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NEET • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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Showing 1-30 of 560 questions

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1 Medium
1 of 560
Remembering +4 | -1
The internal energy of one mole of a gas is:
View answer and explanation
Correct option: A
32RT\frac{3}{2}RT
Option A (32RT\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.
2 Medium
2 of 560
Analyzing +4 | -1
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti{T_i} is the initial temperature and Tf{T_f} is the final temperature, which of the following statements is correct?
View answer and explanation
Correct option: A
(Tf)irrev>(Tf)rev{\left( {{T_f}} \right)_{\mathrm{irrev}}} > {\left( {{T_f}} \right)_{\mathrm{rev}}}
Option A ((Tf)irrev>(Tf)rev{\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.
3 Easy
3 of 560
Analyzing +4 | -1
For which of the process, ∆S is negative?
View answer and explanation
Correct option: C
N2(1 g  atom)→N2(8 g  atom){\mathrm{N}_2}\left( {1\,\mathrm{g}\,\,\mathrm{atom}} \right) \to {\mathrm{N}_2}\left( {8\,\mathrm{g}\,\,\mathrm{atom}} \right)
Option C (N2(1 g  atom)→N2(8 g  atom){\mathrm{N}_2}\left( {1\,\mathrm{g}\,\,\mathrm{atom}} \right) \to {\mathrm{N}_2}\left( {8\,\mathrm{g}\,\,\mathrm{atom}} \right)) is correct because entropy decreases when the number of particles or the volume they occupy decreases.
4 Easy
4 of 560
Understanding +4 | -1
The work done by a weightless piston in causing an expansion ∆V (at constant temperature), when the opposing pressure, P is variable, is given by:
View answer and explanation
Correct option: A
  W=−∫PΔV\;W = - \int P\Delta V
Option A,   W=−∫PΔV\;W = - \int P\Delta V, is correct because it represents the work done by a variable opposing pressure during an expansion.
5 Hard
5 of 560
Analyzing +4 | -1
Assertion :- At constant pressure of the change H2O(s)→H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) work done is negative.

Reason :- During phase transition work done is always negative.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because the assertion is true as the work done is negative during expansion, but the reason is false as work done is not always negative during phase transition.
6 Hard
6 of 560
Application +4 | -1
For the reaction, X2O4(l)⟶2XO2(g),ΔU=2.1 kcal,ΔS=20 cal K−1\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g}), \Delta U=2.1 \mathrm{~kcal}, \Delta S=20 \mathrm{~cal} \mathrm{~K}^{-1} at 300 K300 \mathrm{~K}. Hence, ΔG\Delta G is
View answer and explanation
Correct option: B
−2.7 kcal-2.7 \mathrm{~kcal}
Option B (-2.7 kcal) is correct because it is calculated using the formula ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, where ΔH\Delta H is derived from ΔU\Delta U and the change in the number of moles of gas.
7 Easy
7 of 560
Understanding +4 | -1
In a closed insulated container, a liquid is stirred with a paddle to increase its temperature. In this process, which of the following is true
View answer and explanation
Correct option: A
∆E=W≠0,Q=0
Option A (∆E=W≠0,Q=0) is correct because the process is adiabatic (Q=0) and work is done on the liquid by the paddle, increasing its internal energy.
8 Easy
8 of 560
Understanding +4 | -1
Which of the following statement is true?
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Correct option: D
The heat of neutralization of strong acid with strong base is always the same
Option D (The heat of neutralization of strong acid with strong base is always the same) is correct because the reaction involves the same net ionic equation: H+ + OH- → H2O, releasing a fixed amount of heat.
9 Easy
9 of 560
Remembering +4 | -1
For one mole of a monoatomic gas γ\gamma is equal to
View answer and explanation
Correct option: A
1.66
Option A (1.66) is correct because for a monoatomic gas, the ratio of specific heats, γ\gamma, is equal to 5/3, which is approximately 1.66.
10 Easy
10 of 560
Analyzing +4 | -1
Consider the following spontaneous reaction 3X2(g)→2X3(g)3 X_2(g) \rightarrow 2 X_3(g). What are the sign of ΔH,ΔS\Delta H, \Delta S and ΔG\Delta G for the reaction?
