Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

Work done in reversible adiabatic process by an ideal gas is given by:
[A] 2.303RTlogV1V2
[B] nRγ1(T1T2)
[C] 2.303RTlogV2V1
[D] None of these.

Answer
VerifiedVerified
506.1k+ views
like imagedislike image
Hint: Use the first law of thermodynamics here. Heat released or absorbed is zero for an adiabatic process. Use heat capacities in terms of constant volume and pressure to get the correct answer in terms of gamma.

Complete step by step answer:
We know that the first law of thermodynamics states that-
     ΔU=q+w

We should know that ΔU is the change in internal energy of the system and it is a function of temperature and volume. ‘q’ is the heat absorbed or released and ‘w’ is the work done.
We know that for an adiabatic process, no heat is released from or enters the system. So for an adiabatic process, ‘q’ is zero. Therefore, we can write the above discussion that-
     ΔU=w
Or, we can write that- ΔU=pdv
But we can write that, Cv=(UT)V
The heat capacity of an ideal gas in independent of volume and just depends on temperature, Therefore we can write that- Cv=(dUdT)dU=CvdT for an ideal gas.
So, for a reversible adiabatic process of an ideal gas we can write that- PdV=CvdT

But we know that for a mole of an ideal gas,
     PV=RT=(CpCv)Tor,P=(CpCv)TV
Therefore, (CpCv)TdVV=CvdT

We know that, CPCV=γ therefore, substituting this in the above equation and separating the variables we will get-
     dTT+(γ1)dVV=0
Now, integrating this we will get- TVγ1=constant
Now, eliminating T we can write that- PVγ=constant
So, from these two relations eliminating volume we can write that- P(1γ)Tγ=constant

Now, let us try to calculate the work done - W=V1V2PdV
Now, we have calculated above that PVγ=constant=K.
Therefore, we can write that -
     W=KV1V2VγdV=Kγ1(V1(1γ)V2(1γ))=P1V1P2V2γ1=R(T1T2)γ1

We can understand from the above derivations that work done in a reversible adiabatic process for an ideal gas is R(T1T2)γ1. If we include the number of moles to this, we can write that nR(T1T2)γ1 , where T1 and T2 are the initial and final temperature respectively.
So, the correct answer is “Option B”.

Note: For an ideal gas, the calculated value of γ is 53 . As we discussed above that CPCV=γ and for an ideal gas, CP=52 and Cv=32 which gives us the value of gamma as 53.
Also, R = (CPCV) therefore, we can write the work done as CV(T1T2) .
The work done is directly proportional to the quantity of gas in moles and the temperature difference.
Latest Vedantu courses for you
Grade 9 | CBSE | SCHOOL | English
Vedantu 9 CBSE Pro Course - (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for CBSE students
PhysicsPhysics
Social scienceSocial science
BiologyBiology
ChemistryChemistry
EnglishEnglish
MathsMaths
₹41,000 (9% Off)
₹37,300 per year
Select and buy