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Question

A glass bulb of volume 400cm3 is connected to another of volume 200cm3 by means of a tube of negligible volume. The bulbs contain dry air and both are at a common temperature and pressure of 20C and 1.00 atm respectively. The larger bulb is immersed in steam at 10C and the smaller in melting ice at 0C. Find the final common pressure.

A
6.13 atm
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B
7.13 atm
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C
2.13 atm
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D
1.13 atm
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Solution

The correct option is D 1.13 atm
Since the total number of moles must remain constant,

From n=PVRT,

1(400)293+1(200)293=P(400283+200273)

P=32932283+1273=1.13atm

Answer is 1.13atm

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