JEE Advanced 2021 Paper 2 · Q17 · First Law & Internal Energy
One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown on the U/R (K) vs S (J K$^{-1}$ mol$^{-1}$) plot. Process I is the vertical drop from (p$_1$, V$_1$) at U/R = 2250 to (p$_2$, V$_2$) at U/R = 450 (entropy unchanged). Process II is the horizontal arrow from (p$_2$, V$_2$) to (p$_3$, V$_3$) at constant U/R = 450 (internal energy unchanged). If the work done by the gas in the two processes are same, the value of $\ln \frac{V_3}{V_2}$ is _____.
(U: internal energy, S: entropy, p: pressure, V: volume, R: gas constant.)
(Given: molar heat capacity at constant volume, $C_{V,m}$ of the gas is $\frac{5}{2} R$.)
Reveal answer + step-by-step solution
Correct answer:10
Solution
Process I: $\Delta S = 0$ $\Rightarrow$ reversible adiabatic. Hence $W_I = -\Delta U_I$. $\Delta U_I = n C_{V,m}(T_2 - T_1) = U_2 - U_1 = R(450) - R(2250) = -1800 R$ J. Thus $W_I = +1800 R$ (work done BY the gas, positive).
Find $T_2$ at state 2: $U_2 = n C_{V,m} T_2$ $\Rightarrow$ $450 R = 1 \cdot (\tfrac{5}{2} R) T_2$ $\Rightarrow$ $T_2 = 180$ K.
Process II: $\Delta U = 0$ $\Rightarrow$ isothermal at $T_2 = 180$ K (for an ideal gas). $W_{II} = n R T \ln(V_3/V_2) = (1)(R)(180)\ln(V_3/V_2)$.
Equal work: $1800 R = 180 R \ln(V_3/V_2)$ $\Rightarrow$ $\ln(V_3/V_2) = 1800/180 = 10$.
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