Q. 14

# <span lang="EN-US

Given

Constant pressure p=1.0×105 Pa

Change in volume ΔV=(200-100)×10-6m3=10-4m3

Internal energy U= 1.5pV

So, change in internal energy ΔU=1.5pΔV

=1.5×1.0×105×10-4

=15J

We know that work done by the gas is given as

ΔW=pΔV=1.0×105×10-4=10J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Therefore,

ΔQ=15+10=25J

heat absorbed by the system=25J.

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