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:
\[
W = 2(P_1V_1 - P_2V_2) \quad \text{(Option 1)}
\]
- If the work done on the gas (by external) is asked:
\[
W = 2(P_2V_2 - P_1V_1) \quad \text{(Option 2)}
\]
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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.
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:
