Two thermally insulated vessel 1 and 2 are filled with air at temperature `(T_1T_2), volume (V_1V_2)` and pressure `(P_1P_2)` respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

Options

A. `(T_(1)T_(2)(P_(1)V_(1)+P_(2)V_(2)))/(P_(1)V_(1)T_(2)+P_(2)V_(2)T_(1))`
B. `(T_(1)T_(2)(P_(1)V_(1)+P_(2)V_(2)))/(P_(1)V_(1)T_(1)+P_(2)V_(2)T_(2))`
C. `(T_(1)+T_(2))/2`
D. `T_(1)+T_(2)`

Answer

B

Explanation

In this case, the total number of moles remain constant.
`becausePV=nRT" "thereforen=(PV)/(RT)`
Let T be the equilibrium temperature. `" "thereforen_(1)+n_(2)=n`
`therefore(P_(1)V_(1))/(RT_(1))+(P_(2)V_(2))/(RT_(2))=(P(V_(1)+V_(2)))/(RT)`
`therefore(P_(1)V_(1)T_(2)+P_(2)V_(2)T_(1))/(T_(1)T_(2))=(P(V_(1)+V_(2)))/T`
`thereforeT=(P(V_(1)+V_(2))T_(1)T_(2))/(P_(1)V_(1)T_(2)+P_(2)V_(2)T_(1))" "....(1)`
After the valve is opened, the mixture has a temperature T, volume `(V_(1)+V_(2))` and pressure P.
`therefore` Applying Boyle's law
`P_(1)V_(1)+P_(2)V_(2)=P(V_(1)+V_(2))" "...(2)`
Using (2) in (1), we get
`T=((P_(1)V_(1)+P_(2)V_(2))T_(1)T_(2))/(P_(1)V_(1)T_(2)+P_(2)V_(2)T_(1))" "...(3)`