JEE Advanced 2023 · Paper 1 · Chemistry Q11

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ChemistryInteger+4Question 45 of 51

One mole of an ideal monoatomic gas undergoes two reversible processes ( $\mathrm{A}\to\mathrm{B}$ and $\mathrm{B}\to\mathrm{C}$ ) as shown in the given figure:

$\mathrm{A}\to\mathrm{B}$ is an adiabatic process. If the total heat absorbed in the entire process ( $\mathrm{A}\to\mathrm{B}$ and $\mathrm{B}\to\mathrm{C}$ ) is $RT_2\ln 10$ , the value of $2\log V_3$ is _____.

[Use, molar heat capacity of the gas at constant pressure, $C_{\mathrm{p,m}} = \frac52 R$ ]

Temperature (K) against volume (m^3). Point A at (V1, T1) with V1 = 10 m^3 and T1 = 600 K; a curve falls from A to point B at (V2, T2) with T2 = 60 K; then a horizontal line at 60 K runs from B to point C at (V3, T2). Dashed guides mark V = 10 and T = 600 and 60.

Your answer

Answer and worked solution

Official answer: 7

Heat. A → B is adiabatic, so it absorbs no heat. B → C is isothermal at $T_2$ :
$q=RT_2\ln\frac{V_3}{V_2}=RT_2\ln10\implies V_3=10V_2.$

Adiabatic step. For a monoatomic gas $\gamma=\frac53$ , and $TV^{2/3}$ is constant:
$600\cdot10^{2/3}=60\cdot V_2^{2/3}\implies V_2^{2/3}=10^{5/3}\implies V_2=10^{5/2}\ \text{m}^3.$

Result. $V_3=10^{7/2}$ , so $2\log V_3=7$ .

Topics:First law and enthalpy Entropy

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JEE Advanced 2023 · Paper 1, Chemistry question 11