Question

1 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

Let:

$$P1 = 0.71Pa, V1 = 0.75m^3, V2 = 2 * V1, T1 = 137{}^\circ \text{C}$$ $$P2 - ?, V2 - ?, T2 - ?$$

At isothermal process temperature is constant and according to Boyle’s law the product:

$$P * V = \text{constant}.$$

From this follows:

$$P1 * V1 = P2 * V2, T2 = T1,$$ $$P1 * V1 = P2 * 2 * V1,$$ $$P2 = \frac{P1}{2},$$ $$P2 = \frac{0.71}{2} = 0.355Pa, T2 = 137{}^\circ \text{C}$$

Answer: at final state temperature: $137{}^\circ \text{C}$, pressure: $0.355Pa$.