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

Here is the solution to the problem given above.

Let:
P1 = 0.71 Pa,V1= 0.75m^3, V2 = 2 * V1T1 = 137oC
P2-? V2- ? T2- ?

At isothermal process temperature is constant, and according to Boyle's law the product P * V = constant

From this follows:
P1 * V1 = P2 * V2,T2 = T1,
P1 * V1 = P2 * 2 * V1,
P2 = P1/2
P2 = 0.71/2 = 0.355Pa,T2 = 137oC

Final answer:

At final state, temperature is 137oC, pressure is 0.355Pa