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:
- \( 2(P_1V_1 - P_2V_2) \)
- \( 2(P_2V_2 - P_1V_1) \)
- \( 2(\sqrt{P_1V_1} - \sqrt{P_2V_2}) \)
- \( 2(P_2\sqrt{V_2} - P_1\sqrt{V_1}) \)
The Correct Option is A, B
Approach Solution - 1
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:
\[ W = 2(P_2V_2 - P_1V_1) \quad \text{(Option 2)} \]
Approach Solution -2
The pressure and volume of an ideal gas are related as \( PV^{3/2} = K \) (constant). We need to find the work done when the gas is taken from state A (\( P_1, V_1, T_1 \)) to state B (\( P_2, V_2, T_2 \)).
Concept Used:
For a thermodynamic process, the work done by the gas is given by:
\[ W = \int_{V_1}^{V_2} P \, dV \]
Given the process follows \( PV^{3/2} = K \), we can substitute \( P = \frac{K}{V^{3/2}} \) into the integral and evaluate it.
Step-by-Step Solution:
Step 1: Write the work done integral using the given polytropic relation \( PV^{3/2} = K \).
\[ W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{K}{V^{3/2}} \, dV \]
Step 2: Factor out the constant \( K \) and integrate.
\[ W = K \int_{V_1}^{V_2} V^{-3/2} \, dV \] \[ W = K \left[ \frac{V^{-3/2 + 1}}{-3/2 + 1} \right]_{V_1}^{V_2} = K \left[ \frac{V^{-1/2}}{-1/2} \right]_{V_1}^{V_2} \] \[ W = -2K \left[ V^{-1/2} \right]_{V_1}^{V_2} = -2K \left( \frac{1}{\sqrt{V_2}} - \frac{1}{\sqrt{V_1}} \right) \] \[ W = 2K \left( \frac{1}{\sqrt{V_1}} - \frac{1}{\sqrt{V_2}} \right) \]
Step 3: Express the constant \( K \) in terms of initial state parameters \( P_1 \) and \( V_1 \). Since \( K = P_1 V_1^{3/2} \), substitute this into the expression.
\[ W = 2 (P_1 V_1^{3/2}) \left( \frac{1}{\sqrt{V_1}} - \frac{1}{\sqrt{V_2}} \right) \] \[ W = 2 P_1 V_1^{3/2} \left( V_1^{-1/2} - V_2^{-1/2} \right) \] \[ W = 2 P_1 V_1 \left( 1 - \frac{\sqrt{V_1}}{\sqrt{V_2}} \right) \]
Step 4: Also express \( K \) in terms of final state parameters \( P_2 \) and \( V_2 \) to find a symmetric form. Since \( K = P_1 V_1^{3/2} = P_2 V_2^{3/2} \), we can write the work done as:
\[ W = 2K \left( \frac{1}{\sqrt{V_1}} - \frac{1}{\sqrt{V_2}} \right) = 2 \left( \frac{K}{\sqrt{V_1}} - \frac{K}{\sqrt{V_2}} \right) \] \[ W = 2 \left( \frac{P_1 V_1^{3/2}}{\sqrt{V_1}} - \frac{P_2 V_2^{3/2}}{\sqrt{V_2}} \right) \] \[ W = 2 \left( P_1 V_1 - P_2 V_2 \right) \]
Thus, the work done by the gas is \( 2(P_1V_1 - P_2V_2) \).
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