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.

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