Question

You build a heat engine that takes 1.00 mol of an ideal diatomic gas through the cycle shown in Fig. P20.39. (a) Show that process ab is an isothermal compression. (b) During which process(es) of the cycle is heat absorbed by the gas? During which process(es) is heat rejected? How do you know? Calculate (c) the temperature at points a, b, and c; (d) the net heat exchanged with the surroundings and net work done by the engine in one cycle; (e) the thermal efficiency of the engine.

Short answer

The process ab is an isothermal compression because the pressure is rising at same temperature.

Step-by-step solution

Identification of given data

The number of moles of diatomic gas is\(n = 1\;{\rm{mol}}\)

The pressure of diatomic gas at point a is\({P_a} = 2 \times {10^5}\;{\rm{Pa}}\)

The volume of diatomic gas at point a is\({V_a} = 0.010\;{{\rm{m}}^3}\)

The pressure of diatomic gas at point a is\({P_b} = 4 \times {10^5}\;{\rm{Pa}}\)

The volume of diatomic gas at point a is \({V_b} = 0.005\;{{\rm{m}}^3}\)

Conceptual Explanation

Whether the process is isothermal compression or not is decided by checking pressure and temperature between two positions. The temperature at point a and b is calculated to check isothermal process.

Whether the process ab is isothermal compression or not

The temperature at position a is given as:

\({P_a}{V_a} = nR{T_a}\)

Here,\(R\)is the universal gas constant and its value is\(8.31\;{\rm{J}}/{\rm{mol}} \cdot {\rm{K}}\).

Substitute all the values in the above equation.

\(\begin{aligned}\left( {2 \times {{10}^5}\;{\rm{Pa}}} \right)\left( {0.010\;{{\rm{m}}^3}} \right) = \left( {1\;{\rm{mol}}} \right)\left( {8.31\;{\rm{J}}/{\rm{mol}} \cdot {\rm{K}}} \right){T_a}\\{T_a} = 240.67\;{\rm{K}}\end{aligned}\)

The temperature at position b is given as:

\({P_b}{V_b} = nR{T_b}\)

Substitute all the values in the above equation.

\(\begin{aligned}\left( {4 \times {{10}^5}\;{\rm{Pa}}} \right)\left( {0.005\;{{\rm{m}}^3}} \right) = \left( {1\;{\rm{mol}}} \right)\left( {8.31\;{\rm{J}}/{\rm{mol}} \cdot {\rm{K}}} \right){T_b}\\{T_b} = 240.67\;{\rm{K}}\end{aligned}\)

The temperature at positions a and b of process ab are same and pressure for this process is rising from position a to position b so the process ab is an isothermal compression.

Therefore, the process ab is an isothermal compression because the pressure is rising at same temperature.