Two identical thermally insulated vessels interconnected by a tube with a valve contain one mole of the same ideal gas each. The gas temperature in one vessel is equal to `T_1` and in the other, `T_2`. The molar heat capacity of the gas of constant volume equals `C_v`. the valve having been opened, the gas comes to a new equilibrium state. Find the entropy increment `Delta S` of the gas. Demonstrate that `Delta S gt 0`.

Explanation

For and ideal gas to internal energy depends on temperature only. We can consider the process in question to be one of simultaneous free expansion. Then the total energy `U = U_1 + U_2` Since
`U_1 = C_V T_1, U_2 = C_V T_2, U = 2 C_V (T_1 + T_2)/(2) ` and `(T_1 + T_2)//2` is the final temperature. The entropy change is obtained by considering isochoric processes because in effect, the gas remains confined to its vessel.
`Delta S = int _(T_(1))^((T_1 + T_2)//2) (C_V dT)/(T) - int_((T_1 + T_2)//2)^(T_2) C_V (dT)/(T) = C_V 1n ((T_1 + T_2)^2)/(4 T_1 T_2)`
Since `(T_1 + T_2)^2 = (T_1 - T_2)^2 + 4 T_1 T_2, Delta S gt 0`.