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Question

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 00 C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 620 C. What is the new value of the pressure inside the bulbs ? The volume of the connecting tube is negligible.

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Solution

When the bulbs are maintained at two different temperatures.

The total heat gained by 'B' is the heat lost by 'A'. Let the final temperature be X

So, m1SDt=m2SDt

n1X=62n2n2X

So, X = 310 C = 304K

[Since, Initial temperature = 00 C]

P = 76 cm of Hg, V1=V2Hencen1=n2]

for a single ball

P1V1T1=P2V2T2

76×V273=P2×V304

P2=304×76273

= 84.630 = 840 C


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