Question

A rigid, non-conducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane represents 1/3 of the tank, contains nitrogen gas at 6 bar and 100°C. The other side, represents. 2/3 of the tank, is evacuated. If the membrane ruptures and the gas filled up the tank, describe how the process can be reversed in which the gas returns to its initial state. How much work (in kJ) is done?

Answer 1

In this process, no work done takes place and no heat transfer takes place

Change in internal energy $\Delta$ U = 0

Change in temperature $\Delta$ T = 0

T1 = T2 = 100 °C

The process is not reversible

this is the isothermal process in which the gas returns to its initial state

Work done W = nRT ln (V1/V2)

= P2 x V2 x ln (V1/V2)

Initial volume V1 = 4 m3

Final volume V2 = 4/3 m3

Pressure P2 = 6 bar

W = 6 bar x 10^5 Pa/bar x 4/3 m3 x ln (4*3/4)

W = 878889.83 J x 1kJ/1000 J

W = 878.9 kJ