Bulb A of 4 lit contains a gas at 3 atm and 27° C . It is connected through a closed tap to an evacuated bulb B of 3 lit that is kept at 177° C . The tap is opened, bulb A stays at 27° C and bulb B st

Question

Bulb A of 4 lit contains a gas at 3 atm and 27° C . It is connected through a closed tap to an evacuated bulb B of 3 lit that is kept at 177° C . The tap is opened, bulb A stays at 27° C and bulb B stays at 177° C . The final pressure in the system is

TestHub · Accepted Answer

Total moles stay the same. Initially only bulb A holds gas: $n R$ = 3 x 4/300 = 0.04.

After the tap is opened both bulbs are at the same pressure $P$ : $n R$ = $P$ (4/300 + 3/450), with volumes in lit and temperatures in K.

0.04 = $P$ (0.01333 + 0.00667) = 0.02 $P$ , so $P$ = 2.0 atm.

Treating both bulbs as isothermal gives 12/7 = 1.71 atm, which is the trap.

Answer: (B).

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