Skip to main content
Back to THERMODYNAMICS

Your Progress

0 / 500 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 301-330 of 500 questions

Preparing practice and saved progress…

301 Medium
301 of 500
Application +4 | -1
During isothermal expansion of one mole of an ideal gas from 10 atm10 {~atm} to 1 atm1 {~atm} at 273 K273 {~K}, thework done is (gas constant =2=2 ) :
View answer and explanation
Correct option: C
−1257.43cal-1257.43 {cal}
Option C (-1257.43 cal) is correct because it's the work done calculated using the formula for isothermal expansion of an ideal gas.
302 Medium
302 of 500
Application +4 | -1
A gas expands against a variable pressure given by P=20VP=\frac{20}{V} (where PP in atm and VV in L\mathrm{L} ). During expansion from volume of 1 litre to 10 litre, the gas undergoes a change in internal energy of 400 J400 \mathrm{~J}. How much heat is absorbed by the gas during expansion?
View answer and explanation
Correct option: C
5065.8 J5065.8 \mathrm{~J}
Option C (5065.8 J) is correct because the heat absorbed is calculated using the first law of thermodynamics, where q = ΔU - w, and the work done is found by integrating P dV.
303 Easy
303 of 500
Understanding +4 | -1
The heat of neutralisation for strong acid and strong base forming 2 moles of water is
View answer and explanation
Correct option: A
−2×57.1 kJ-2 \times 57.1 \mathrm{~kJ}
Option A is correct because the heat of neutralization for a strong acid and strong base is -57.1 kJ per mole of water formed, so for 2 moles, it's -2 * 57.1 kJ.
304 Easy
304 of 500
Remembering +4 | -1
The law of Lavoisier and Laplace is based on :
View answer and explanation
Correct option: A
The principle of conservation of energy
Option A (The principle of conservation of energy) is correct because the law of Lavoisier and Laplace is based on the conservation of energy.
305 Medium
305 of 500
Application +4 | -1
At 298 K298 \mathrm{~K} what will be the change in standard internal energy change for the given reaction OF2( g)+H2O(g)→O2( g)+2HF(g);\mathrm{OF}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{O}_2(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) ;
View answer and explanation
Correct option: C
−310 kJ-310 \mathrm{~kJ}
Option C (-310 kJ) is correct because it is calculated using the formula ΔU = ΔH - Δn_gRT, accounting for the change in the number of moles of gas.
306 Easy
306 of 500
Understanding +4 | -1
The gas in a refrigerator causes cooling on expansion because:
View answer and explanation
Correct option: B
Heat of the gas is lost as work is done by the gas
Option B is correct. When a gas expands in a refrigerator, it does work, which causes it to lose heat and cool down.
307 Easy
307 of 500
Understanding +4 | -1
Energy equivalent to one erg, one joule and one calorie are in order:
View answer and explanation
Correct option: C
1  cal>1  J>1  erg1\mathrm{\;cal} > 1\;J > 1\;erg
Option C (1  cal>1  J>1  erg1\mathrm{\;cal} > 1\;J > 1\;erg) is correct because 1 calorie is greater than 1 joule, which is greater than 1 erg.
308 Medium
308 of 500
Remembering +4 | -1
Work done in reversible adiabatic process is given by:
View answer and explanation
Correct option: B
nR(γ−1)(T2−T1)\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)
Option B (nR(γ−1)(T2−T1)\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)) is correct because it represents the work done in a reversible adiabatic process in terms of the number of moles, gas constant, adiabatic index, and temperature change.
309 Medium
309 of 500
Understanding +4 | -1
For an isothermal reversible expansion process, the value of qq can be calculated by the expression
View answer and explanation
