Two adiabatic containers have volumes `V_(1)` and `V_(2)` respectively. The first container has monoatomic gas at pressure `p_(1)` and temperature `T_(1)`. The second container has another monoatomic gas at pressure `p_(2)` and temperature `T_(2)`. When the two containers are connected by a narrow tube, the final temperature and pressure of the gases in the containers are P and T respectively. ThenA. `T=(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))/(p_(1)V_(1)+p_(2)V_(2))`B. `T=(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))/(p_(1)V_(2)+p_(2)V_(1))`C. `p=(p_(1)V_(2)+p_(2)V_(1))/(V_(1)+V_(2))`D. `p=(p_(1)V_(1)+p_(2)V_(2))/(V_(1)+V_(2))`

Two adiabatic containers have volumes `V_(1)` and `V_(2)` respectively

1.	Two adiabatic containers have volumes `V_(1)` and `V_(2)` respectively. The first container has monoatomic gas at pressure `p_(1)` and temperature `T_(1)`. The second container has another monoatomic gas at pressure `p_(2)` and temperature `T_(2)`. When the two containers are connected by a narrow tube, the final temperature and pressure of the gases in the containers are P and T respectively. ThenA. `T=(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))/(p_(1)V_(1)+p_(2)V_(2))`B. `T=(p_(1)V_(1)T_(2)+p_(2)V_(2)T_(1))/(p_(1)V_(2)+p_(2)V_(1))`C. `p=(p_(1)V_(2)+p_(2)V_(1))/(V_(1)+V_(2))`D. `p=(p_(1)V_(1)+p_(2)V_(2))/(V_(1)+V_(2))`
Answer» Correct Answer - D

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