InterviewSolution
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InterviewSolution
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One mol of an ideal diatomic gas underwent an adiabatic expansion form `298K, 15.00atm`, and `5.25L` to `2.5atm` against a constant external pressure of `1.00atm`. What is the final temperature of the system? |
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Answer» This is an isobaric adiabatic expansion against constant externla pressure, but overall pressure decreases (volume increases, gas expands). Final temperature `T_(2)` is given by `P-V-T` relation as: `T_(2)=T_(1)((C_(v)+P_(ex)(R )/(P_(1)))/(C_(v)+P_(ex)(R )/(P_(2))))` For diatomic gas, `C_(v) = (5)/(2) R, T_(1) = 298 K, T_(2) = ?` `P_(2) = 2.50atm, P_(1) = 15.00atm. P_(ex) = 1.900 atm` `:. T_(2) = 298 (((5)/(2)R+(R)/(15))/((5)/(2)R+(R)/(2.5))) = 263.7 K` |
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