An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume `V_1` and contains ideal gas at pressure `P_1` and temperature `T_1`.The other chamber has volume `V_2` and contains ideal gas at pressure `P_2` and temperature `T_2`. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container 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. `(P_(1) V_(1)T_(1) + P_(2)V_(2)T_(2))/(P_(1)V_(1) +P_(2)V_(2))`
C. `(P_(1) V_(1)T_(2) + P_(2)V_(2)T_(1))/(P_(1)V_(1) +P_(2)V_(2))`
D. `(T_(1)T_(2) (P_(1)V_(1) + P_(2)V_(2)))/(P_(1) V_(1) T_(1) + P_(2)V_(2)T_(2))`

Answer

A

Explanation

`P_(1)V_(1)T_(1) (##BSL_CHM_THRM_E01_026_S01##) P_(2), V_(2), T_(2)`
As container is insulated
`q = 0 `
`q_(1) + q_(2) = 0`
`n_(1)C_(m) (T- T_(1)) + n_(2) C_(m) (T - T_(2)) = 0`
`T =(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+ n_(2))=((P_(1)V_(1) + P_(2)V_(2)) T_(1)T_(2))/(P_(1)V_(1)T_(2) + P_(2)V_(2)T_(1))`