Question

Three moles of an ideal gas are taken around the cycle abc shown in the figure (Figure 1) . For this gas, Cp=29.1 J/(mol?K). Process ac is at constant pressure, process ba is at constant volume, and process cb is adiabatic. The temperatures of the gas in states a, c, and b are Ta=300K, Tc= 492 K, and Tb= 600 K.

Calculate the total work W for the cycle

Answer 1

The value of work done along the path \(a\) to \(c\) is, \(W_{1}=p \Delta V\)

\(=n R\left(T_{c}-T_{a}\right)\)

\(=(3)(8.314 \mathrm{~kJ} / \mathrm{mole})(492 \mathrm{k}-300 \mathrm{k})\)

\(=4789 \mathrm{~J}\)

The value of work done along the path \(c\) to \(b\) is, \(W_{2}=\frac{n R\left(T_{c}-T_{b}\right)}{(\gamma-1)}\)

\(=\frac{(3)(8.314 \mathrm{~kJ} / \mathrm{mole})(492-600)}{(1.4-1)}\)

\(=-6734 \mathrm{~J}\)

The value of work done along the path \(b\) to \(a\) is, \(W_{3}=p \Delta v\)

\(=0\)

Therefore the total value of work done is \(1945 \mathrm{~J}\).