Question

ideal gas occupying a volume of 2dm3 and a pressure of 5 bar undergoes isothermal and irreversible expansion against external pressure of 1 bar final volume of the system of work involved in the process is

Answer 1

Given:

• Initial volume of the gas (Vi) = 2L
• Initial pressure of the gas (Pi) = 5bar
• External pressure (Pext) = 1bar
• The expansion is isothermal and irreversible.

To find:

The work involved in the process.

Solution:

• At equilibrium, the pressure of the gas will be equal to the external pressure. So, Pf = Pext = 1bar
• Since the process is isothermal, PiVi = PfVf ⇒ Vf = Pi.Vi/Pf = 10L
• The change in volume during the process (ΔV) = Vf-Vi = 8L
• For an irreversible process, work done = -Pext.ΔV = -1.8 = -8barL

Answer:

The work involved in the process is equal to -8barL