An engine operates by taking $n$ moles of an ideal gas through the cycle $ABCDA$ shown in figure. The thermal efficiency of the engine is : (Take $C_v = 1.5 \, R,$ where $R$ is gas constant)

Question

An insulated cylinder of volume $60\,\text{cm}^3$ is filled with a gas at $27^\circ\text{C}$ and $2$ atmospheric pressure. The gas is then compressed making the final volume $20\,\text{cm}^3$ while allowing the temperature to rise to $77^\circ\text{C}$. The final pressure is ____________ atmospheric pressure.

Answer 1

To determine the thermal efficiency of the engine operating on a cycle with $n$ moles of an ideal gas, we first need to understand that the thermal efficiency for a cyclic process is given by the formula:

\[\eta = 1 - \frac{Q_C}{Q_H}\]

where $Q_C$ is the heat rejected and $Q_H$ is the heat absorbed.

Given isothermal and adiabatic processes within the cycle, let's find expressions for $Q_H$ and $Q_C$ using applicable thermodynamic processes:

For an isothermal process, the heat exchanged is given by: \[Q = nRT \ln\left(\frac{V_f}{V_i}\right)\] where V_f and V_i are final and initial volumes, respectively.
For an adiabatic process, the heat exchange is zero, as it does not involve heat transfer.

Assume from segment AB there is isothermal expansion and CD isothermal compression, while BC and DA are adiabatic.

Given $C_v = 1.5 R$, we can use the equations for work done in an isothermal process:

$Q_H = nRT_B \ln\left(\frac{V_B}{V_A}\right)$ for AB.
$Q_C = nRT_D \ln\left(\frac{V_D}{V_C}\right)$ for CD.

Since in the cycle V_B = V_C and V_A = V_D, the expressions simplify:

$Q_H = nRT_B \ln\left(\frac{V_C}{V_A}\right)$
$Q_C = nRT_D \ln\left(\frac{V_A}{V_C}\right)$

The efficiency is:

\[\eta = 1 - \frac{T_D}{T_B}\]

If T_D/T_B = 0.85, plug this value into the formula:

\[\eta = 1 - 0.85 = 0.15\]

Thus, the thermal efficiency of this engine is 0.15.

Answer 2