutralization of a weak acid (HCN) with a strong base (NaOH) involves the reaction:\nHCN + NaOH → NaCN + H_2O\nThis can be considered as the sum of two processes:
(i) Dissociation of HCN: $HCN \rightarrow H^+ + CN^-; \Delta H = \Delta H_{dissociation}$
(ii) Neutralization of $H^+$ by $OH^-$ : $H^+ + OH^- \rightarrow H_2O; \Delta H = -55.9 \text{ KJ/mol}$

Step 4: Therefore, the heat of neutralization of HCN is:
$\Delta H_{neutralization} = \Delta H_{dissociation} + (-55.9 \text{ KJ/mol})$
-12.1 KJ/mol = $\Delta H_{dissociation}$ - 55.9 KJ/mol
$\Delta H_{dissociation} = -12.1 + 55.9 = 43.8 \text{ KJ/mol}$

Why other options are wrong:
- Option A (– 43.8 KJ): This is incorrect because it has the wrong sign. Dissociation is an endothermic process.
- Option C (68 KJ): This is incorrect because it's not the correct difference between the two neutralization enthalpies.
- Option D (– 68 KJ): This is incorrect because it has the wrong sign and is not the correct difference between the two neutralization enthalpies.

Key Concepts:
- Heat of neutralization
- Heat of dissociation
- Strong acids and weak acids

Memory Aids:
- Remember that the heat of neutralization of a strong acid and strong base is a constant value.

Quick Tips:
- Understand the difference between the neutralization of strong acids/bases and weak acids/bases.
- Dissociation of a weak acid is an endothermic process.

21 Medium

21 of 244Question permalink

Application +4 | -1

The heat of combustion of methane is

- \;880\;\mathrm{kJ\;mo}{\mathrm{l}^{ - 1}}.

. If 3.2 g of methane is burnt:

View answer and explanation

Correct option: A

- 176 kJ of heat is evolved

Option A (- 176 kJ of heat is evolved) is correct because the heat evolved is directly proportional to the moles of methane burnt.

Solution approach

Correct Answer + Core Reason: - 176 kJ of heat is evolved. Heat evolved is proportional to the moles of methane burnt.

Step-by-Step Solution:

Step 1: Calculate the number of moles of methane ( $CH_4$ ) burnt.

Step 2: Molar mass of $CH_4$ = 12 + 4(1) = 16 g/mol

Step 3: Moles of $CH_4$ = $\frac{3.2 \text{{ g}}}{16 \text{{ g/mol}}} = 0.2 \text{{ mol}}$

Step 4: Heat evolved when 1 mole of $CH_4$ is burnt = 880 kJ

Step 5: Heat evolved when 0.2 moles of $CH_4$ is burnt = $0.2 \times 880 \text{{ kJ}} = 176 \text{{ kJ}}$

Step 6: Since heat is evolved, the sign is negative. Therefore, heat evolved = -176 kJ

Why other options are wrong:
- Option B: This option suggests heat is absorbed, which is incorrect as combustion is an exothermic process where heat is evolved.
- Option C: This option calculates the heat evolved for 0.1 mol instead of 0.2 mol of methane.
- Option D: This option is incorrect because option A is the correct answer.

Key Concepts:
- Heat of combustion
- Stoichiometry
- Exothermic reactions

Memory Aids:
- Combustion is always exothermic (releases heat).

Quick Tips:
- Always convert mass to moles before using the heat of combustion value.
- Remember to include the negative sign for heat evolved in exothermic reactions.

22 Medium

22 of 244Question permalink

Remembering +4 | -1

The second law of thermodynamics introduced the concept of:

View answer and explanation

Correct option: C

Entropy

Option C (Entropy) is correct because the second law of thermodynamics introduces the concept of entropy, which describes the degree of disorder in a system.

Solution approach

Correct Answer + Core Reason: Entropy. The second law of thermodynamics is fundamentally about the increase of entropy in an isolated system.

Step-by-Step Solution:

Step 1: Recall the definition of the second law of thermodynamics.

Step 2: The second law states that the total entropy of an isolated system can only increase over time or remain constant in ideal cases where the system is in a steady state or undergoing a reversible process.

Step 3: Entropy is a measure of the disorder or randomness of a system.

Step 4: Therefore, the second law introduces the concept of entropy.

