Question

0.05 m3 of a gas at 6.9 bar expands reversibly in a cylinder behind a piston according to the law pv1.2= Constant, until the volume is 0.08 m3. Calculate the work done by the gas and sketch the process on a p-v diagram.

Answer 1

For PV$^{\gamma}$ = constant

work done by the gas = $\frac{(P_2V_2-P_1V_1)}{(1-\gamma)}$ where $\gamma = 1.2$

for calculation of $P_2 = P_1(\frac{V_1}{V_2})^{1.2}$

$P_2= 6.9(\frac{0.05}{0.08})^{1.2}$

P₂= 3.92bar

hence work done=$\frac{3.92\times0.08-6.9\times0.05}{1-1.2}$

work done by gas = 0.0157 bar m³

and the PV is given below: