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A sample of argon gas at `1atm` pressure and `27^(@)C` expands reversibly and adiabatically from `1.25 dm^(3)` to `2.50 dm^(3)`. Calculate the enthalpy change in this process. `C_(vm)` for orgon is `12.48J K^(-1) mol^(-1)`. |
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Answer» Number of moles of argon present in the sample `=(PV)/(RT)=(1.25xx1)/(0.0821xx300)=0.05075` For adiabatic expansion, `(T_(1))/(T_(2))=((V_(2))/(V_(1)))^(r-1)` or `(300)/(T_(2))=((2.50)/(1.25))^(1.66-1)` or `T_(2)=188.55K` `C_P=C_(V)+R` `=12.48+8.314` `=20.794JK^(-1)" "mol^(-1)` `DeltaH=nxxC_(P)xxDeltaT` `=0.05075xx20.794xx(300-188.55)` `=117.6J` |
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