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Question

Two glass bulbs of internal volume 0.5 and 0.2 L respectively are connected by a narrow tube of negligible volume. The pressure of air in the vessel is 75 cm at 17C. The smaller bulb is immersed in melting ice and the larger bulb in boiling water. Calculate final pressure in the bulbs neglecting the expansion of glass.

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Solution


At first at 170C or 290K,
the moles in 1st bulb, n1=75×0.5R×290
the moles in 2nd bulb, n2=75×0.2R×290
after immersing smaller bulb (2) in ice water and larger bulb (1) in boiling water, same gas will be transferred between the bulbs, until pressure of both bulb become equal.
In final conditions,
n11=P×0.5373R and n12=P×0.2273R
now, n1+n2=n11+n12
75×0.5R×290+75×0.2R×290=P×0.5373R+P×0.2273R
0.18=P(1.34+0.73)×103
P=86.95 cm of Hg

1082315_771376_ans_6070d7cd6ce1479ca323f9243d2b20bf.jpg

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