anonymous
  • anonymous
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.
Physics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Pl explain
anonymous
  • anonymous
It is 9/7 PV/RT as per the printed question paper in front of me..
experimentX
  • experimentX
hold on ... I may be wrong.

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experimentX
  • experimentX
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
  • experimentX
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
  • experimentX
I guess the last option is correct.
anonymous
  • anonymous
I think that's the correct answer. Thanks

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