For adiabatic condition, dq=0
Thus by first law of thermodynamics.
dU=dW+dq
⇒dU=dW⟶(1) as dq=0 for adiabatic
⇒dW=−Pext(V2−V1)
⇒dU=CvdT
For monoatomic gas
Degree of freedom= 3M=3×1=3⟶ All transitional
Cv=32K or for 1 mole K=R
Cv=32R
Putting all values in equation 1, we have
CvdT=−Pext(V2−V1)
⇒32(T2−T1)=−1(2L−1L)
⇒(T2−T1)=−1×23R
⇒T2=T1−23R
Thus, final temperature is T−23R .