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 181-210 of 500 questions

Preparing practice and saved progress…

181 Medium
181 of 500
Understanding +4 | -1
During isothermal expansion of an ideal gas, its
View answer and explanation
Correct option: C
Enthalpy remains unaffected
Option C (Enthalpy remains unaffected) is correct because during isothermal expansion of an ideal gas, the change in temperature is zero, leading to no change in enthalpy.
182 Medium
182 of 500
Analyzing +4 | -1
Equilibrium constant for a hypothetical
reaction, A(g)f2B(g)A\left( g \right)f2B\left( g \right) at 400 K is 102M.10^2M. At 500 K, for the same reaction,
equilibrium constant is 2×102M2 \times {10^2}M. According
to Lechatelier’s principle, favourable
conditions for the formation of 'A' are
View answer and explanation
Correct option: A
Low T, High P
Option A (Low T, High P) is correct because low temperature and high pressure favor the backward reaction, which leads to the formation of 'A' according to Le Chatelier's principle.
183 Medium
183 of 500
Application +4 | -1
The W/QW / Q of a carnot engine is 1/61 / 6. When temperature of sink is reduced by 62∘C62^{\circ} {C}, this ratio becomes twice, therefore the initial temperature of the sink and source are:
View answer and explanation
Correct option: B
37∘C,99∘C37^{\circ} {C}, 99^{\circ} {C}
Option B (37∘C,99∘C37^{\circ} {C}, 99^{\circ} {C}) is correct because solving the Carnot engine efficiency equations with the given conditions yields these temperatures for the sink and source.
184 Easy
184 of 500
Remembering +4 | -1
Heat of formation of compound is defined as :
View answer and explanation
Correct option: A
Heat evolved to form one mole of the compound from its elements
Option A (Heat evolved to form one mole of the compound from its elements) is correct because the heat of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
185 Hard
185 of 500
Understanding +4 | -1
STATEMENT-1 : All combustion reactions are exothermic.