View answer and explanation
Correct option: D
−ve,−ve,−ve-ve,-ve,-ve
Option D (-ve,-ve,-ve) is correct because for a spontaneous reaction with a decrease in the number of gas molecules, Δ H, Δ S, and Δ G are all negative.
11 Easy
11 of 560
Application +4 | -1
If one mole of a gas initially at 300 K300 \mathrm{~K} is heated to 500 K500 \mathrm{~K}. The Gibb's free energy change, ΔG\Delta \mathrm{G} for this gas is given by 3.38×10XJ3.38 \times 10^{\mathrm{X}} \mathrm{J}.

Given : S=1.5+3×10−3 T(JK−1 mol−1)\mathrm{S}=1.5+3 \times 10^{-3} \mathrm{~T}\left(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right) and Cv=1.5R\mathrm{C}_{\mathrm{v}}=1.5 \mathrm{R}. Then the value X\mathrm{X} is :
View answer and explanation
Correct option: B
3377
Option B (3377) is correct because it is the calculated value of the Gibb's free energy change based on the given entropy and heat capacity.
12 Medium
12 of 560
Remembering +4 | -1
The enthalpy of a monoatomic gas at TT Kelvin is:
View answer and explanation
Correct option: D
52RT\frac{5}{2} \mathrm{RT}
Option D (52RT\frac{5}{2} \mathrm{RT}) is correct because it represents the enthalpy of a monoatomic gas, considering both translational and electronic energy.
13 Easy
13 of 560
Understanding +4 | -1
For a chemical reaction, ΔG\Delta G will always be negative if:
View answer and explanation
Correct option: C
ΔH\Delta H is negative and TΔST \Delta S is positive
Option C (ΔH\Delta H is negative and TΔST \Delta S is positive) is correct because a negative Δ H and a positive TΔ S will always result in a negative Δ G, indicating spontaneity.
14 Hard
14 of 560
Understanding +4 | -1
During complete combustion of one mole of butane, 2658 kJ2658 \mathrm{~kJ} of heat is released. The thermochemical reaction for above change is
View answer and explanation
Correct option: C
C4H10(g)+132O2(g)→4CO2(g)+5H2O(l);ΔCH=−2658.0 kJ mol−1\mathrm{C}_4 \mathrm{H_{10}(g})+\frac{13}{2} \mathrm{O_2(g}) \rightarrow 4 \mathrm{CO_2(g})+ 5 \mathrm{H}_2 \mathrm{O}(l) ; \Delta_C H =-2658.0 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option C (C4H10(g)+132O2(g)→4CO2(g)+5H2O(l);ΔCH=−2658.0 kJ mol−1\mathrm{C}_4 \mathrm{H_{10}(g})+\frac{13}{2} \mathrm{O_2(g}) \rightarrow 4 \mathrm{CO_2(g})+ 5 \mathrm{H}_2 \mathrm{O}(l) ; \Delta_C H =-2658.0 \mathrm{~kJ} \mathrm{~mol}^{-1}) is correct because it represents the complete combustion of one mole of butane with the correct stoichiometric coefficients and the given enthalpy change.
15 Easy
15 of 560
Understanding +4 | -1
Which of the following statement is incorrect regarding third law of thermodynamics.
View answer and explanation
Correct option: B
There is imperfect order in a crystal at absolute zero
The correct answer is 'There is imperfect order in a crystal at absolute zero' because the third law states that the entropy of a perfectly crystalline solid at absolute zero is zero, implying perfect order.

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CHEMICAL BONDING & MOLECULAR STRUCTURE 595 MCQs STRUCTURE OF ATOM 499 MCQs CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 429 MCQs Equilibrium 412 MCQs STATES OF MATTER 410 MCQs ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES 397 MCQs
16 Easy
16 of 560
Application +4 | -1
What is the change in internal energy when a gas contracts from 377 mL377 \mathrm{~mL} to 177 mL177 \mathrm{~mL} under a constant pressure of 1520 torr, while at the same time being cooled by removing 124 J124 \mathrm{~J} heat?
View answer and explanation
Correct option: B
−83.48 J-83.48 \mathrm{~J}
Option B (-83.48 J) is correct because the change in internal energy is calculated using the first law of thermodynamics, considering both the heat removed and the work done on the gas during contraction.
