InterviewSolution
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InterviewSolution
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A sample of argon gas at 1 atm pressure and `27^(@)C` expands reversibly and adiabatically from `1.25dm^(3)` to `2.50 dm^(3)`. Calculatethe enthalpy changein this process. `C_(v,m)` for argonis `12.48 JK^(-1) mol^(-1)` |
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Answer» For adiabatic process , `T_(1)V_(1)^(gamma-1)= T_(2)V_(2)^(gamma-1) ` or `((V_(1))/(V_(2)))^(gamma-1) = (T_(2))/(T_(1))` But `gamma=( C_(p))/(C_(v))` `:gt gamma -1 = (C_(p))/(C_(v))-1= ( C_(p)-C_(v))/(C_(v))= (R)/(C_(v))` ,brgt Hence `((V_(1))/(V_(2)))^(R//C_(v))= (T_(2))/(T_(1))` or `ln.(T_(2))/(T_(1))=(R)/(C_(v))ln.(V_(1))/(V_(2)) ` or `log.(T_(2))/(T_(1))= (R)/(C_(v)) log. (V_(1))/(V_(2))` i.e., `log. (T_(2))/(300) = ( 8.314)/( 12.48) log. ( 12.5)/(25)` or ` logT_(2) = log 300 + 0.666 log . ( 1)/(2)` `= 2.4771 + 0.666 ( - 0.3010) = 2.2766` or `T_(2) =` Antilog 2.2766 `=189 K` `C_(p)=(DeltaH)/( DeltaT)` or `DeltaH = C_(p) Delta T ` or for n moles `Delta H = n C_(p) Delta T`. Taking the gas as ideal. `PV = nRT ` or ` 1 xx 1.25 = n xx 0.0821 xx 300` or `n = 0.05` mole Further, `C_(p) = C_(v) + R = 12.48 + 8.314 = 20.794J` `:. Delta H = 0.05 xx 20.794 xx ( 189 - 300) = - 115 . 4 J` |
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