Question

An ideal gas is in equilibrium at initial state with temperature T=137oC, pressure P = 0.75Pa and volume V = 0.75 m3. If there is a change in state in which the gas undergoes an isothermal process to a final state of equilibrium during which the volume is doubled. Calculate the temperature and pressure of the gas at this final state.

Answer 1

In as isothermal process, \(T_{before}=T_{after}=137^{o}C\).
More, in an isothermal process, \(P_1V_1=P_2V_2\). Therefore, \(P_2=\frac{P_1V_1}{V_2}\)
Since the volume at the final state is doubled, \(\frac{V_1}{V_2}=\frac{1}{2}\). So
\[P_2=\frac{ P_1V_1 }{ V_2 }=P_1\frac{ 1 }{ 2 }0=0.75\times \frac{ 1 }{ 2 }=.375(Pa)\]