A heat engine using a diatomic ideal gas goes through the following closed cycle:

• Isothermal compression until the volume is halved.
• Isobaric expansion until the volume is restored to its initial value.

Isochoric cooling until the pressure is restored to its initial value. What are the thermal efficiencies of ($a$) this heat engine and

($b$) a Carnot engine operating between the highest and lowest temperatures reached by this engine?

The volume of the gas is halved by compressing it isothermally. This translates to a two-fold increase in pressure. since,

$$\begin{gathered} pV=p^{\text{'}}(V/2) \\ p\text{'}=2p \end{gathered}$$

This indicates that the gas's work will be equal to the heat received by the system, and it will be provided by the system.

$Q=W=nRT\ln (0.5)$

The effort done by the gas while expanding to a first value of the volume will be,

$$\begin{gathered} W=2p\frac{V}{2} \\ =pV \end{gathered}$$

while the heat consumption can be expressed as,

$Q=n{C}_p\Delta T$

It's obvious that as the volume increased, the temperature doubled as well. This means that the temperature difference is simply $T$. As a result, the heat

$Q=n{C}_{p}T$

We now have the work done by the gas, the work done on the gas, and the two heat intakes. This means that we can discover the efficiency as quickly as possible.

$$\begin{gathered} \eta =\frac{W}{Q} \\ =\frac{nRT\ln (0.5)+pV}{n{C}_{p}T} \end{gathered}$$

Because the heat exchanged in the isotherm is negative, we don't account for it. Let us additionally deduce from the ideal gas law that

$pV=nRT$

Substitute in the equation,

$\eta =\frac{nRT(1+\ln (0.5\left)\right)}{n{C}_{p}T}$

The isobaric heat capacity is calculated as follows:

$C_p=\frac{\gamma R}{\gamma -1}$

substituting this efficiency as,

$$\begin{gathered} \eta =\frac{nRT(1+\ln (0.5\left)\right)}{nT\left(\gamma R\right)/(\gamma -1)} \\ =\frac{(\gamma -1)(1+\ln (0.5\left)\right)}{\gamma } \end{gathered}$$

Apply the numerical values, we get

$$\begin{gathered} \eta =\frac{(1.67-1)(1+\ln (0.5\left)\right)}{1.67} \\ =12.3\% \end{gathered}$$

Leonard Carnot proposed the Carnot engine as a speculative thermodynamic cycle.

It calculates the maximum efficiency that a heat engine can achieve during the conversion of heat into work and, conversely, when operating between two reservoirs.

We are attempting to determine the efficiency of a Carnot engine operating at a temperature ratio of 2:1. This Efficiency of the engine will be,

$$\begin{gathered} \eta =\frac{T_h-T_c}{T_h} \\ =\frac{2T-T}{2T} \\ =50\% \end{gathered}$$