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

Since pressure, temperature, and volume are state functions, we can relate the two states, with the ideal gas law, as\[{P_1 V_1 \over n_1T_1} = {P_2 V_2 \over n_2T_2}\]R cancels out because it is the same for both states.

We will assume that the system is closed. \(\therefore n_1 = n_2 = n = \rm const.\)
We know the process is isothermal. \(\therefore T = {\rm const.} \therefore T_1 = T_2\)

The final expression becomes\[P_1 V_1 = P_2 V_2\]