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Question

16 g of oxygen gas expand at STP to occupy double of its original volume. The work during the process is:
(1 L.atm = 101.3 J)

A
260 kcal
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B
180 kcal
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C
130 kcal
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D
271.17 kcal
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Solution

The correct option is D 271.17 kcal
At STP, 16 g of O2 or 12 mole of O2 will occupy 11.2 L volume.
Now, if the volume is doubled, it means V2 = 22.4 L
So, (V2V1) = 22.4 - 11.2 = 11.2 L
At STP = P = 1 atm
Now, w=P×(V2V1)=1×11.2 L. atm = -11.2 L.atm = 11.2×101.3 J =
w=1134.56 J
Since, 1 cal = 4.184 J
w = 1134.564.184 = 271.17 kcal

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