Correct option: A
q=2.303  nRTlog⁡V2V1q = 2.303\,\,nRT\log \frac{{{V_2}}}{{{V_1}}}
Option A (q=2.303  nRTlog⁡V2V1q = 2.303\,\,nRT\log \frac{{{V_2}}}{{{V_1}}}) is correct because it represents the heat absorbed during an isothermal reversible expansion.
310 Hard
310 of 500
Application +4 | -1
The heat of combustion of ethanol into carbon dioxides and water is −327  kcal- 327\,\,kcal at constant pressure. The heat evolved (in cal) at constant volume and 27o  C{27^\mathrm{o}}\,\,C (if all gases behave ideally) is (R=2R = 2 cal mol−1 K−1mo{l^{ - 1}}\,{K^{ - 1}} )……
View answer and explanation
Correct option: A
-326400
The correct answer is A (-326400) because it is calculated by applying the formula relating heat of combustion at constant pressure and constant volume, accounting for the change in the number of moles of gas.
311 Medium
311 of 500
Understanding +4 | -1
At 27∘C27^{\circ} \mathrm{C} the reaction, C6H6(l)+15/2O2(g)→6CO2(g)+3H2O(l){C}_6 {H}_{6}({l})+15 / 2 {O}_{2}({g}) \rightarrow 6 {CO}_{2}({g})+3 {H}_2 {O}{({l})} proceeds spontaneously because the magnitude of
View answer and explanation
Correct option: B
ΔH>TΔS\Delta {H} > {T} \Delta {S}
Option B (ΔH>TΔS\Delta H > T \Delta S) is correct because for a spontaneous reaction, the change in Gibbs free energy (ΔG\Delta G) must be negative, and ΔG=ΔH−TΔS\Delta G = \Delta H - T \Delta S. Therefore, ΔH\Delta H must be less than TΔST \Delta S for the reaction to be spontaneous.
312 Medium
312 of 500
Understanding +4 | -1
In an isobaric process, when temperature change from T1\mathrm{T}_1 to T2,ΔS\mathrm{T}_2, \Delta \mathrm{S} is equal to:
View answer and explanation
Correct option: A
2.303Cplog⁡(T2/T1)2.303 \mathrm{C}_{\mathrm{p}} \log \left(\mathrm{T}_2 / \mathrm{T}_1\right)
Option A (2.303Cplog⁡(T2/T1)2.303 \mathrm{C}_{\mathrm{p}} \log (\mathrm{T}_2 / \mathrm{T}_1)) is correct because it represents the entropy change in an isobaric process derived from the general entropy change formula.
313 Medium
313 of 500
Understanding +4 | -1
An ideal monoatomic gas initially at 300 K300 \mathrm{~K} expands adiabatically into vacuum to double its volume. The final temperature of gas is
View answer and explanation
Correct option: A
300 K300 \mathrm{~K}
Option A (300 K300 \mathrm{~K}) is correct because adiabatic expansion into a vacuum is a free expansion, which is an isothermal process for an ideal gas.
314 Medium
314 of 500
Application +4 | -1
Choose the reaction in which ΔH\Delta {H} is not equal to ΔU\Delta {U} ?
View answer and explanation
Correct option: B
C2H4( g)+H2( g)→C2H6( g){C}_2 {H}_4({~g})+{H}_2({~g}) \rightarrow {C}_2 {H}_6({~g})
Option B (C2H4( g)+H2( g)→C2H6( g){C}_2 {H}_4({~g})+{H}_2({~g}) \rightarrow {C}_2 {H}_6({~g})) is correct because the change in the number of gaseous moles (Δng\Delta n_g) is not zero, making ΔH\Delta H not equal to ΔU\Delta U.
315 Medium
315 of 500
Remembering +4 | -1
Which of the following represents the first law
of thermodynamics
View answer and explanation
Correct option: A
ΔU=q+W\Delta U=q+W
Option A (ΔU=q+W\Delta U=q+W) is correct because it represents the first law of thermodynamics, stating that the change in internal energy equals heat plus work.
316 Medium
316 of 500
Understanding +4 | -1
For the reaction 2Cl(g)→Cl2(g),2Cl\left( g \right) \to C{l_2}\left( g \right), the signs of ΔH\Delta H and ΔS\Delta S respectively are
View answer and explanation
Correct option: C
−,  −- ,\,\, -
Option C (−,−-, -) is correct because bond formation releases energy (negative Δ H) and decreases randomness (negative Δ S).
317 Hard
317 of 500
Application +4 | -1
Calculate the standard internal energy change for the following reaction at 25∘C25^{\circ} \mathrm{C}.

