A sample of argon gas at `1atm` pressure and `27^@C` expands reversibly and adiabatically from `1.25dm^3` to `2.5dm^3`. Calculate the enthalpy change in the process. Given that `C_(v(m))` for `Ar` is `12.45JK^(-1)mol^(-1)` and antilog `(0.199)=1.58`.

Explanation

We have to find out `DeltaH_(P)=nC_(P)dT`
No. of moles of `Ar`
`n=(PV)/(RT)=(1xx1.25)/(0.082xx300)=0.05`
For adiabatic expansion
`TV^((gamma-1))=`constant
`:' C_(P)=C_(v)+R=12.48+8.314=20.8 JK^(-1) mol^(-1)`
`:. (C_(P))/(C_(v))=1+R/(C_(v))=gamma=1+R/(C_(v))=gamma=1+R/(C_(v))`
`impliesTV^((R-C_(v))=` constant `implies T_(1)V_(1)^((R//C_(v)))=T_(2)V_(2)^((R//C_(v)))`
`implies (T_(1))/(T_(2))=((V_(2))/(V_(1)))^(R//C_(v))`
`implies "ln"(T_(1))/(T_(2))=R/(C_(v))"ln"(V_(2))/(V_(1))=8.314/12.48 "ln"(2.50)/(1.25)`
antilog `0.199=1.58`
`implies "log" (300)/(T_(2))=(8.314)/(12.48)"log"2implies"log"(300)/(T_(2))=0.199`
`implies300/(T_(2))=1.58impliesT_(2)=300/1.58`
`T_(2)=189.87 K`
`DeltaT=T_(2)-T_(1)=189.87-300=-110.13`
`DeltaH=0.05xx20.8xx(-110.13)=-114.43 J`