Name: Macroscopic and Microscopic Variables
Uploaded: 2019-07-19
Duration: 5 min 3 s
Description: This video enables students to understand the definition and examples of the macroscopic and microscopic variables of the thermodynamic system.

Question

Derive the expression for Carnot engine efficiency.

Accepted Answer

The Carnot efficiency is the theoretical maximum efficiency when the heat engine is operating between two temperatures: The temperature $T_{H}$ at which the high temperature reservoir operates and the temperature $T_{L}$ at which the low temperature reservoir operates.

The efficiency of carnot engine, $η$ is given by

$η = \frac{Work done}{Heat extracted} = \frac{W}{Q_{H}} = \frac{Q_{H} - Q_{L}}{Q_{H}} = 1 - \frac{Q_{L}}{Q_{H}}$

Applying isothermal conditions, we get

$Q_{H} = μ R T_{H} \ln (\frac{V_{2}}{V_{1}}), Q_{L =} μ R T_{L} \ln (\frac{V_{3}}{V_{4}})$

$η = 1 - \frac{μ R T_{L} \ln (\frac{V_{2}}{V_{1}})}{μ R T_{H} \ln (\frac{V_{3}}{V_{4}})} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot 1$

Applying adiabatic condition, we get

$T_{H} {V_{2}}^{γ - 1} = T_{L} {V_{3}}^{γ - 1}, T_{H} {V_{1}}^{γ - 1} = T_{L} {V_{4}}^{γ - 1} \Rightarrow {(\frac{V_{2}}{V_{1}})}^{γ - 1} = {(\frac{V_{3}}{V_{4}})}^{γ - 1} \Rightarrow \frac{V_{2}}{V_{1}} = \frac{V_{3}}{V_{4}}$

Substituting in equation $1$

$η = 1 - \frac{T_{L}}{T_{H}}$

The efficiency of Carnot engine is independent of working substance. It depends only on temperatures of source and sink.

Derive the expression for Carnot engine efficiency.

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