2H2O2(l)⟶2H2O(l)+O2(g)H2O2(l)=−188.0 kJ mol−1H2O(l)=−286.0 kJ mol−1(R=8.314 J K−1 mol−1)\begin{aligned} & 2 \mathrm{H}_2 \mathrm{O}_2(\mathrm{l}) \longrightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{O}_2(\mathrm{g}) \\ & \mathrm{H}_2 \mathrm{O}_2(\mathrm{l})=-188.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{H}_2 \mathrm{O}(\mathrm{l})=-286.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \left(\mathrm{R}=8.314 \mathrm{~J~K}^{-1} \mathrm{~mol}^{-1}\right) \end{aligned}
View answer and explanation
Correct option: A
−193.58 kJ-193.58 \mathrm{~kJ}
Option A (−193.58 kJ-193.58 \mathrm{~kJ}) is correct because it is calculated by using the given enthalpies of formation and the formula relating Δ H and Δ E.
318 Medium
318 of 500
Application +4 | -1
1 g H21 \mathrm{~g~H}_2 gas at S.T.P. is expanded so that volume is doubled. Find the work done in Joules. (1 litre atm = 101.3 J101.3 \mathrm{~J})
View answer and explanation
Correct option: A
1134.56 J1134.56 \mathrm{~J}
Option A (1134.56 J) is correct because the work done during the expansion is calculated using the formula W = P(V2 - V1), where P is the pressure and V is the volume.
319 Easy
319 of 500
Remembering +4 | -1
The enthalpies of elements under the following conditions are assumed to be
View answer and explanation
Correct option: A
Zero at 298 K and 1 atm
Option A (Zero at 298 K and 1 atm) is correct because, by convention, the standard enthalpy of formation of an element in its standard state is defined as zero at 298 K and 1 atm.
320 Medium
320 of 500
Understanding +4 | -1
Considering entropy(s) as the thermodynamic parameter, the criterion for the spontaneity of any process is:
View answer and explanation
Correct option: A
ΔSsys +ΔSsurr >0\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>0
Option A (ΔSsys +ΔSsurr >0\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>0) is correct because the spontaneity of a process is determined by the total entropy change of the system and surroundings, which must be greater than zero for a spontaneous process.
321 Easy
321 of 500
Understanding +4 | -1
For the dissociation reaction, H2( g)→2H(g),ΔH=162Kcal\mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}), \Delta \mathrm{H}=162 \mathrm{Kcal}, heat of atomization of H\mathrm{H} is
View answer and explanation
Correct option: A
81Kcal81 \mathrm{Kcal}
Option A (81 Kcal) is correct because the heat of atomization of H is half of the dissociation energy of H2.
322 Medium
322 of 500
Analyzing +4 | -1
The standard molar enthalpies of formations of H2O(l)\mathrm{H}_2 \mathrm{O}({l}) and H2O2(l)\mathrm{H}_2 \mathrm{O}_2({l}) are −286-286 and −188 kJ/mol,-188 \mathrm{~kJ} / \mathrm{mol}, respectively. Molar enthalpies of vaporization of H2O\mathrm{H}_2 \mathrm{O} and H2O2\mathrm{H}_2 \mathrm{O}_2 are 4444 and 53 kJ,53 \mathrm{~kJ}, respectively. The bond dissociation enthalpy of O2(g)\mathrm{O}_2(\mathrm{g}) is 498 kJ/mol.498 \mathrm{~kJ} / \mathrm{mol}. Calculate the bond dissociation enthalpy of O−O\mathrm{O}-\mathrm{O} bond in H2O2,\mathrm{H}_2 \mathrm{O}_2, assuming that the bond dissociation enthalpy of O−H\mathrm{O}-\mathrm{H} bond is same in both H2O\mathrm{H}_2 \mathrm{O} and H2O2.\mathrm{H}_2 \mathrm{O}_2.
View answer and explanation
Correct option: A
142 kJ/mol142 \mathrm{~kJ} / \mathrm{mol}
Option A (142 kJ/mol) is correct because by using Hess's Law and the given enthalpies of formation and vaporization, along with the bond dissociation enthalpy of O2, we can calculate the O-O bond dissociation enthalpy.
323 Hard
323 of 500
Application +4 | -1
Determine ΔG∘\Delta G^{\circ} for the following reaction

