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Two bulbs of equal volume are connected ...

Two bulbs of equal volume are connected by a narrow tube of negligible volume and filled with a gas at STP. IF one of the bulbs is kept in melting ice and the other in the water bath at `62^@C`,what will be the new pressure of the gas?

Text Solution

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Let the volume of each bulb=`Vcm^3` and each contains n number of moles of the gas.
`therefore` Total number of moles contained in the bulb initially
`=2n=(76 times 2V)/(R times 273)`[from [pV=nRT] ..........(1)
Let the final pressure in both the bulbs=p.
Number of moles in one of the bulbs `=(pV)/(R times 273)`
and that in the other bulb `=(pV)/(T(273+62))=(pV)/(R times 335)`
`therefore` Total number of moles contained in the bulbs
`=(pV)/(Rtimes273)=(pV)/(R times 335)=(pV)/R(1/273+1/335)`........(2)
`therefore` From (1) and (2) we get
`(76times2V)/(R times273)=(pV)/R[1/273+1/335]`
or,`(pV)/R(0.0037+0.0030)=(76timesVtimes2)/(R times 273)`
or, `p times 0.0067=0.5568 ` or ,`p=0.5568/0.0066=83.10 cmHg`
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