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Mohanlal Andra
Asked: 2022-11-06T00:14:17+05:30 In:

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)`

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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  1. 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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