Question

The pressure and volume of an ideal gas are related as $PV^{3/2} = K$ (constant). The work done when the gas is taken from state A ($P_1, V_1, T_1$) to state B ($P_2, V_2, T_2$) is:

• A. $2(P_1V_1 - P_2V_2)$
• B. $2(P_2V_2 - P_1V_1)$
• C. $2(\sqrt{P_1V_1} - \sqrt{P_2V_2})$
• D. $2(P_2\sqrt{V_2} - P_1\sqrt{V_1})$

Answer 1

Answer: (A), (B)

$2(P_1V_1 - P_2V_2)$

Answer 2

For $PV^{3/2} = \text{constant}$, we know that:

$W = \int P \, dV$

Since $P = \frac{K}{V^{3/2}}$:

$W = \int_{V_1}^{V_2} \frac{K}{V^{3/2}} \, dV$

Integrating, we get:

$W = \left[ -\frac{2K}{V^{1/2}} \right]_{V_1}^{V_2} = 2(P_1V_1 - P_2V_2)$

- If the work done by the gas is asked:

$W = 2(P_1V_1 - P_2V_2) \quad \text{(Option 1)}$

- If the work done on the gas (by external) is asked:

$W = 2(P_2V_2 - P_1V_1) \quad \text{(Option 2)}$