Why other options are wrong:
- Option A: The third law of thermodynamics deals with the behavior of systems at absolute zero temperature, not introduced by the second law.
- Option B: Work is a concept related to energy transfer, but not specifically introduced by the second law.
- Option D: Internal energy is a state function related to the energy content of a system, but not the central concept introduced by the second law.

Key Concepts:
- Second law of thermodynamics
- Entropy

Memory Aids:
- Entropy increases (or remains constant) in any real process.

Quick Tips:
- Remember that the second law is about the direction of spontaneous processes and the increase of disorder.

23 Medium

23 of 244Question permalink

Application +4 | -1

At 27℃ one mole of an ideal gas is compressed isothermally and reversible from a pressure of 2 atm to 10 atm. The value of ∆E and q are (R=2 cal)

View answer and explanation

Correct option: A

0, -965.84 cal

Option A (0, -965.84 cal) is correct because for an isothermal process, ∆E = 0, and q = -W. The work done is calculated using the formula for reversible isothermal compression.

Solution approach

Correct Answer + Core Reason: 0, -965.84 cal. For an isothermal process, the change in internal energy is zero. The heat is equal to the negative of the work done in a reversible process.

Step-by-Step Solution:

Step 1: Identify the given information:
- Temperature (T) = 27°C = 300 K
- Number of moles (n) = 1
- Initial pressure (P1) = 2 atm
- Final pressure (P2) = 10 atm
- R = 2 cal/mol.K

Step 2: For an isothermal process, ∆E = 0.

Step 3: Calculate the work done (W) for a reversible isothermal compression:
- $W = -nRT \ln{\frac{{P_2}}{{P_1}}}$

Step 4: Substitute the values:
- $W = -(1 \text{{ mol}}) \times (2 \text{{ cal/mol.K}}) \times (300 \text{{ K}}) \times \ln{\frac{{10 \text{{ atm}}}}{{2 \text{{ atm}}}}}$

Step 5: Calculate the natural logarithm:
- $\ln{\frac{{10}}{{2}}} = \ln{5} \approx 1.609$

Step 6: Calculate the work done:
- $W = - (1) \times (2) \times (300) \times (1.609) = -965.4 \text{{ cal}}$

Step 7: Since the process is isothermal and reversible, $q = -W$
- $q = -(-965.4) = -965.4 \text{{ cal}}$ (approximately -965.84 cal due to rounding)

Step 8: Final Answer: ∆E = 0, q = -965.84 cal

Why other options are wrong:
- Option B: This option incorrectly assigns a non-zero value to ∆E and an incorrect value to q.
- Option C: This option incorrectly assigns a positive value to ∆E and an incorrect value to q.
- Option D: This option incorrectly assigns positive values to both ∆E and q.

Key Concepts:
- Isothermal process
- Reversible process
- Change in internal energy (∆E)
- Work done in isothermal reversible process
- Heat (q)

Memory Aids:
- Isothermal: ∆T = 0, ∆E = 0
- Reversible: q = -W
- $W = -nRT \ln{\frac{{P_2}}{{P_1}}}$ for isothermal reversible process

Quick Tips:
- Remember that for an isothermal process, the change in internal energy is always zero.
- Pay attention to the sign conventions for work done and heat.

24 Medium

24 of 244Question permalink

Analyzing +4 | -1

At the same conditions of pressure, volume and
temperature, work done is maximum for which
gas if all gases have equal masses?

View answer and explanation

Correct option: A

\mathrm{NH}_3

Option A (NH3) is correct because, with equal masses, the gas with the smallest molar mass (NH3) will have the most moles, leading to maximum work done.

Solution approach

Correct Answer + Core Reason: $\mathrm{NH}_3$ . With equal masses, the gas with the smallest molar mass will have the most moles, leading to maximum work.

Step-by-Step Solution:

Step 1: Understand the problem: We need to find which gas does the maximum work at the same conditions (P, V, T) with equal masses.

Step 2: Recall the relationship between work and moles: For a given P, V, and T, the work done is proportional to the number of moles (n) of the gas. More moles mean more work.

Step 3: Determine the molar masses of the gases:
- $\mathrm{NH}_3$ : 14 + 3(1) = 17 g/mol
- $\mathrm{N}_2$ : 2(14) = 28 g/mol
- $\mathrm{Cl}_2$ : 2(35.5) = 71 g/mol
- $\mathrm{H}_2\mathrm{S}$ : 2(1) + 32 + 16 = 34 g/mol

Step 4: Compare the molar masses: $\mathrm{NH}_3$ has the smallest molar mass.