CO(g)+12O2( g)→CO2( g)\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})

ΔH∘=−282.84 kJ given, S∘CO2=213.8\Delta H^{\circ}=-282.84 \mathrm{~kJ} \text { given, } \mathrm{S}^{\circ} \mathrm{CO}_2=213.8

 S∘O2=205.8JK/mol\mathrm{~S}^{\circ} \mathrm{O}_2=205.8 \mathrm{JK} / \mathrm{mol}
View answer and explanation
Correct option: C
−257.033 kJ-257.033 \mathrm{~kJ}
Option C (−257.033 kJ-257.033 \mathrm{~kJ}) is correct because it is calculated by using the Gibbs-Helmholtz equation, considering the change in enthalpy and entropy for the given reaction.
324 Medium
324 of 500
Application +4 | -1
A gas, while expanding absorbs 25 J25 \mathrm{~J} of heat and does 243 J243 \mathrm{~J} of work. Thus, ΔE\Delta E (change in internal energy) for the gas is
View answer and explanation
Correct option: D
−218 J-218 \mathrm{~J}
Option D (−218 J-218 \mathrm{~J}) is correct because the change in internal energy is calculated using the first law of thermodynamics, considering the signs of heat absorbed and work done.
325 Medium
325 of 500
Application +4 | -1
For a perfectly crystalline solid Cp,m=aT3C_{p, m}=a T^3, where aa is constant. If Cp,mC_{p, m} is 0.42 J/K mol at10 K0.42 \mathrm{~J} / \mathrm{K} \mathrm{~mol} \mathrm{~at} 10 \mathrm{~K}, molar entropy at 20 K20 \mathrm{~K} is :
View answer and explanation
Correct option: C
1.12 J/K mol1.12 \mathrm{~J} / \mathrm{K} \mathrm{~mol}
Option C (1.12 J/K mol1.12 \mathrm{~J} / \mathrm{K} \mathrm{~mol}) is correct because it is calculated by integrating Cp,m/TC_{p,m}/T from 0 to 20 K, using the given relationship Cp,m=aT3C_{p, m}=a T^3.
326 Medium
326 of 500
Analyzing +4 | -1
Consider the following conversions:

(i) O(g)+e−→O−(g),ΔH1\mathrm{O}_{(g)}+e^{-} \rightarrow O^{-}(g), \Delta H_1

(ii) F(g)+e−→F−(g),ΔH2\mathrm{F}_{(g)}+e^{-} \rightarrow F^{-}(g), \Delta H_2

(iii) Cl(g)+e−→Cl−(g),ΔH3\mathrm{Cl}_{(\mathrm{g})}+\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}(\mathrm{g}), \Delta \mathrm{H}_3

(iv) O(g)−+e−→O2−(g),ΔH4\mathrm{O}_{(g)}^{-}+e^{-} \rightarrow O^{2-}(g), \Delta H_4

That according to given information the incorrect statements is:
View answer and explanation
Correct option: D
ΔH1\Delta H_1 and ΔH3\Delta H_3 are negative whereas ΔH2\Delta H_2 and ΔH4\Delta H_4 are positive
Option D is correct because the first electron gain enthalpy for O, F, and Cl are negative, but the second electron gain enthalpy for oxygen is positive due to electron-electron repulsion.
327 Easy
327 of 500
Application +4 | -1
A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If ∆H is the enthalpy change and ∆E is the change in internal energy, then
View answer and explanation
Correct option: B
∆H<∆E
Option B (∆H<∆E) is correct because the number of moles of gas decreases during the reaction, leading to a negative ∆n and thus ∆H < ∆E.
328 Hard
328 of 500
Application +4 | -1
Standard entropy of X2,Y2X_2, Y_2 and XY4X Y_4 are 30, 60 and 110 J/K mol110 \mathrm{~J} / \mathrm{K} \mathrm{~mol} respectively. For the reaction 12X2+2Y2→XY4,ΔH=−40 kJ\frac{1}{2} \mathrm{X}_2+2 \mathrm{Y}_2 \rightarrow \mathrm{XY}_4, \Delta \mathrm{H}=-40 \mathrm{~kJ}. To be at equilibrium, the temperature will be:
View answer and explanation
Correct option: B
1600 K1600 \mathrm{~K}
The correct answer is 1600 K1600 \mathrm{~K} because at equilibrium, ΔG=0\Delta G = 0, and the temperature is calculated using ΔH=TΔS\Delta H = T\Delta S.
329 Medium
329 of 500
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) is correct because in an irreversible adiabatic expansion, less work is done, resulting in a smaller decrease in internal energy and a higher final temperature compared to a reversible process.
330 Easy
330 of 500
Understanding +4 | -1
Which is an extensive property of the system?
View answer and explanation
Correct option: B
Volume
Option B (Volume) is correct because volume is an extensive property, meaning it depends on the amount of substance present.
Revision notes and key takeaways

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

Showing 301-330 of 500 questions

Your practice

0 / 500 answered · 500 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.