Calculate change in internal energy for a gas under going from state-I (300K,2xx10^(-2)m^(3)) to state - II (400K,4xx10^(-2)m^(3)) for one mol. Of vanderwal gas. (a) If gas is ideal [C_(v)=12J//Kmol] (b) If gas is real {"Given":((delU)/(delV))_(T)=T((delP)/(delT))_(V)-P" "C_(V)=12J//k//mol" "a=2J.m.//mol^(2)}

Calculate change in internal energy for a gas under going from state-I

1.	Calculate change in internal energy for a gas under going from state-I (300K,2xx10^(-2)m^(3)) to state - II (400K,4xx10^(-2)m^(3)) for one mol. Of vanderwal gas. (a) If gas is ideal [C_(v)=12J//Kmol] (b) If gas is real {"Given":((delU)/(delV))_(T)=T((delP)/(delT))_(V)-P" "C_(V)=12J//k//mol" "a=2J.m.//mol^(2)}
Answer» Solution :(a) `DeltaU=nC_(V)(T_(2)-T_(1))=1xx12xx100=1200J` (b) `du=((delU)/(DELT))_(v)DT+((delU)/(delV))_(T)DV` `[[(P+a/V^(2))(V-b)=RT],[P=(RT)/(V-b)-a/V^(2)],[rArr(delP)/(delT)=R/(V-b)]]` = `((delU)/(delT))_(v)dT+[T((delP)/(delT))_(V)-P]dV` = `C_(V)dT+a/V^(2)dV` `dU=C_(V)(T_(2)-T_(1))+a(1/V_(1)-1/V_(2))` = `C_(V)(100)+a(1/4)xx10^(2)` = `12xx100+2(1/4)xx100=1250`