Q. 5

# The pressure of gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc.(a) Calculate the work done by the gas.(b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?

Given

Initial pressure P1=10kPa=10×103Pa

Final pressure P2=50kPa=50×103Pa

Initial volume V1=200cc=200×10-6m3

Final volume V2=50cc= 50×10-6m3

a) We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V2-V1

Pressure =P

Thus, work done by the gas

ΔW=PΔV

Here we have given two values of pressure. So, we will take the average value of pressure

Average pressure P

Therefore, ΔW = 30×103× (50-200)×10-6

ΔW= -4.5J

Work done by the gas is -4.5J.

b) Given that no heat is supplied or extracted from the gas.

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

Since ΔQ=0

Therefore ΔU = -ΔW = -(-4.5) J=4.5J

the change in internal energy of the gas is 4.5J.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
JEE Advanced Level Questions of NLM48 mins
Work done in Thermodynamic ProcessFREE Class
Microsporogenesis & Megasporogenesis49 mins
Meselson and Stahl experiment49 mins
NEET 2021 | Transcription - An important Topic62 mins
DNA Fingerprinting42 mins
MCQs of Ecology for NEET52 mins
Interactive Quiz on Sexual Reproduction in flowering plants44 mins
Pedigree chart58 mins
Different high order thinking questions based on DNA structure39 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses