Question

A liter of air, initially at room temperature and atmospheric pressure, is heated at constant pressure until it doubles in volume. Calculate the increase in its entropy during this process.

Short answer

The increase in entropy is$816.51J{K}^{-1}$.

Step-by-step solution

Given Information

Pressure $=P=101325Pa$

Temperature $=T=300K$

Volume $=V=1L$

Heat capacity at constant pressure$=C_p=29J{K}^{-1}$

Calculation

For an ideal gas, as the volume doubles, the temperature must also double.

The change in entropy is given as:

$$\begin{gathered} \Delta S={\int }_{T_i}^{T_f}\frac{Q}{T} \\ \Delta S={\int }_{T_i}^{T_f}\frac{C_{p}dT}{T} \\ \Delta S=C_p\ln [T{]}_{T_i}^{T_f} \\ \Delta S=C_p\ln \left[\frac{T_f}{T_i}\right] \\ \Delta S=C_p\ln \left[\frac{2T_i}{T_i}\right] \\ \Delta S=C_p\ln 2 \end{gathered}$$

For the number of molecules in one liter of air,

$$\begin{gathered} n=\frac{PV}{RT} \\ n=\frac{101325\times 1}{8.314\times 300} \\ n=40.62\mathrm{mol} \end{gathered}$$

Hence, change in entropy for one liter of air can be calculated as:

$$\begin{gathered} \Delta S=n{C}_p\ln 2 \\ \Delta S=40.62\times 29\times \ln 2 \\ \Delta S=816.51J{K}^{-1} \end{gathered}$$

Final answer

Hence, the entropy of air will be $816.51J{K}^{-1}$when air is heated at constant pressure and its volume doubles.