One of the two thermally insulated vessels inferconnected by a tube with a valve contains `v = 2.2` moles of an ideal gas. The other vessel if evacuated. The valve having been opened, the gas increased its volume `n = 3.0` times. Find the entropy increment of the gas.

Explanation

In this case, called free expansion ono work is done and no heat is exchanged. So inernal energy must remain inchanged `U_f = U_i`. For an ideal gas this implies constant temperature `T_f = T_i`. The process is irreversible but the entropy change can be calculated by considering a reversible insothermal process. Then, as before
`Delta S = int (dQ)/(T) = int_(V_1)^(V_2) (pdV)/(T) = vR 1n n = 20.1 J//K`.