Step 5: Conclusion: Since the masses are equal, $\mathrm{NH}_3$ will have the largest number of moles, and therefore, it will do the maximum work.

Why other options are wrong:
- Option B: $\mathrm{N}_2$ has a higher molar mass than $\mathrm{NH}_3$ , so it will have fewer moles.
- Option C: $\mathrm{Cl}_2$ has the highest molar mass, so it will have the fewest moles.
- Option D: $\mathrm{H}_2\mathrm{S}$ has a higher molar mass than $\mathrm{NH}_3$ , so it will have fewer moles.

Key Concepts:
- Work done by a gas
- Molar mass
- Relationship between moles and work

Memory Aids:
- Smaller molar mass means more moles for the same mass.

Quick Tips:
- When comparing gases with equal masses, focus on the molar mass to determine the number of moles.

25 Medium

25 of 244Question permalink

Understanding +4 | -1

A gas expands isothermally and reversibly. The work done by the gas is:

View answer and explanation

Correct option: C

Maximum

Option C (Maximum) is correct because a reversible isothermal expansion allows the system to perform the maximum possible work as it maintains equilibrium throughout the process.

Solution approach

Correct Answer + Core Reason: Maximum. A reversible process is one that occurs infinitesimally slowly, allowing the system to remain in equilibrium at all times, thus maximizing the work done.

Step-by-Step Solution:

Step 1: Recall that for a reversible process, the system is always in equilibrium with its surroundings.

Step 2: In an isothermal and reversible expansion, the gas expands slowly, allowing it to perform the maximum possible work.

Step 3: This is because the pressure of the gas is always infinitesimally close to the external pressure.

Why other options are wrong:
- Option A: Zero is incorrect because the gas is expanding and doing work.
- Option B: Minimum is incorrect because the work done is maximum for a reversible process.
- Option D: Equal to work done is vague and doesn't specify the nature of the work (maximum, minimum, etc.).

Key Concepts:
- Reversible Process
- Isothermal Process
- Work Done by a Gas

Memory Aids:
- Reversible = Maximum work.

Quick Tips:
- Remember that reversible processes are ideal and perform the maximum possible work.

26 Medium

26 of 244Question permalink

Understanding +4 | -1

\Delta \,U

is equal to

View answer and explanation

Correct option: C

Adiabatic work

Option C (Adiabatic work) is correct because in an adiabatic process, there is no heat exchange with the surroundings, so the change in internal energy is equal to the work done.

Solution approach

Correct Answer + Core Reason: Adiabatic work. In an adiabatic process, no heat is exchanged, so all the change in internal energy is due to work.

Step-by-Step Solution:

Step 1: Define the terms. $\Delta U$ is the change in internal energy. Adiabatic process means $q = 0$ .

Step 2: Apply the first law of thermodynamics: $\Delta U = q + w$

Step 3: Since $q = 0$ for an adiabatic process, $\Delta U = w_{adiabatic}$

Step 4: Conclusion: The change in internal energy is equal to the adiabatic work.

Why other options are wrong:
- Option A: Isochoric work is work done at constant volume, which is not necessarily equal to the change in internal energy.
- Option B: Isobaric work is work done at constant pressure, which is not necessarily equal to the change in internal energy.
- Option D: Isothermal work is work done at constant temperature, which is not necessarily equal to the change in internal energy.

Key Concepts:
- First Law of Thermodynamics
- Adiabatic Process
- Isochoric Process
- Isobaric Process
- Isothermal Process

Memory Aids:
- Adiabatic: \

27 Medium

27 of 244Question permalink

Understanding +4 | -1

Vibrational energy is:

View answer and explanation

Correct option: A

Partially potential and partially kinetic

Option A (Partially potential and partially kinetic) is correct because vibrational energy involves both the movement of atoms (kinetic energy) and the restoring forces between them (potential energy).

Solution approach

Correct Answer + Core Reason: Partially potential and partially kinetic. Vibrational energy involves both movement and restoring forces.

Step-by-Step Solution:

Step 1: Define vibrational energy.\nVibrational energy is the energy associated with the vibration of atoms or molecules around their equilibrium positions.

Step 2: Explain the components of vibrational energy.\nKinetic energy: Arises from the movement of the atoms or molecules as they vibrate.\nPotential energy: Arises from the restoring forces that pull the atoms or molecules back towards their equilibrium positions.

Step 3: Conclusion: Vibrational energy is partially potential and partially kinetic.

Why other options are wrong:
- Option B: Only potential is incorrect because vibration involves movement, so there's kinetic energy as well.
- Option C: Only kinetic is incorrect because vibration involves restoring forces, so there's potential energy as well.
- Option D: None of the above is incorrect because option A is the correct description.

Key Concepts:
- Vibrational energy
- Kinetic energy
- Potential energy

Memory Aids:
- Vibration involves both movement and restoring forces.

Quick Tips:
- Remember that vibration is a form of motion, so it involves kinetic energy. Also, restoring forces are involved, so it involves potential energy.

28 Medium

28 of 244Question permalink

Analyzing +4 | -1

Consider the following reaction occurring in an automobile

2 {C}_8 {H}_{18}({~g})+25 {O}_2({~g}) \longrightarrow 16 {CO}_2({~g})+18 {H}_2 {O}({g})

the sign of

\Delta {H}, \Delta {S}

and

\Delta {G}

would be

View answer and explanation

Correct option: B

-,+,-

Option B (-,+,-) is correct because combustion reactions are exothermic (negative Δ H), increase disorder (positive Δ S), and are spontaneous (negative Δ G).

Solution approach

Correct Answer + Core Reason: -,+,-. Combustion reactions release heat (exothermic), increase entropy due to gas formation, and are spontaneous.

Step-by-Step Solution:

Step 1: Analyze the reaction: $2 {C}_8 {H}_{18}({~g})+25 {O}_2({~g}) \longrightarrow 16 {CO}_2({~g})+18 {H}_2 {O}({g})$

Step 2: Determine the sign of $\Delta H$ : Combustion reactions are exothermic, meaning they release heat. Therefore, $\Delta H$ is negative (-).

Step 3: Determine the sign of $\Delta S$ : The number of gas molecules increases from 27 (2 + 25) to 34 (16 + 18). An increase in the number of gas molecules leads to an increase in entropy. Therefore, $\Delta S$ is positive (+).

Step 4: Determine the sign of $\Delta G$ : For a spontaneous reaction, $\Delta G$ must be negative. Combustion reactions are spontaneous. Therefore, $\Delta G$ is negative (-).

Step 5: Summarize the signs: $\Delta H$ (-), $\Delta S$ (+), $\Delta G$ (-).

Why other options are wrong:
- Option A (+,-,+): Incorrect because combustion is exothermic (negative Δ H), not endothermic (positive Δ H).
- Option C (-,+,+): Incorrect because combustion is spontaneous (negative Δ G), not non-spontaneous (positive Δ G).
- Option D (+,+,-): Incorrect because combustion is exothermic (negative Δ H), not endothermic (positive Δ H).

Key Concepts:
- Enthalpy (Δ H): Exothermic vs. Endothermic reactions
- Entropy (Δ S): Increase in disorder
- Gibbs Free Energy (Δ G): Spontaneity of reactions
- Combustion reactions

Memory Aids:
- Exothermic = Heat Exits = Negative Δ H
- More gas molecules = Higher Entropy = Positive Δ S
- Spontaneous = Δ G is Gone = Negative Δ G

Quick Tips:
- Remember that combustion reactions are always exothermic and spontaneous.
- Count the number of gas molecules on both sides of the equation to determine the sign of Δ S.

29 Medium

29 of 244Question permalink

Application +4 | -1

The standard entropies of

{N_2}(g),\,\,{H_2}(g)

and

N{H_3}(g)

are

191.5,\,\,130.5,\,\,192.6\,\,\,\mathrm{J}{\mathrm{K}^{\mathrm{ - 1}}}\,\mathrm{mo}{\mathrm{l}^{\mathrm{ - 1}}}

. The value of

\Delta S^\circ

formation of ammonia is

View answer and explanation

Correct option: A

- 98.9\,J {K^{ - 1}} \,mo{l^{ - 1}}

Option A (-98.9 J K⁻¹ mol⁻¹) is correct because it is calculated using the standard entropies of the reactants and products for the formation of ammonia.

Solution approach

Correct Answer + Core Reason: $- 98.9\,J {K^{ - 1}} \,mo{l^{ - 1}}$ . The standard entropy change for the formation of ammonia is calculated using the formula $\Delta S^\circ = \sum S^\circ_{products} - \sum S^\circ_{reactants}$ .

