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III. Long Answer Questions · Q22

Q.Derive the expression for Carnot engine efficiency.

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Step 1. For the two isothermal legs of the Carnot cycle, QH=μRTHln⁡(V2/V1)Q_H=\mu RT_H\ln(V_2/V_1) (A→B) and QL=μRTLln⁡(V3/V4)Q_L=\mu RT_L\ln(V_3/V_4) (C→D).

Step 2. For the two adiabatic legs, applying TVγ−1=constantTV^{\gamma-1}=\text{constant}: THV2γ−1=TLV3γ−1T_HV_2^{\gamma-1}=T_LV_3^{\gamma-1} (B→C) and THV1γ−1=TLV4γ−1T_HV_1^{\gamma-1}=T_LV_4^{\gamma-1} (D→A).

Step 3. Dividing these two adiabatic relations gives (V2V1)γ−1=(V3V4)γ−1\left(\dfrac{V_2}{V_1}\right)^{\gamma-1}=\left(\dfrac{V_3}{V_4}\right)^{\gamma-1}, so V2V1=V3V4\dfrac{V_2}{V_1}=\dfrac{V_3}{V_4}. …

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