17 Easy
17 of 560
Remembering +4 | -1
The exchange of heat energy during chemical reaction at constant temperature and pressure occurs in form of:
View answer and explanation
Correct option: C
Enthalpy
Option C (Enthalpy) is correct because enthalpy change (∆H) represents the heat exchanged at constant temperature and pressure.
18 Easy
18 of 560
Remembering +4 | -1
In exothermic reaction, heat is
View answer and explanation
Correct option: A
given out
Option A (given out) is correct because exothermic reactions release heat to the surroundings.
19 Hard
19 of 560
Application +4 | -1
What is the internal energy (kJ)({kJ}) change occurs when 36 g36 {~g} of H2O(l){H}_2 {O}(l) converted to H2O(g){H}_2 {O}({g}) ? ΔH∘\Delta {H}^{\circ} (vapourisation) == 40.79 kJ/mol40.79 {~kJ} / {mol}
View answer and explanation
Correct option: A
75.38
Option A (75.38) is correct because it is calculated by finding the change in internal energy using the enthalpy of vaporization and accounting for the work done during the phase change.
20 Easy
20 of 560
Analyzing +4 | -1
A container has hydrogen and oxygen mixture in ratio of 1 : 1 by weight, then
View answer and explanation
Correct option: C
Entropy of the mixture increases
Option C is correct. When hydrogen and oxygen mix, the entropy increases due to the increase in randomness and disorder of the system.
21 Medium
21 of 560
Remembering +4 | -1
For an ideal monoatomic gas, molar heat capacity at constant volume (Cv)\left( {{C_v}} \right) is
View answer and explanation
Correct option: B
32R\frac{3}{2}R
Option B, 32R\frac{3}{2}R, is the correct molar heat capacity at constant volume (Cv)\left( {{C_v}} \right) for an ideal monoatomic gas.
22 Hard
22 of 560
Application +4 | -1
What is the change in internal energy when a gas contracts from 325 ml325 \mathrm{~ml} to 125 ml125 \mathrm{~ml} at a constant pressure of 2 bar,2 \mathrm{~bar}, while at the same time being cooled by removing 124 J124 \mathrm{~J} heat?
View answer and explanation
Correct option: B
−84 J-84 \mathrm{~J}
Option B (-84 J) is correct because it is calculated using the first law of thermodynamics, considering both the heat removed and the work done due to contraction.
23 Easy
23 of 560
Remembering +4 | -1
The criterion for a spontaneous process is
View answer and explanation
Correct option: B
ΔG<0\Delta \mathrm{G}<0
Option B (ΔG<0) is correct because a spontaneous process occurs when the change in Gibbs free energy (ΔG) is negative.
24 Medium
24 of 560
Analyzing +4 | -1
Statement-I : At equilibrium ΔG0\Delta {G^0} always zero.
Statement-II : At STP active masses of 11.2 L of O2(g){O_2}\left( g \right) and 5.6 L of N2(g){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 ΔG0\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.
25 Medium
25 of 560
Understanding +4 | -1
The process of evaporation of a liquid is accompanied by:
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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).
26 Easy
26 of 560
Understanding +4 | -1
Decomposition of ozone is exothermic reaction
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Correct option: C
Spontaneous at all temperatures
Option C, Spontaneous at all temperatures, is correct because the decomposition of ozone is exothermic and increases entropy, making ∆G negative at all temperatures.
27 Easy
27 of 560
Understanding +4 | -1
Internal energy is sum of
View answer and explanation
Correct option: B
All types of energy of the system
Option B (All types of energy of the system) is correct because internal energy is the sum of all kinetic and potential energies of all the particles within the system.
28 Medium
28 of 560
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.
29 Easy
29 of 560
Understanding +4 | -1
If a process is both endothermic and spontaneous, then:
View answer and explanation
Correct option: A
ΔS>0\Delta \mathrm{S}>0
Option A (ΔS>0\Delta \mathrm{S}>0) is correct because for a spontaneous endothermic process, the increase in entropy must be large enough to overcome the positive enthalpy change, making Δ G negative.
30 Medium
30 of 560
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.
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