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) = (1 xx 1.25)/(0.082 xx 300) = 0.05`
For adiabatic expansion,
`TV^(y-1)` = Constant
`C_(p) = C_(v) + R = 12.48 + 8.314 = 20.8` Joules
`(C_(p)) = C_(v) + R = 12.48 + 8.314 = 20.8` joules
`therefore (C_(p))/C_(v) =1 + R/C_(v)`
`rArr gamma = 1 + R/(C_(v)) rArr y^(-1) = R/(C_(v)) rArr TV^(R/C_(v))` = Constant
`rArr T_(1)V_(1) ^(T//C_(v)) = T_(2)V_(2)^(R//C_(v)) rArr T_(1)/T_(2) = (V_(2)/V_(1))^(R//C_(v))`
`rArr 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`
`rArr log (300)/T_(2) = (8.314)12.48 log_(2) rArr log (300)/T_(2) = 0.199 rArr 300/T_(2) = 1.58 rArr T_(2) = 300/(1.58)`
`T_(2) = 189.87` K
`therefore DeltaT = T_(2)-T_(1) = 189.87 - 300 = -110.13`
`DeltaH = 0.05 xx 20.8 xx (-110.13) = -114.53` Joules