The correct option is A (T−4) K
Given, γ=53
Specific heat at constant volume is given by,
Cv=Rγ−1=3R2
From first law of thermodynamics,
ΔQ=ΔU+ΔW
For an adiabatic process, ΔQ=0
⇒ΔU=−ΔW
From the data given in the question, we can write that
CVΔT=−6R
⇒ΔT=−6RCV
⇒ΔT=−4 K
This gives the final temperature Tf as (T−4) K.
Thus, option (a) is the correct answer.
Alternate method:
Work done by gas during an adiabatic process is given by
W=Rγ−1[T1−T2]
⇒6R=R53−1[T−Tf]
⇒T−Tf=4 K
⇒Tf=(T−4) K
Hence, option (a) is the correct answer.