Q. 155.0( 1 Vote )

# A gas is enclosed in a cylindrical vessel fitted with a frictionless piston. The gas is slowly heated for some time. During the process, 10 J of heat is supplied, and the piston is found to move out 10 cm. Find the increase in the internal energy of the gas. The area of cross section of the cylinder = 4 cm2 and the atmospheric pressure = 100 kPa.

Given

Heat supplied to system ΔQ = 10J

Atmospheric pressure P= 100kPa =100×103Pa

Displacement of the piston d=10cm=0.1m

Area of cross section of cylinder A=4cm2 =410-4m2

Gas will expand when is heat to the system. Therefore, the volume of gas expanded ΔV

ΔV=area ×displacement

=A ×d

ΔV=4×0.1×10-4=40×10-6m3

We know that work done by the gas is given as

ΔW=PΔV

=100×103×40×10-6

ΔW =4J

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

ΔU=ΔQ-ΔW = 10-4 =6J

Thus, the increase in internal energy is 6J.

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