Answer :

Given


Heat extracted from the system ΔQ=-70cal=-70×4.2=-294J


The negative sign is because heat is extracted from the system.


From graph


VA=250cc=250×10-6m3


VB=100cc=100×10-6m3


PA=200kPa=200×103Pa


PB=500kPa=500×103Pa


We know that work done by the gas is given as


ΔW=PΔV


Here since we have two values of pressure we will take average pressure.



ΔW=Pavg(VA-VB)


= 350×103× (250-100) ×10-6


=-52.5J


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,


ΔU=-294-(-52.5) =-241.5J


change in internal energy is -241.5J.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
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
caricature
view all courses
RELATED QUESTIONS :

<span lang="EN-USHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

A thermally insulHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

50 cal of hHC Verma - Concepts of Physics Part 2

<span lang="EN-USHC Verma - Concepts of Physics Part 2

Consider two procHC Verma - Concepts of Physics Part 2