Step-by-Step Solution:
Step 1: Write the balanced chemical equation for the formation of ammonia: ${N_2}(g) + 3{H_2}(g) \rightarrow 2N{H_3}(g)$ .
Step 2: Identify the standard entropies: $S^\circ(N_2) = 191.5\,J K^{-1} mol^{-1}$ , $S^\circ(H_2) = 130.5\,J K^{-1} mol^{-1}$ , $S^\circ(NH_3) = 192.6\,J K^{-1} mol^{-1}$ .
Step 3: Calculate the standard entropy change: $\Delta S^\circ = 2 \times S^\circ(NH_3) - [S^\circ(N_2) + 3 \times S^\circ(H_2)]$ .
Step 4: Substitute the values: $\Delta S^\circ = 2 \times 192.6 - [191.5 + 3 \times 130.5]$ .
Step 5: Calculate: $\Delta S^\circ = 385.2 - [191.5 + 391.5] = 385.2 - 583 = -197.8\,J K^{-1} mol^{-1}$ for 2 moles of ammonia.
Step 6: Divide by 2 to get the value for 1 mole of ammonia: $\Delta S^\circ = -197.8 / 2 = -98.9\,J K^{-1} mol^{-1}$ .

Why other options are wrong:
- Option B: Zero. This is incorrect because the entropy change is not zero for this reaction.
- Option C: $+ 129.4\,J \cdot {K^{ - 1}} \cdot \,mo{l^{ - 1}}$ . This is incorrect; the calculation leads to a negative value.
- Option D: $- 29.4\,J \cdot {K^{ - 1}} \cdot \,mo{l^{ - 1}}$ . This is incorrect; the calculation leads to -98.9 J K⁻¹ mol⁻¹.

Key Concepts:
- Standard Entropy
- Entropy Change of Reaction
- Stoichiometry

Memory Aids:
- Remember the formula: $\Delta S^\circ = \sum S^\circ_{products} - \sum S^\circ_{reactants}$ .

Quick Tips:
- Always balance the chemical equation before calculating entropy changes.
- Pay attention to the stoichiometric coefficients.

30 Medium

30 of 244Question permalink

Application +4 | -1

For an adiabatic expansion of a perfect gas ∆P/P is equal to:

View answer and explanation

Correct option: C

\; - \gamma \frac{{\Delta V}}{V}

Option C (

\; - \gamma \frac{{\Delta V}}{V}

) is correct because for an adiabatic process,

PV^\gamma

is constant, and differentiating this gives the relationship between changes in pressure and volume.

Solution approach

Correct Answer + Core Reason: $\; - \gamma \frac{{\Delta V}}{V}$ . This relationship is derived from the adiabatic process equation $PV^\gamma = constant$ .

Step-by-Step Solution:

Step 1: Start with the adiabatic process equation: $PV^\gamma = constant$

Step 2: Take the natural logarithm of both sides: $\ln(P) + \gamma \ln(V) = \ln(constant)$

Step 3: Differentiate both sides: $\frac{{dP}}{P} + \gamma \frac{{dV}}{V} = 0$

Step 4: Rearrange to solve for $\frac{{dP}}{P}$ : $\frac{{dP}}{P} = - \gamma \frac{{dV}}{V}$

Step 5: Replace $dP$ with $\Delta P$ and $dV$ with $\Delta V$ : $\frac{{\Delta P}}{P} = - \gamma \frac{{\Delta V}}{V}$

Why other options are wrong:
- Option A: $\Delta V/V$ . This is incorrect because it doesn't account for the adiabatic index $\gamma$ .
- Option B: $\gamma \frac{{\Delta V}}{V}$ . This is incorrect because it has the wrong sign; pressure decreases as volume increases in an adiabatic expansion.
- Option D: $- {\gamma ^2}\frac{{\Delta V}}{V}$ . This is incorrect because the relationship is linear with respect to $\gamma$ .

Key Concepts:
- Adiabatic process
- Adiabatic index ( $\gamma$ )
- Differentiation

Memory Aids:
- Remember that in adiabatic expansion, pressure and volume are inversely related, hence the negative sign.

Quick Tips:
- Know the adiabatic process equation and how to differentiate it.

Your Progress

Thermodynamics practice · Medium

Explore Other Chapters in CHEMISTRY

Key Takeaways

Continue Your Practice

Your practice