Q.4.5

Question

Prove directly (by calculating the heat taken in and the heat expelled) that a Carnot engine using an ideal gas as the working substance has an efficiency of $\frac{1 - t_{c}}{t_{h}}$

Step-by-Step Solution

Verified

Answer

Hence we proved that

Carnot engine using an ideal gas as the working substance has the efficiency of $1 - \frac{T_{c}}{T_{h}}$ .

1Step 1: To Prove

Carnot engine using an ideal gas as the working substance has the efficiency of $1 - \frac{T_{c}}{T_{h}}$

2Step 2: Explanation

The Carnot cycle begins with the isothermal expansion of 1 mol of gas, which changes its state from $(P_{1}, V_{1}, T_{h}) t o (P_{2}, V_{2}, T_{c})$ .The heat absorbed $(Q_{H})$ by the gas from the source at constant temperature

$T_{h}$ is given by:

$Q_{h} = W_{1} = R T_{h} \log_{e} (\frac{V_{2}}{V_{1}}) - - - (1)$

The Carnot cycle's second stage is the adiabatic expansion of 1 mol of gas taking its state from $(P_{2}, V_{2}, T_{h})$ to $(P_{3}, V_{3}, T_{c})$ The work done $(W_{2})$ by the gas is given by:

$W_{2} = \frac{R (T_{h} - T_{c})}{γ - 1} - - - (2)$

The Carnot cycle's third stage involves isothermal compression of 1 mol of gas taking its state from $(P_{3}, V_{3}, T_{c}) t o (P_{4}, V_{4}, T_{h})$ by the gas to the sink at constant temperature $T_{c}$ is given by

$Q_{c} = W_{3} = R T_{c} \log_{e} (\frac{V_{3}}{V_{4}}) - - - (3)$

The fourth stage of the Carnot cycle is adiabatic compression, which involves compressing 1 mol of gas to its original condition.

$(P_{4}, V_{4}, T_{c})$ to $(P_{1}, V_{1}, T_{h})$ The work done $(W_{4})$ on the gas is given by:

$W_{4} = \frac{R [T_{h} - T_{c}]}{γ - 1} - - - (4)$

The efficiency of Carnot engine is given by

$\begin{aligned} η = \frac{output work}{heat supplied} \\ η = \frac{Q_{h} - Q_{c}}{Q_{h}} = 1 - \frac{Q_{c}}{Q_{h}} - - - (5) \end{aligned}$

Substitute (1) and (3) in (5)

$η = 1 - \frac{R T_{c} \log_{e} (\frac{V_{3}}{V_{4}})}{R T_{h} \log_{e} (\frac{V_{2}}{V_{1}})} - - - (6)$

3Step 3: Further Continuation to the proof

For an adiabatic expansion:

$T V^{γ - 1} = Constant$

In the second stage, for an adiabatic expansion:

$\begin{aligned} T_{h} V_{2}^{γ - 1} = T_{c} V_{3}^{γ - 1} \\ \frac{T_{c}}{T_{h}} = \frac{V_{2}^{r - 1}}{{V_{3}}^{r - 1}} = {(\frac{V_{2}}{V_{3}})}^{r - 1} \\ (\frac{V_{2}}{V_{3}}) = {(\frac{T_{c}}{T_{h}})}^{\frac{1}{r - 1}} - - - - - - (7) \end{aligned}$

In the fourth stage, for an adiabatic compression:

$\begin{aligned} T_{c} V_{4}^{γ - 1} = T_{h} V_{1}^{γ - 1} \\ \frac{T_{2}}{T_{1}} = \frac{V_{1}^{γ - 1}}{V_{4}^{γ - 1}} = {(\frac{V_{1}}{V_{4}})}^{γ - 1} \\ (\frac{V_{1}}{V_{4}}) = {(\frac{T_{c}}{T_{h}})}^{\frac{1}{γ - 1}} - - - (8) \end{aligned}$