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JEE Advanced 2024 Paper 2 · Chemistry · Chemical Thermodynamics · medium

JEE Advanced 2024 Paper 2 Chemistry: Paragraph 1

A sample of 1.001.00 mol of a monatomic ideal gas (Cv=32RC_v=\tfrac32R, R=8.314 J K−1 mol−1R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}) is at 300 K300\ \mathrm{K} and 6.006.00 bar. It is allowed to expand adiabatically against a constant external pressure of 1.001.00 bar until the gas pressure equals 1.001.00 bar (the process is irreversible). Take ln⁡2=0.6931\ln2=0.6931, ln⁡3=1.0986\ln3=1.0986.

Question 1

Numerical answer

The final temperature of the gas (in K) is
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Answer: 200

For an adiabatic process q=0q=0, so ΔU=w=−Pext(V2−V1)\Delta U=w=-P_{ext}(V_2-V_1). Also ΔU=nCv(T2−T1)\Delta U=nC_v(T_2-T_1). Final volume V2=RT2/P2V_2=RT_2/P_2 and initial volume V1=RT1/P1V_1=RT_1/P_1, with Pext=P2=1P_{ext}=P_2=1 bar. Then −Pext(V2−V1)=−R(T2−T1P2P1)=−R(T2−50)-P_{ext}(V_2-V_1)=-R\left(T_2-T_1\dfrac{P_2}{P_1}\right)=-R\left(T_2-50\right). Equate: 1.5R(T2−300)=−R(T2−50)1.5R(T_2-300)=-R(T_2-50), so 2.5T2=450+50=5002.5T_2=450+50=500, giving T2=200 KT_2=200\ \mathrm{K}. (Checking: it is lower than 300 K, as expected for an expansion that does work at the expense of internal energy.)

Question 2

Numerical answer

The entropy change of the gas (in J K−1\mathrm{J\,K^{-1}}) for this process, correct to two decimal places, is
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Answer: 6.47

Entropy is a state function, so evaluate ΔS\Delta S along a reversible path between the same end states (300 K,6 bar)→(200 K,1 bar)(300\ \mathrm{K},6\ \mathrm{bar})\to(200\ \mathrm{K},1\ \mathrm{bar}): ΔS=nCpln⁡T2T1−nRln⁡P2P1\Delta S=nC_p\ln\dfrac{T_2}{T_1}-nR\ln\dfrac{P_2}{P_1} with Cp=52RC_p=\tfrac52R. Then ΔS=2.5Rln⁡23+Rln⁡6=R[2.5(ln⁡2−ln⁡3)+(ln⁡2+ln⁡3)]=R[3.5ln⁡2−1.5ln⁡3]\Delta S=2.5R\ln\tfrac23+R\ln6=R\left[2.5(\ln2-\ln3)+(\ln2+\ln3)\right]=R\left[3.5\ln2-1.5\ln3\right]. Numerically 3.5(0.6931)−1.5(1.0986)=2.4259−1.6479=0.77803.5(0.6931)-1.5(1.0986)=2.4259-1.6479=0.7780, so ΔS=8.314×0.7780≈6.47 J K−1\Delta S=8.314\times0.7780\approx6.47\ \mathrm{J\,K^{-1}}. It is positive, as it must be for an irreversible adiabatic process (no heat flows to the surroundings).
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