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*Simulator*

**Previous Years JEE Advance Questions**

(A) Internal energy

(B) Irreversible expansion work

(C) Reversible expansion work

(D) Molar enthalpy

**[JEE 2009]**

**Sol.**(A,D)

**[JEE 2010]**

**Sol.**2

$\mathrm{W}_{\mathrm{d}}=-(4 \times 1.5)+(0)-(1 \times 1)-\left(\frac{2}{3} \times 2.5\right)$

$\mathrm{W}_{\mathrm{S}}=-\mathrm{P}_{1} \mathrm{V}_{1} \ln \frac{\mathrm{V}_{2}}{\mathrm{V}_{1}}$

$=-4 \times 05 \times \ln \frac{5.5}{0.5}$

**[JEE 2011]**

**Sol.**$A \rightarrow(P, R, S) B \rightarrow(R, S) C \rightarrow T, D \rightarrow(P, Q, T)$

(A) $\mathrm{T}_{1}=\mathrm{T}_{2}$

(B) $\mathrm{T}_{3}>\mathrm{T}_{1}$

(C) $\mathrm{W}_{\text {isothermal }}>\mathrm{W}_{\text {adiabatic }}$

(D) $\Delta \mathrm{U}_{\text {isothemal }}>\Delta \mathrm{U}_{\text {adibaic }}$

**[JEE 2012]**

**Sol.**(A,D)

(A) $\Delta \mathrm{S}_{\mathrm{x} \rightarrow \mathrm{z}}=\Delta \mathrm{S}_{\mathrm{x} \rightarrow \mathrm{y}}+\Delta \mathrm{S}_{\mathrm{y} \rightarrow \mathrm{z}}$

(B) $\mathrm{W}_{\mathrm{x} \rightarrow \mathrm{z}}=\mathrm{W}_{\mathrm{x} \rightarrow \mathrm{y}}+\mathrm{W}_{\mathrm{y} \rightarrow \mathrm{z}}$

(C) $\mathrm{W}_{\mathrm{x} \rightarrow \mathrm{y} \rightarrow \mathrm{z}}=\mathrm{W}_{\mathrm{x} \rightarrow \mathrm{y}}$

(D) $\Delta \mathrm{S}_{\mathrm{x} \rightarrow \mathrm{y} \rightarrow \mathrm{z}}=\Delta \mathrm{S}_{\mathrm{x} \rightarrow \mathrm{y}}$

**[JEE 2012]**

**Sol.**(A,C)

**Paragraph for Question 6 and 7**

A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure.

(A) K to L and L to M

(B) L to M and N to K

(C) L to M and M to N

(D) M to N nd N to K

**[JEE 2013]**

**Sol.**(B)

Isochoric $\Rightarrow \mathrm{V}-$ constant

(A) Heating, cooling, heating, cooling

(B) cooling, heating, cooling, heating

(C) Heating, cooling, cooling, heating

(D) Cooling, heating, heating, cooling

**[JEE 2013]**

**Sol.**(C) Isochoric $\Rightarrow \mathrm{V}-$ constant

(A) q = 0

(B) $\mathrm{T}_{2}=\mathrm{T}_{1}$

(C) $\mathrm{P}_{2} \mathrm{V}_{2}=\mathrm{P}_{1} \mathrm{V}_{1}$

$(\mathrm{D}) \mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}=\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}$

**[JEE 2014]**

**Sol.**1,2,3

$\mathrm{q}=0 ; \mathrm{w}=0 ; \Delta \mathrm{U}=0 ; \mathrm{T}_{1}=\mathrm{T}_{2} ; \mathrm{PV}=\mathrm{constant}$

(1 L atm = 101.3 J)

(A) 5.763 (B) 1.013 (C) –1.013 (D) –5.763

**[JEE – Advanced – 2016]**

**Sol.**(C)

(A) With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive

(B) With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases

(C) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases

(D) With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system negative

**[JEE – Adv. 2017]**

**Sol.**(B,C)

(A) The work done on the gas is maximum when it is compressed irreversibly from ($\mathrm{p}_{2}$ , $\mathrm{V}_{2}$) to ($\mathrm{p}_{1}, \mathrm{V}_{1}$) against constant pressure $\mathrm{p}_{1}$

(B) The work done on the gas is less when it is expanded reversibly from $\mathrm{V}_{1}$ to $\mathrm{V}_{2}$ under adiabatic conditions as compared to that when expanded reversibly from $\mathrm{V}_{1}$ to $\mathrm{V}_{2}$ under isothermal conditions.

(C) The change in internal energy of the gas (i) zero, if it is expanded reversibly with $\mathrm{T}_{1} \mathrm{T}_{2}$, and (ii) positive, if it is expanded reversibly under adiabatic conditions with $\mathrm{T}_{1} \neq \mathrm{T}_{2}$

(D) If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic

**[JEE – Adv. 2017]**

**Sol.**1,2,4

(3) (i) $\left.\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=0 \text { (isothermal hence } \Delta \mathrm{T}=0\right)$

(ii) $\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}=-\mathrm{ve}(\mathrm{q}=0, \mathrm{w}<0)$

$\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T} \Rightarrow \Delta \mathrm{T}<0$

$\begin{aligned}(\mathbf{4}) \mathrm{q} &=0(\text { adiabatic }), \mathrm{w}=0(\text { free expansion }) \\ \Delta \mathrm{U} &=0 \Rightarrow \Delta \mathrm{T}=0 \text { (isothermal) } \end{aligned}$

The correct option(s) is (are)

(A) $\mathrm{q}_{\mathrm{AC}}=\Delta \mathrm{U}_{\mathrm{BC}}$ and $\mathrm{w}_{\mathrm{AB}}=\mathrm{P}_{2}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$

(B) $\mathrm{w}_{\mathrm{BC}}=\mathrm{P}_{2}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$ and $\mathrm{q}_{\mathrm{BC}}=\Delta \mathrm{H}_{\mathrm{AC}}$

(C) $\Delta \mathrm{H}_{\mathrm{CA}}<\Delta \mathrm{U}_{\mathrm{CA}}$ and $\mathrm{q}_{\mathrm{AC}}=\Delta \mathrm{U}_{\mathrm{BC}}$

(D) $\mathrm{q}_{\mathrm{BC}}=\Delta \mathrm{H}_{\mathrm{AC}}$ and $\Delta \mathrm{H}_{\mathrm{CA}}>\Delta \mathrm{U}_{\mathrm{CA}}$

**[JEE – Adv. 2018**

**Sol.**(B,C)

If $\mathrm{T}_{2}>\mathrm{T}_{1},$ the correct statement(s) is (are)

(Assume $\Delta \mathrm{H}^{\theta}$ and $\Delta \mathrm{S}^{\theta}$ are independent of temperature and ratio of $\ln \mathrm{K}$ at $\mathrm{T}_{1}$ to $\ln \mathrm{K}$ at $\mathrm{T}_{2}$ is greater than $\mathrm{T}_{2 / \mathrm{T}_{\mathrm{T}}}$. Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

(A) $\Delta \mathrm{H}^{\theta}<0, \Delta \mathrm{S}^{\theta}<0$

(B) $\Delta \mathrm{G}^{\theta}<0, \Delta \mathrm{H}^{\theta}>0$

(C) $\Delta \mathrm{G}^{\theta}<0, \Delta \mathrm{S}^{\theta}<0$

(D) $\Delta \mathrm{G}^{\theta}<0, \Delta \mathrm{S}^{\theta}>0$

**[JEE – Adv. 2018]**

**Sol.**(A,C)

idiot

Good