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Question

When 1gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, the volume increase from 1 cm3 to 1671 cm3. The heat of vaporisation at this pressure is 540 cal/gm. Find (a) the work done (in J) in change of phase and (b) increase in internal energy of water.(1 atmosphere = 1.01×106dyne/cm2)

A
1683.67J, 299.33J
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B
168.67J, 2099.33J
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C
168.67J, 1099.33J
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D
128.67J, 2099.33J
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Solution

The correct option is B 168.67J, 2099.33J
(a) As the process is isobaric
ΔW=PdV=P[VFVI]i.e.=1.01×106[16711]=1686.7×106ergs
or, = 168.67J (= 40 cal)
[as 1 erg = 107J]
(b) According to first law of thermodynamics
ΔQ=ΔU+ΔW
Here =mL=1 × 540cal=2268J
[as 1 cal = 4.2J]
So, ΔU=ΔQ+ΔW
= 2268 - 168.67
= 2099.33J




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