Two thermally insulated vessels 1 and 2 are filled with air at temperatures $$\left( {{T_1},{T_2}} \right),$$  volume $$\left( {{V_1},{V_2}} \right)$$  and pressure $$\left( {{P_1},{P_2}} \right)$$  respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

Solution :
Here $$Q = 0$$  and $$W = 0.$$  Therefore from first law of thermodynamics $$\Delta U = Q + W = 0$$
∴ Internal energy of the system with partition = Internal energy of the system without partition.
$$\eqalign{ & {n_1}{C_v}{T_1} + {n_2}{C_v}{T_2} = \left( {{n_1} + {n_2}} \right){C_v}\,T \cr & \therefore \,\,T = \frac{{{n_1}{T_1} + {n_2}{T_2}}}{{{n_1} + {n_2}}} \cr & {\text{But }}{n_1} = \frac{{{P_1}{V_1}}}{{R{T_1}}}\,{\text{and }}{n_2} = \frac{{{P_2}{V_2}}}{{R{T_2}}} \cr & \therefore \,\,T = \frac{{\frac{{{P_1}{V_1}}}{{R{T_1}}} \times {T_1} + \frac{{{P_2}{V_2}}}{{R{T_2}}} \times {T_2}}}{{\frac{{{P_1}{V_1}}}{{R{T_1}}} + \frac{{{P_2}{V_2}}}{{R{T_2}}}}} \cr & = \frac{{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)}}{{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}} \cr} $$