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

According to standard gas equation
`(P_(1)V_(1))/(T_1) = mu_(1)R` and `(P_(2)V_(2))/(T_2) = mu^(2)R` ...(i)
As no work is done in removing the partition, total energy remains conserved. Therefore, ltbr. `3/2 (P_(1)V_(1)+P_(2)V_(2)) = 3/2P(V_(1)+V_(2))`
`:. P = (P_(1)V_(1)+P_(2)V_(2))/(V_(1)+V_(2))`..(ii)
For mixuture of two gases
`(mu_(1)+mu_(2))RT = P(V_(1)+V_(2))`
Using (i) and (ii)
`((P_(1)V_(1))/(RT_(1))+(P_(2)V_(2))/(T_2)) RT = ((P_(1)V_(1)+P_(2)V_(2))(V_(1)V_(2)))/((V_(1)+V_(2)))`
`((P_(1)V_(1))/(T_(1))+(P_(2)V_(2))/(T_2))T = (P_(1)V_(1)+P_(2)V_(2))`
On solving `T = ((P_(1)V_(1)+P_(2)V_(2))T_(1)T_(2))/((P_(1)V_(1)T_(1)+P_(2)V_(2)T_(2))`.