2 mole of an ideal gas is expanded adiabatically and reversibly from 10 L to 80 L. The initial temperature of gas was 47°C. ∆U for this process is ____. Cv, of gas =  3 R/2.

For adiabatic process, q = 0 For gas, CV=32R,CP=CV+R=32R+R=52R and γ=CP/CV=53 Now for adiabatic reversible process, T1⋅V1γ−1 or, T2=T1⋅V1V2γ−1=320×108053−1=80K Now, ΔU=W=nCVT2−T1=2×32R×(80−320)=−1440 cal