STATEMENT-2 : Enthalpies of products are greater than enthalpies of reactants (ΣvpΔfH(P)>ΣvRΔfH(R))\left(\Sigma v_p \Delta_f H(P)>\Sigma v_R \Delta_f H(R)\right)
View answer and explanation
Correct option: C
If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE
The correct answer is D (If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE) because exothermic reactions release heat, meaning the enthalpy of products is less than that of reactants.
186 Medium
186 of 500
Application +4 | -1
What is ∆n for combustion of 1 mole of benzene, when both the reactants and products are gas at 298 K
View answer and explanation
Correct option: C
0.5
Option C (0.5) is correct because the change in the number of moles (∆n) is calculated by subtracting the moles of gaseous reactants from the moles of gaseous products in the balanced combustion equation, resulting in +0.5.
187 Medium
187 of 500
Application +4 | -1
If the enthalpy change for the reaction is 2NH3( g)→N2( g)+3H2( g)2 \mathrm{NH}_{3(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} is 92 kJ mol−192 \mathrm{~kJ} \mathrm{~mol}^{-1} the enthalpy of formation of ammonia is
View answer and explanation
Correct option: B
−46 kJ mol−1-46 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option B (−46 kJ mol−1-46 \mathrm{~kJ} \mathrm{~mol}^{-1}) is correct because the enthalpy of formation is half the negative of the given enthalpy change for the reverse reaction.
188 Medium
188 of 500
Understanding +4 | -1
The change in internal energy equals
View answer and explanation
Correct option: D
Adiabatic work
Option D (Adiabatic work) is correct because in an adiabatic process, there is no heat exchange with the surroundings (Q=0), so the change in internal energy is equal to the work done.
189 Easy
189 of 500
Analyzing +4 | -1
Which of the following gas will liberate maximum heat on combustion per gram?
View answer and explanation
Correct option: A
Methane
Option A (Methane) is correct because it has the highest hydrogen-to-carbon ratio, leading to maximum heat release per gram during combustion.
190 Medium
190 of 500
Understanding +4 | -1
Select the correct statements.
View answer and explanation
Correct option: C
q=+q=+ve means heat is transferred from the surroundings to the system
Option C is correct because a positive q value indicates that heat is transferred from the surroundings to the system, meaning the system gains heat.
191 Medium
191 of 500
Understanding +4 | -1
The correct relation for KfK_f is/are:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the given expressions are valid representations of the cryoscopic constant, KfK_f.
192 Medium
192 of 500
Application +4 | -1
If at 298 K the bond energies of C−H, C−C, C=CC - H,\,C - C,\,C = C and H−HH - H are respectively, 414, 347, 615 and 435 kJ mol−1,{\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}, the value of enthalpy change for the reaction H2C=CH2(g)+H2(g)→H3C−CH3(g){H_2}C = C{H_2}\left( g \right) + {H_2}\left( g \right) \to {H_3}C - C{H_3}\left( g \right)
at 298 K will be:
View answer and explanation
Correct option: D
– 125 kJ
Option D (– 125 kJ) is correct because the enthalpy change is calculated by subtracting the bond energies of the products from the bond energies of the reactants.
193 Medium
193 of 500
Application +4 | -1
The temperature of 15 ml15 \mathrm{~ml} of a strong acid increases by 2∘C2^{\circ} \mathrm{C} when 15 ml15 \mathrm{~ml} of a strong base is added to it. If 5 ml5 \mathrm{~ml} of each are mixed, temperature should increase by
View answer and explanation
Correct option: C
2∘C2^{\circ} \mathrm{C}
Option C (2∘C2^{\circ} \mathrm{C}) is correct because the temperature change depends on the concentration and the heat of neutralization, which remains constant under the same conditions.
194 Medium
194 of 500
Application +4 | -1
Temperature of 5 moles of a gas is decreased
by 2K at constant pressure. Indicate the correct
statement
View answer and explanation
Correct option: B
Work done over the gas is = 10 R
Option B (Work done over the gas is = 10 R) is correct because the work done at constant pressure is calculated as -PΔV, and since the temperature decreases, the volume decreases, resulting in work being done *on* the gas.
195 Hard
195 of 500
Application +4 | -1
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)C{H_3}OH\left( l \right) + \frac{3}{2}{O_2}\left( g \right) \to C{O_2}\left( g \right) + 2{H_2}O\left( l \right) At 298 K, standard Gibb’s energies of formation for CH3OH(l), H2O(l)  and  CO2(g)C{H_3}OH\left( l \right),\,{H_2}O\left( l \right)\,\,and\,\,C{O_2}\left( g \right) are -166.2, -237.2 and −394.4kJmol−1- 394.4\mathrm{ }kJ\mathrm{ }mo{l^{ - 1}} respectively. If standard enthalpy of combustion of methanol is −726kJmol−1- 726kJ\mathrm{ }mo{l^{ - 1}} efficiency of the fuel cell will be:
