Two identical containers joned by a small pipe initially contain the same gas at pressue `p_(o)` and abosolute temperature `T_(o^.)` One container is now mantained at the same temperature while the other is heated to `2T_(0^.)` The commmon pressure of the gases will be

Options

A. `3/2p_(o)`
B. `4/3p_(o)`
C. `3/5p_(o)`
D. `2p_(o)`

Answer

B

Explanation

`(n_(1)+n_(2))=(n_(1)+n_(2))_(f)`
`(p_(o)V)/(RT_(o))+(p_(o)V)/(RT_(o))=(pV)/(RT_(o))+(pV)/(2RT_(o))`
`" "p=(4//3)p_(o)`