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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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haridas_mandalBest ResponseYou've already chosen the best response.0
It is 9/7 PV/RT as per the printed question paper in front of me..
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
hold on ... I may be wrong.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
the volume is same and since the tubes are connected the pressure is also same. the total no of moles is \[ {3 PV \over RT}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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 \( {P_1 V \over R 2T}\) \[ {P_1 V \over R 2T} = {18 P V \over 7 *2 RT} = {9 \over 7 }.. \]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
I guess the last option is correct.
 one year ago

haridas_mandalBest ResponseYou've already chosen the best response.0
I think that's the correct answer. Thanks
 one year ago
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