View answer and explanation
Correct option: C
0.97
Option C (97%) is correct because the efficiency of the fuel cell is calculated by comparing the change in Gibbs free energy to the standard enthalpy of combustion.
196 Medium
196 of 500
Application +4 | -1
The latent heat of fusion of ice at 0∘C0^{\circ} \mathrm{C} being 1.44 k cal mol−11.44 \mathrm{~k~cal} \mathrm{~mol}^{-1} and difference in heat capacity of ice and water being 9.3 cal mol−19.3 \mathrm{~cal} \mathrm{~mol}^{-1}, thus latent heat of fusion of ice at −27∘C-27^{\circ} \mathrm{C} is
View answer and explanation
Correct option: B
1.254 k cal mol−11.254 \mathrm{~k~cal} \mathrm{~mol}^{-1}
The correct answer is B (1.254 k~cal mol⁻¹) because the latent heat of fusion changes with temperature according to Kirchhoff's law.
197 Easy
197 of 500
Understanding +4 | -1
Which of the following is an intensive property?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because temperature, viscosity, and surface tension are all intensive properties, meaning they do not depend on the amount of substance.
198 Medium
198 of 500
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 (If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I) is correct because irreversible adiabatic compression involves more work done, leading to a higher final temperature compared to reversible adiabatic compression.
199 Hard
199 of 500
Application +4 | -1
A gas present in a cylinder fitted with a frictionless piston expands against a constant pressure of 1 atm1 {~atm} from a volume of 2 litre to a volume of 6 litre. In doing so, it absorbs 800 J800 {~J} heat from surroundings. Determine increase in internal energy of process.
View answer and explanation
Correct option: A
394.95
Option A (394.95 J) is correct because it is the increase in internal energy calculated using the first law of thermodynamics.
200 Medium
200 of 500
Application +4 | -1
H2S(g)→HS−(g)+H+(g),ΔH∘=x1,\mathrm{H}_2 \mathrm{S}(\mathrm{g}) \rightarrow \mathrm{HS}^{-}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{g}), \Delta \mathrm{H}^{\circ}=\mathrm{x}_1,
ΔHf∘(H2S(g))=x2,ΔHf∘(H+(g))=x3\Delta \mathrm{H_f^\circ}\left(\mathrm{H}_2 \mathrm{S}\right. (\mathrm{g})) =\mathrm{x}_2, \Delta \mathrm{H_f^\circ}\left(\mathrm{H}^{+}(\mathrm{g})\right)=\mathrm{x}_3
Hence, ΔHf∘(HS−)\Delta \mathrm{H_f^\circ}\left(\mathrm{HS}^{-}\right) is:
View answer and explanation
Correct option: A
x1+x2−x3\mathrm{x_1+x_2-x_3}
Option A (x1+x2−x3\mathrm{x_1+x_2-x_3}) is correct because it correctly rearranges the enthalpy equation to solve for the heat of formation of HS-.
201 Medium
201 of 500
Application +4 | -1
The heat of combustion of carbon monoxide at constant pressure and at constant volume at 27∘C27^{\circ} \mathrm{C} will differ from one another by:
View answer and explanation
Correct option: C
300 cal300 \mathrm{~cal}
Option C (300 cal) is correct because it is the difference between heat of combustion at constant pressure and constant volume, calculated using the formula ΔH - ΔU = Δn_gRT.
202 Hard
202 of 500
Application +4 | -1
The standard enthalpies of formation of CO2(g),  H2O(l)C{O_2}\left( g \right),\,\,{H_2}O\left( l \right) and glucose (s) at 25o  C{25^\mathrm{o}}\,\,C are −400  KJ⋅mol−1,  −300  KJ⋅mol−1- 400\,\,KJ \cdot mo{l^{ - 1}},\,\, - 300\,\,KJ \cdot mo{l^{ - 1}} and −1300  KJ/mol- 1300\,\,KJ/mol respectively. The standard enthalpy of combustion per gram of glucose at 25o  C{25^\mathrm{o}}\,\,C is
View answer and explanation
Correct option: C
- 16.11 KJ
Option C (- 16.11 KJ) is correct because it is the standard enthalpy of combustion per gram of glucose calculated using the given enthalpies of formation.
203 Medium
203 of 500
Analyzing +4 | -1
A: Dissolution of sugar in water proceed via increases in entropy.
R: Entropy decreases when egg is hard boiled.
View answer and explanation
Correct option: C
If A is correct but R is incorrect
Option C is correct because the dissolution of sugar in water increases entropy, but boiling an egg also increases entropy, making the reason incorrect.
204 Medium
204 of 500
Application +4 | -1
Cdiamond+O2→CO2;ΔH{\mathrm{C}_{\mathrm{diamond}}} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2};\Delta H=-395.3 kJ/mol
Cgraphite+O2→CO2;ΔH{\mathrm{C}_{\mathrm{graphite}}} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2};\Delta H=-393.4 kJ/mol
Cgraphite→Cdiamond;ΔH{\mathrm{C}_{\mathrm{graphite}}} \to {\mathrm{C}_{\mathrm{diamond}}};\Delta \mathrm{H}=?
View answer and explanation
Correct option: D
+1.9+ 1.9
Option D (+1.9+ 1.9) is correct because it is calculated by subtracting the enthalpy of formation of CO2 from diamond from that of graphite.
205 Medium
205 of 500
Analyzing +4 | -1
Given that C+O2→CO2,ΔH∘=−x kJ\mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2, \Delta \mathrm{H}^{\circ}=-\mathrm{x} \mathrm{~kJ}

