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Question

A cylinder of ideal gas is closed by an 8 kg movable piston (area 60 cm2). Atmospheric pressure is 105 Pa. When the gas is heated from 30°C to 100°C, the piston rises by 20 cm. The piston is then fixed in its place and the gas is cooled back to 30°C. Let ΔQ1 be the heat added to the gas in the heating process and |ΔQ2| the heat lost during cooling. Then the value of ΔQ1|ΔQ2| will be

A
0
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B
136 J
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C
136 J
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D
68 J
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Solution

The correct option is B 136 J

In case 1: Temperature from 30oC100oC
ΔQ1=ΔU1+ΔW1ΔQ1=ΔU1+PΔV....(1)

Total pressure at piston


P=P0+mgA
Using the given data,
P=105+8×1060×104=1.13×105 Pa

Putting this value in eq.(1)

ΔQ!=ΔU1+(1.13×105)(60×104×0.2) =ΔU1+136

Case 2: 100oC30oC
ΔQ2=ΔU2+ΔW2
ΔW2=0 (given)
ΔQ2=ΔU2=ΔU1
(because ΔU depends only on initial and final state)
|ΔQ2|=ΔU1

So,
ΔQ1|ΔQ2|=ΔU1+136ΔU1ΔQ1|ΔQ2|=136 J

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