Question

Repeat Problem 57, with the pressure now kept constant.

Short answer

The change in the entropy of the gas at constant pressure is 5.98 J/K

Step-by-step solution

The given data

The number of moles is n = 1.00 mol .

The gas ismonatomic.

The initial temperature is${\mathrm{T}}_{\mathrm{i}}=300\mathrm{K}$.

The final temperature is${\mathrm{T}}_f=400\mathrm{K}$.

The pressure is constant.

Understanding the concept of entropy change

By finding the equation for heat absorbed at constant pressure and using Equation 20-1, we can find the change in the entropy of the gas at constant pressure.

Formula:

The energy absorbed as heat by the gas at constant pressure, $\mathrm{dQ}=\frac{5}{2}\mathrm{nRdT}$ (1)

The change in entropy of the gas, $∆\mathrm{S}=\int \frac{\mathrm{dQ}}{\mathrm{T}}$ (2)

Calculation of the entropy change

Since the pressureis kept constant, substituting the value of equation (1) in equation (2), we can get the change in entropy as given:

$$\begin{gathered} ∆\mathrm{S}={\int }_{{\mathrm{T}}_{\mathrm{i}}}^{{\mathrm{T}}_{\mathrm{f}}}\frac{\left({\displaystyle \frac{5\mathrm{nR}}{2}}\right)\mathrm{dT}}{\mathrm{T}} \\ =\frac{5}{2}\mathrm{nRln}\left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}\right) \\ =\frac{5}{2}\times 1.00\mathrm{mol}\times 8.31\frac{\mathrm{J}}{\mathrm{mol}·\mathrm{K}}\times \left(\frac{400\mathrm{K}}{300\mathrm{K}}\right) \\ =5.98\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change is 5.98 J/K