two glass bulbs A &B of volume V , 2V respectively are connected by a narrow capilary tube. The bulbs contain gas at temp T and Pr P. Now the temp of bulb A is doubled and that of B is tripled. The final number of moles of gas in bulb A is : 9 options 3/2 PV/RT, 5/7 PV/RT, 12/7 PV/RT & 9/7 PV /RT.

Question

anonymous · Answer

Pl explain

anonymous · Answer

It is 9/7 PV/RT as per the printed question paper in front of me..

experimentx · Answer

hold on ... I may be wrong.

experimentx · Answer

the volume is same and since the tubes are connected the pressure is also same. the total no of moles is 
$$ {3 PV \over RT}$$

experimentx · Answer

now the gases are heated with temperature $ 2T $ and $ 3T $, the new pressure is be $ P_1$
Now,
$$ {P_1 V \over R 2T} + {P_1 2V \over R 3T} = {3PV \over RT}\
P_1  = {18P \over 7} $$
The number of moles in tube A is given by $ {P_1 V \over R 2T}$
$$ {P_1 V \over R 2T} = {18 P V \over  7 *2 RT} = {9 \over 7 }.. $$

experimentx · Answer

I guess the last option is correct.

anonymous · Answer

I think that's the correct answer. Thanks