Question

The CORRECT expression(s) for isothermal expansion of 1 mol of an ideal gas is(are)

• A. \( \Delta A = RT \ln \frac{V_{initial}}{V_{final}} \)
• B. \( \Delta G = RT \ln \frac{V_{initial}}{V_{final}} \)
• C. \( \Delta H = RT \ln \frac{V_{final}}{V_{initial}} \)
• D. \( \Delta S = R \ln \frac{V_{final}}{V_{initial}} \)

Hint:

In an isothermal process, the change in entropy for an ideal gas is given by \( \Delta S = R \ln \frac{V_{final}}{V_{initial}} \), where \( R \) is the gas constant.

Answer 1

The Correct Option is A, B, D

1. Change in Internal Energy ($\Delta U$) and Enthalpy ($\Delta H$)

The internal energy ($\Delta U$) and enthalpy ($\Delta H$) of an ideal gas depend only on temperature. Since the process is isothermal ($\Delta T = 0$):

$$\Delta U = 0$$

$$\Delta H = \Delta U + \Delta(PV) = 0 + \Delta(nRT) = nR\Delta T = 0$$

Thus, the expression for $\Delta H$ in option (C) must be incorrect:

(C) $\Delta H = RT \ln \frac{V_{final}}{V_{initial}}$ [INCORRECT]

2. Change in Entropy ($\Delta S$)

For any reversible process (and $\Delta S$ is a state function, so $\Delta S$ is the same as for a reversible path between the same states), the change in entropy for 1 mole of an ideal gas is given by:

$$\Delta S = n C_V \ln \frac{T_{final}}{T_{initial}} + n R \ln \frac{V_{final}}{V_{initial}}$$

Since the process is isothermal ($T_{final} = T_{initial}$ and $n=1$):

$$\Delta S = (1) C_V \ln(1) + (1) R \ln \frac{V_{final}}{V_{initial}}$$

$$\Delta S = R \ln \frac{V_{final}}{V_{initial}}$$

Comparing this with option (D):

(D) $\Delta S = R \ln \frac{V_{final}}{V_{initial}}$ [CORRECT]

3. Change in Helmholtz Free Energy ($\Delta A$) and Gibbs Free Energy ($\Delta G$)

The changes in Helmholtz free energy ($\Delta A$) and Gibbs free energy ($\Delta G$) are defined as:

$$\Delta A = \Delta U - T \Delta S$$

$$\Delta G = \Delta H - T \Delta S$$

Since $\Delta U = 0$ and $\Delta H = 0$ for an isothermal process of an ideal gas:

$$\Delta A = 0 - T \Delta S = -T \Delta S$$

$$\Delta G = 0 - T \Delta S = -T \Delta S$$

Therefore, $\Delta A = \Delta G = -T \Delta S$.

Substituting the expression for $\Delta S$:

$$\Delta A = \Delta G = -T \left( R \ln \frac{V_{final}}{V_{initial}} \right)$$

$$\Delta A = \Delta G = -RT \ln \frac{V_{final}}{V_{initial}}$$

Using the property of logarithms that $-\ln(x) = \ln(1/x)$:

$$\Delta A = \Delta G = RT \ln \left( \frac{V_{final}}{V_{initial}} \right)^{-1}$$

$$\Delta A = \Delta G = RT \ln \frac{V_{initial}}{V_{final}}$$

Comparing this with options (A) and (B):

(A) $\Delta A = RT \ln \frac{V_{initial}}{V_{final}}$ [CORRECT]

(B) $\Delta G = RT \ln \frac{V_{initial}}{V_{final}}$ [CORRECT]

Conclusion

The correct expressions are (A), (B), and (D)