A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from

Answer: \(\Delta U = -106.5\ \text{J}\) (internal energy decreases by 106.5 J)

Given: \(P = 1\ \text{atm} = 1.013\times10^{5}\ \text{Pa}\), \(V_1 = 5\ \text{L}\), \(V_2 = 10\ \text{L}\), \(Q = +400\ \text{J}\) (absorbed)

\(\Delta V = 10 - 5 = 5\ \text{L} = 5\times10^{-3}\ \text{m}^3\)

Work done by the gas (expansion against constant external pressure):

\(W = P\Delta V = 1.013\times10^{5}\times5\times10^{-3} = 506.5\ \text{J}\)

First law of thermodynamics:

\(\Delta U = Q - W = 400 - 506.5 = -106.5\ \text{J}\)

The negative sign shows that the gas does more work (506.5 J) than the heat it absorbs (400 J), so its internal energy decreases by 106.5 J.