IIT JEE Advanced 2008 Paper 2

(A)→(Q) When the value is opened, it is a case of free expansion under a diabetic condition: ΔA = 0, ΔW = 0, ΔU = 0 where T = constant. (B)→(P), (R) We have P = $\frac{k}{V^2}$ However, PV = nRT ⇒ $\frac{k}{V}$ = nRT If V → 2V, then T → $\frac{T}{2}$. Therefore. ΔV = $n\left(\frac{3}{2R}\right)\Delta T$ = $n\left(\frac{3}{2}R\right)\left(-\frac{k}{nRV^2}\right)\Delta V$ ΔU = $-\frac{3k}{2V}$ Now, ΔW = $\int\limits_{V}^{2V}P\,dV$ = $k\int\limits_{V}^{2V}\frac{dV}{V^2}$ = $-k\left(\frac{1}{2V}-\frac{1}{V}\right)$ = $\frac{k}{2V}$ Therefore, ΔQ = ΔU + ΔW = $\frac{3k}{2V}+\frac{k}{2V}$ = $\left(-\frac{k}{V}\right)$ < 0 (C)→ (P), (S) P = $\frac{k}{V^{4/3}}$ However, PV = nRT Therefore, $\frac{k}{V^{1/3}}$ = nRT If V → 2V, then T = $\left[\frac{T}{(2)^{1/3}}\right]$ < T. ΔV = $n\left(\frac{2}{3}R\right)T\left(\frac{1}{2^{1/3}}-1\right)$ = $\frac{3k}{2V^{1/3}}\left(\frac{1}{2^{1/3}}-1\right)$ ΔW = $\int\limits_{V}^{2V}$ pdV = $k\int\limits_{V}^{2V}\frac{dV}{V^{4/3}}$ = $-3k\int\limits_{V}^{2V}V^{-1/3}$ = - 3k $\left(\frac{1}{(2V)^{1/3}}-\frac{1}{V^{1/3}}\right)$ Now, ΔQ = ΔU + ΔW = $\frac{9k}{2V^{1/3}}\left(1-\frac{1}{2^{1/3}}\right)$ > 0 (D)→(Q), (S) We have PV = nRT That is, P′(2V ) = nRT ′ Therefore, $\frac{T'}{T}\left(\frac{2P'}{P}\right)$ > 1 as P′ > P and T ′ > T Now, ΔW = (Area under PV graph) > 0 ΔU = $n\left(\frac{3}{2}R\right)\Delta T$ = $\frac{3}{2}nR\left(T'-T\right)$ > 0 ΔQ = ΔU + ΔW > 0