2CO+O2→2CO2,ΔH∘=−y kJ2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2, \Delta \mathrm{H}^{\circ}=-\mathrm{y} \mathrm{~kJ}

What is heat of formation of CO\mathrm{CO}?
View answer and explanation
Correct option: A
y−2x2\frac{\mathrm{y-2 x}}{2}
Option A is correct because the heat of formation of CO can be calculated by manipulating the given equations using Hess's Law.
206 Medium
206 of 500
Application +4 | -1
For an adiabatic expansion of a perfect gas ∆P/P is equal to:
View answer and explanation
Correct option: C
  −γΔVV\; - \gamma \frac{{\Delta V}}{V}
Option C (−γΔVV- \gamma \frac{{\Delta V}}{V}) is the correct expression for an adiabatic process relating pressure and volume.
207 Medium
207 of 500
Analyzing +4 | -1
In the following reaction,

HF(aq)+NaOH(aq)⟶NaF(aq)+H2O(l);ΔH∘=−68.6 kJ3mol−1\begin{aligned} & \mathrm{HF}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaF}(a q)+\mathrm{H}_2 \mathrm{O}(l) ; \Delta H^\circ=-68.6 \mathrm{~kJ}^3 \mathrm{mol}^{-1} \\ \end{aligned}

This value is much higher than the heat of neutralisation of strong acid with strong base (−57.3 kJ mol−1)\left(-57.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\right). This is because
View answer and explanation
Correct option: B
there is a hydration of F−F^-ion and the process is
exothermic
Option B is correct because the higher heat of neutralization for HF is due to the exothermic hydration of the F- ion, which releases additional energy.
208 Medium
208 of 500
Application +4 | -1
A quantity of 1 g1 \text{ g} of water on evaporation at
atmospheric pressure forms 1671 cm31671 \, \text{cm}^3
of steam.
Heat of vaporization is 540 cal/g.540 \text{ cal/g}. The approximate
increase in internal energy is (1 L – atom = 24 cal)
View answer and explanation
Correct option: B
500 cal
Option B (500 cal) is correct because the increase in internal energy is calculated using the first law of thermodynamics, considering heat absorbed and work done.
209 Medium
209 of 500
Application +4 | -1
Calculate the number of kJ\mathrm{kJ} of heat necessary to raise the temperature of 60 g60 \mathrm{~g} of aluminium from 35∘C35^{\circ} \mathrm{C} to 55∘C55^{\circ} \mathrm{C}. Molar heat capacity of Al{Al} is 24 J mol−1 K−124 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}.
View answer and explanation
Correct option: A
1.067 kJ
The correct answer is 1.067 kJ. The heat required is calculated using the formula q = m * C * ΔT, where m is the number of moles, C is the molar heat capacity, and ΔT is the change in temperature.
210 Medium
210 of 500
Understanding +4 | -1
The enthalpy of formation of HI is 30.4 kJ. Which statement is false according to this observation?
View answer and explanation
Correct option: C
HI is a stable compound
Option C (HI is a stable compound) is correct because a positive enthalpy of formation indicates that HI is an endothermic and therefore unstable compound.
Revision notes and key takeaways

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

Showing 181-210 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.