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haridas_mandal

  • 3 years 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.

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  1. haridas_mandal
    • 3 years ago
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    Pl explain

  2. haridas_mandal
    • 3 years ago
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    It is 9/7 PV/RT as per the printed question paper in front of me..

  3. experimentX
    • 3 years ago
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    hold on ... I may be wrong.

  4. experimentX
    • 3 years ago
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    the volume is same and since the tubes are connected the pressure is also same. the total no of moles is \[ {3 PV \over RT}\]

  5. experimentX
    • 3 years ago
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    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 \( {P_1 V \over R 2T}\) \[ {P_1 V \over R 2T} = {18 P V \over 7 *2 RT} = {9 \over 7 }.. \]

  6. experimentX
    • 3 years ago
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    I guess the last option is correct.

  7. haridas_mandal
    • 3 years ago
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    I think that's the correct answer. Thanks

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