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The change in entropy when the pressure ...

The change in entropy when the pressure of perfect gas is changed isothermally from `P_1` to `P_2` is

A

`/_\S=nRln(P_1+P_2)`

B

`/_\S=nRln(P_2//P_1)`

C

`/_\S=nRln(P_1//P_2)`

D

`/_\S=nRln((P_1+P_2)/(P_2))`

Text Solution

AI Generated Solution

The correct Answer is:
To find the change in entropy (ΔS) when the pressure of a perfect gas is changed isothermally from \( P_1 \) to \( P_2 \), we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Conditions**: The process is isothermal, meaning the temperature (T) remains constant. For a perfect gas, we can use the ideal gas law, which states that \( PV = nRT \). 2. **Formula for Change in Entropy**: The change in entropy for a reversible process can be expressed as: \[ \Delta S = \frac{dQ}{T} \] where \( dQ \) is the heat transfer. 3. **Relate Heat Transfer to Work Done**: For an ideal gas undergoing an isothermal process, the heat transfer \( dQ \) can be related to the work done \( dW \): \[ dQ = dW \] In terms of pressure and volume, this becomes: \[ dW = -P dV \] Therefore, we can write: \[ dQ = P dV \] 4. **Substituting Ideal Gas Law**: Using the ideal gas law \( P = \frac{nRT}{V} \), we substitute for \( P \): \[ dQ = \frac{nRT}{V} dV \] 5. **Integrate to Find Change in Entropy**: Now we can substitute \( dQ \) into the entropy formula: \[ \Delta S = \int_{V_1}^{V_2} \frac{dQ}{T} = \int_{V_1}^{V_2} \frac{nRT}{V} \frac{dV}{T} \] Since T is constant, we can take it out of the integral: \[ \Delta S = \frac{nR}{T} \int_{V_1}^{V_2} \frac{dV}{V} \] The integral \( \int \frac{dV}{V} \) evaluates to \( \ln \frac{V_2}{V_1} \): \[ \Delta S = \frac{nR}{T} \ln \frac{V_2}{V_1} \] 6. **Relate Volume to Pressure**: From the ideal gas law, we know that \( P_1 V_1 = P_2 V_2 \). Rearranging gives: \[ \frac{V_2}{V_1} = \frac{P_1}{P_2} \] Substituting this back into the entropy change equation: \[ \Delta S = \frac{nR}{T} \ln \left(\frac{P_1}{P_2}\right) \] 7. **Final Result**: Thus, the change in entropy when the pressure of a perfect gas is changed isothermally from \( P_1 \) to \( P_2 \) is: \[ \Delta S = nR \ln \left(\frac{P_1}{P_2}\right) \]
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