JEE Advanced 2021 · Paper 2 · Chemistry Q17

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ChemistryInteger+4Question 36 of 57

One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are same, the value of $\ln\dfrac{V_3}{V_2}$ 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 $\frac52R$ )

U/R (in K, vertical axis) against S (J K^-1 mol^-1, horizontal axis). Process I is a vertical drop at constant S from (p1, V1) at U/R = 2250 to (p2, V2) at U/R = 450. Process II is a horizontal line at constant U/R = 450 (isothermal) from (p2, V2) to (p3, V3) at larger S.

Your answer

Answer and worked solution

Official answer: 10

Read the diagram. $U=C_VT=\frac52RT$ , so $\frac UR=2250$ K means $T_1=900$ K, and $\frac UR=450$ K means $T_2=180$ K.

Process I is at constant $S$ : a reversible adiabatic step.
Process II is at constant $U$ : an isothermal step at 180 K.

Work in each process.

I: $W=-\Delta U=R(2250-450)=1800R$ .
II: $W=RT_2\ln\dfrac{V_3}{V_2}=180R\ln\dfrac{V_3}{V_2}$ .

Equal work. $\ln\dfrac{V_3}{V_2}=\dfrac{1800}{180}=10$ .

Topics:First law and enthalpy Entropy

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