InterviewSolution
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InterviewSolution
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A sample of organ 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» `DeltaH = n xx C_(P) xx DeltaT` and `C_(P) = C_(V) +R = 12.48 + 8.134 = 20.794 J K^(-1)` `n = (PV)/(RT) = (1xx1.25)/(0.0821xx300) = 0.05` Also for reversible adiabatic change `(TV)^(gamma -1) =` constant or `T_(2)V_(2)^(gamma -1) = T_(1)V_(1)^(gamma-1)` or `T_(2) = T_(1) xx ((V_(1))/(V_(2)))^(gamma -1) (gamma = 1.66` for argon) `= 300 xx ((1.25)/(2.50))^(1.66-1) = 300 xx ((1)/(2))^(0.66)` `T_(2) = 189.85K` `DeltaT = T_(2) - T_(1) = 189.85 - 300 =- 110.15` Thus, `DeltaH = 0.05 xx 20.794 xx (-110.15) =- 114.52J` |
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