Answer to Question #95480 in Electrical Engineering for Anthony

Question

0.05m^3 of a gas at 6.9 bar expand reversibly in a cylinder behind a piston according to the law of pv^1.2=constant until the volume is 0.08m^3. Calculate the workdone by the gas and sketch the process on p-v diagram.

Answer 1

1. Calculate the pressure when the gas expanded to 0.08 cubic meters:

$p_aV_a^{1.2}=p_bV_b^{1.2},\\ \space\\ p_b=p_a\Bigg(\frac{V_a}{V_b}\Bigg)^{1.2}=3.926\text{ bar, or }3.926\cdot10^5\text{ Pa}.$

2. The work in the process is

$W=\int_a^b pdV=\int_a^b \frac{\text{const}}{V^{1.2}}dV=-\int_a^b\frac{\text{const}}{0.2V^{0.2}}=\\ \space\\ =-\frac{\text{const}}{0.2V_b^{0.2}}+\frac{\text{const}}{0.2V_a^{0.2}}=\\ \space\\ =-\frac{p_bV_b^{1.2}}{0.2V_b^{0.2}}+\frac{p_aV_a^{1.2}}{0.2V_a^{0.2}}=\frac{p_aV_a-p_bV_b}{0.2}=15476\text{ J}.$

3. The process on p-V diagram looks like