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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Medium

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241 Medium

241 of 244Question permalink

Analyzing +4 | -1

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

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

View answer and explanation

Correct option: A

If both Statement-I and Statement-II are true, and Statement-II is the correct
explanation of Statement-I

Option A is correct because in reversible adiabatic expansion, the gas does more work, leading to a greater decrease in temperature compared to irreversible expansion, and Statement II correctly explains Statement I.

Solution approach

Correct Answer + Core Reason: If both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement-I. Reversible adiabatic expansion results in more work done and thus a lower final temperature compared to irreversible expansion.

Step-by-Step Solution:

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

Step 2: Recall the first law of thermodynamics: $\Delta U = q + w$ . In adiabatic process, $\Delta U = w$ .

Step 3: Compare reversible and irreversible adiabatic expansion:
- Reversible: Expansion occurs in infinitesimally small steps, allowing the system to remain in equilibrium at all times. More work is done.
- Irreversible: Expansion occurs rapidly against a constant external pressure. Less work is done.

Step 4: Since $\Delta U = w$ and $\Delta U = nC_v\Delta T$ , a more negative $w$ (more work done by the system) means a more negative $\Delta T$ (greater decrease in temperature).

Step 5: In reversible adiabatic expansion, more work is done, so the temperature decreases more. Thus, the final temperature is lower.

Step 6: Statement I is true: The final temperature is less in reversible expansion than in irreversible expansion.

Step 7: Statement II is true: The magnitude of work done is more in reversible process than in irreversible process.

Step 8: Statement II explains Statement I: Because more work is done in the reversible process, the internal energy decreases more, leading to a lower final temperature.

Why other options are wrong:
- Option B: Statement II is the correct explanation of Statement I, so this option is wrong.
- Option C: Statement II is true, so this option is wrong.
- Option D: Statement I is true, so this option is wrong.

Key Concepts:
- Adiabatic process
- Reversible and irreversible processes
- First law of thermodynamics
- Internal energy and temperature relationship

Memory Aids:
- Reversible processes are more efficient (do more work).
- Adiabatic: No heat exchange.

Quick Tips:
- Remember that reversible processes do more work than irreversible processes.
- In adiabatic expansion, work is done at the expense of internal energy, leading to a decrease in temperature.

242 Medium

242 of 244Question permalink

Understanding +4 | -1

When 1 mole of gas is heated at constant volume. Temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct?

View answer and explanation

Correct option: B

q=∆U=500 J,W=0

The correct answer is B (q=∆U=500 J,W=0) because at constant volume, no work is done (W=0), and all the heat supplied goes into increasing the internal energy (q=∆U).

Solution approach

Correct Answer + Core Reason: q=∆U=500 J,W=0. At constant volume, heat supplied equals the change in internal energy, and no work is done.

Step-by-Step Solution:

Step 1: Recall the first law of thermodynamics: $\Delta U = q + W$

Step 2: Apply the condition of constant volume: At constant volume, no work is done, so $W = 0$ .

Step 3: Simplify the equation: $\Delta U = q + 0 = q$

Step 4: Substitute the given value: $q = 500\mathrm{J}$ , so $\Delta U = 500\mathrm{J}$

Why other options are wrong:
- Option A: q=-W=500 J,∆U=0: This is incorrect because at constant volume W=0 and ∆U=q.
- Option C: q=W=500 J,∆U=0: This is incorrect because at constant volume W=0 and ∆U=q.
- Option D: ∆U=0,q=W=-500 J: This is incorrect because at constant volume W=0 and ∆U=q.

Key Concepts:
- First law of thermodynamics
- Internal energy (U)
- Heat (q)
- Work (W)
- Constant volume process

Memory Aids:
- Constant volume = No work
- $\Delta U = q$ (at constant volume)

Quick Tips:
- Remember that at constant volume, all heat absorbed goes into increasing internal energy.

243 Medium

243 of 244Question permalink

Application +4 | -1

Heat of combustion of

\mathrm{C}{\mathrm{H}_4},{\mathrm{C}_2}{\mathrm{H}_4},{\mathrm{C}_2}{\mathrm{H}_6}

are - 890,-1411 and -1560 kJ/mol respectively. Which has the lowest calorific fuel value in kJ/g?

View answer and explanation

Correct option: B

{\mathrm{C}_2}{\mathrm{H}_4}

Option B (

{\mathrm{C}_2}{\mathrm{H}_4}

) is correct because it has the lowest calorific value when the heat of combustion is divided by its molar mass.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_4}$ . The calorific value is the heat of combustion per unit mass. We need to calculate this for each option and compare.

Step-by-Step Solution:

Step 1: Calculate the molar masses:
- $\mathrm{C}{\mathrm{H}_4}$ : 12 + 4(1) = 16 g/mol
- ${\mathrm{C}_2}{\mathrm{H}_4}$ : 2(12) + 4(1) = 28 g/mol
- ${\mathrm{C}_2}{\mathrm{H}_6}$ : 2(12) + 6(1) = 30 g/mol

Step 2: Calculate the calorific value (kJ/g) for each compound:
- $\mathrm{C}{\mathrm{H}_4}$ : |-890 kJ/mol| / 16 g/mol = 55.625 kJ/g
- ${\mathrm{C}_2}{\mathrm{H}_4}$ : |-1411 kJ/mol| / 28 g/mol = 50.39 kJ/g
- ${\mathrm{C}_2}{\mathrm{H}_6}$ : |-1560 kJ/mol| / 30 g/mol = 52 kJ/g

Step 3: Compare the calorific values: ${\mathrm{C}_2}{\mathrm{H}_4}$ has the lowest calorific value (50.39 kJ/g).

Step 4: Final Answer: ${\mathrm{C}_2}{\mathrm{H}_4}$

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{H}_4}$ has a calorific value of 55.625 kJ/g, which is higher than that of ${\mathrm{C}_2}{\mathrm{H}_4}$ .
- Option C: ${\mathrm{C}_2}{\mathrm{H}_6}$ has a calorific value of 52 kJ/g, which is higher than that of ${\mathrm{C}_2}{\mathrm{H}_4}$ .
- Option D: The calorific values are not the same for all the given compounds.

Key Concepts:
- Heat of combustion
- Molar mass
- Calorific value

Memory Aids:
- Calorific value = (Heat of combustion) / (Molar mass)

Quick Tips:
- Always calculate the molar mass correctly before calculating the calorific value.
- Remember to take the absolute value of the heat of combustion when calculating the calorific value.

244 Medium

244 of 244Question permalink

Analyzing +4 | -1

Assertion : For a reaction

2 {NH}_3({~g}) \longrightarrow {N}_2({~g})+3 {H}_2({~g}) ; \Delta H>\Delta E

Reason : Enthalpy change is always greater than internal energy change.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

Option B is correct because both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion. The correct explanation involves the change in the number of moles of gas.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. The assertion is correct because the change in the number of moles of gas is positive. The reason is incorrect because enthalpy change is not always greater than internal energy change; it depends on the sign of $\Delta n_g$ .

Step-by-Step Solution:
Step 1: Analyze the Assertion: For the reaction $2 {NH}_3({~g}) \longrightarrow {N}_2({~g})+3 {H}_2({~g})$ , $\Delta H>\Delta E$ . This is correct because $\Delta n_g = (1+3) - 2 = 2 > 0$ , so $\Delta H = \Delta E + \Delta n_g RT$ , which means $\Delta H > \Delta E$ .
Step 2: Analyze the Reason: Enthalpy change is always greater than internal energy change. This is incorrect because it depends on the sign of $\Delta n_g$ . If $\Delta n_g$ is negative, then $\Delta H < \Delta E$ .
Step 3: Determine the relationship: Both the assertion and reason are correct, but the reason does not explain the assertion. The correct explanation involves the change in the number of moles of gas.

Why other options are wrong:
- Option A: If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion. This is incorrect because the reason is not the correct explanation.
- Option C: If Assertion is correct but Reason is incorrect. This is incorrect because the reason is also correct, but not the correct explanation.
- Option D: If both the Assertion and Reason are incorrect. This is incorrect because both are correct.

Key Concepts:
- Enthalpy Change
- Internal Energy Change
- Change in Number of Moles of Gas
- Thermodynamics

Memory Aids:
- Remember the formula: $\Delta H = \Delta E + \Delta n_g RT$

Quick Tips:
- Pay attention to the sign and magnitude of $\Delta n_g$ when comparing $\Delta H$ and $\Delta E$ .

Revision notes and key takeaways

Review THERMODYNAMICS in your textbook or class notes, then use this set to check your understanding.

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