Question

A metallic container of fixed volume of 2.5 x 10^-3 m³ immersed in a large tank of temperature 27 degrees celsius contains two compartments separated by a freely movable wall. Initially, the wall is kept in place by a stopper so that there are 0.02 mol of the nitrogen gas on one side and 0.03 mol of the oxygen gas on the other side, each occupying half the volume. When the stopper is removed, the wall moves and comes to a final position. The movement of the wall is controlled so that the wall moves in infinitesimal quasi-static steps. (a) Find the final volumes of the two sides assuming the ideal gas behavior for the two gases. (b) How much work does each gas do on the other? (c) What is the change in the internal energy of each gas? (d) Find the amount of heat that enters or leaves each gas.

Answer 1

a) Finally, when seprator is at rest, tempreture and pressure on both sides is same.
As PV = nRT,
V is proportional to n for same P and T. Hence
V_nirogen / V_oxygen = 0.02/0.03 = 2/3   ....1
V_nirogen + V_oxygen = 2.5x10^-3 .....2
From 1 and 2 , we get
V_nirogen   = 1x10^-3 m³ and V_oxygen = 1.5x10^-3 m³

b) Throughout the process , tempreture on both sides remains constant, 27 degC / 300 K, hence process on both side is Isothermal.
Work done by gas in isothermal process = nRT (ln Vf/.Vi)
Work done by oxygen = 0.03*8.3*300 ln(1.5/1.25) = 13.62 J
Work done by nitrogen = 0.02*8.3*300 ln(1/1.25) = -11.11 J

c) As tempreure of boht sides does not change, change in internal energy on both sides is zero.

d) In isothermal process Q = W
13.62 J of heat enters, oxygen and 11.11 J of heat leaves Nitrogen