Answer :

Given


Heat extracted from the system ΔQ=-1200J


(negative sign is because heat is extracted from the system)


Number of moles of the gas n=2.0



From the graph we can write


TA=300K


TB=500K


VA=VC


We know that work done by the gas is given as


ΔW=PΔV


Where ΔV =change in volume


P =pressure


So, work done along line CA will be zero, as VA=VC.


Thus, total work done will be


ΔW=WAB+WBC


WAB = P(VB-VA)


But we know that ideal gas equation is


PV=nRT


Where n= number of moles


R=gas constant


T=temperature


PΔV=nRΔT


Thus, we can write WAB = P(VB-VA) = nR(TB-TA)


ΔW= nR(TB-TA) + WBC


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


Process ABCA is a cyclic process. The system is brought back to its initial state. Since internal energy is a state function, change in internal energy will be zero.


So, ΔU=0.


So, first law becomes


ΔQ=ΔW


ΔQ= nR(TB-TA) + WBC


We know that R=8.31J/K mol


WBC =ΔQ- nR(TB-TA)


= -1200 - 2×8.31× (500-300)


=-1200 – 3324


=-4524J


Thus, work done along path BC =-4524J


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