Question

Suppose 0.550 molof an ideal gas is isothermally and reversibly expanded in the four situations given below. What is the change in the entropy of the gas for each situation?

Short answer

The change in entropy of the gas for different situations of part (a), part (b), part (c), and part (d) are 6.34 J/K, 6.34 J/k, 6.34 J/k and 6.34 J/K respectively.

Step-by-step solution

The given data

The number of moles of an ideal gas, n = 0.550 mol

$$\begin{gathered} Forcasea,T=250K \\ Forcasea,V_f=0.8c{m}^3 \\ Forcasea,V_i=0.2c{m}^3 \\ Forcaseb,T=350K \\ Forcaseb,V_f=0.8c{m}^3 \\ Forcaseb,V_i=0.2c{m}^3 \\ Forcasec,T=400K \\ Forcasec,V_f=1.2c{m}^3 \\ Forcasec,V_i=0.3c{m}^3 \\ Forcased,T=450K \\ Forcased,V_f=1.2c{m}^3 \\ Forcased,V_f=0.3c{m}^3 \end{gathered}$$

Understanding the concept of entropy change

We use the equation for change in entropy related to work and temperature. The heat for an isothermal process can be expressed in terms of the final and initial volume.

Formulae:

The entropy change of gas,$∆\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{T}}$ (1)

The heat transferred by the gas,$\mathrm{Q}=\mathrm{nRTln}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right)$ (2)

Calculation of entropy change of four situations

So, the equation of change in entropy by substituting equation (ii) in equation (i) can be given as:

$$\begin{gathered} ∆\mathrm{S}=\frac{\mathrm{nRTln}\left({\displaystyle \frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}}\right)}{\mathrm{T}} \\ =\mathrm{nRTln}\left(\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\right) \end{gathered}$$…………………………………………………(3)

Now, let us consider each situation.

Entropy change for part (a) by using the given values in equation (3) as given:

$$\begin{gathered} ∆\mathrm{S}=0.550\mathrm{mol}\times 8.314\mathrm{J}/\mathrm{mol}.\mathrm{K}\times \ln \left(\frac{0.8{\mathrm{m}}^3}{0.2{\mathrm{m}}^3}\right) \\ =6.34\mathrm{J}/\mathrm{K} \end{gathered}$$

Entropy change for part (b) by using the given values in equation (3) as given:

$$\begin{gathered} ∆\mathrm{S}=0.550\mathrm{mol}\times 8.314\mathrm{J}/\mathrm{mol}.\mathrm{K}\times \ln \left(\frac{0.8{\mathrm{m}}^3}{0.2{\mathrm{m}}^3}\right) \\ =6.34\mathrm{J}/\mathrm{K} \end{gathered}$$

Entropy change for part (c) by using the given values in equation (3) as given:

$$\begin{gathered} ∆\mathrm{S}=0.550\mathrm{mol}\times 8.314\mathrm{J}/\mathrm{mol}.\mathrm{K}\times \ln \left(\frac{1.2{\mathrm{m}}^3}{0.2{\mathrm{m}}^3}\right) \\ =6.34\mathrm{J}/\mathrm{K} \end{gathered}$$

Entropy change for part d by using the given values in equation (3) as given:

$$\begin{gathered} ∆\mathrm{S}=0.550\mathrm{mol}\times 8.314\mathrm{J}/\mathrm{mol}.\mathrm{K}\times \ln \left(\frac{1.2{\mathrm{m}}^3}{0.2{\mathrm{m}}^3}\right) \\ =6.34\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the values of the entropy change for all the cases (a), (b), (c), and (d) are found to be 6.34 J/K in each case.