Answer :

According to First law of thermodynamics,


ΔQ=ΔU+ΔW


Where ΔQ=heat supplied to the system


ΔU=change in internal energy


ΔW=work done by the system


For an isothermal process ΔT=0, therefore ΔU=0.


Then,


ΔQ=ΔW


When heat is supplied to the system gas expands i.e. the volume of the gas increases.


We know that,


Work done = force ×displacement



Volume = area ×displacement


Therefore,


Work done=pressure ×volume


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


Pressure =P


Thus, work done by the object W


W=PΔV


During expansion volume increases. Hence, work done will be positive and equals PV.

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