GATE 2021 CH – Question 33

Thermodynamics · Laws of thermodynamics, open and closed systems, entropy and chemical potential · 1 mark · Numerical answer

The molar heat capacity at constant pressure $C_p$ (in $\mathrm{J\,mol^{-1}\,K^{-1}}$) for n-pentane as a function of temperature (T in K) is given by $C_p/R=2.46+45.4\times10^{-3}T-14.1\times10^{-6}T^2$. Take $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$. At 1000 K, the rate of change of molar entropy of n-pentane with respect to temperature at constant pressure is _____ $\mathrm{J\,mol^{-1}\,K^{-2}}$ (round off to 2 decimal places).

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Correct answer: 0.27 to 0.29

Explanation

At constant pressure, the thermodynamic identity is $dS=C_p\,dT/T$, so $(\partial S/\partial T)_P=C_p/T$. This asks for the entropy derivative, **not** the derivative of $C_p$.

At 1000 K, $C_p/R=2.46+45.4-14.1=33.76$, giving $C_p=33.76(8.314)=280.68064\ \mathrm{J\,mol^{-1}K^{-1}}$.

Divide by 1000 K: $(\partial S/\partial T)_P=0.28068064\ \mathrm{J\,mol^{-1}K^{-2}}$. **Answer: 0.28.**