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A sample of 3.0 mole of perfect gas at 2...

A sample of `3.0` mole of perfect gas at `200 K` and `2.0 atm` is compressed reversibly and adiabatically until the temperature reaches `250 K`. Given that molar heat capacity at `27.5 J K^(-1) mol^(-1)` at constant volume calculate `q,W,DeltaU,DeltaH` and the final pressure and volume.

Text Solution

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Initially n=3 moles,
`T_(1) = 200 R, P_(1) = 2.0 atm, C_(v) = 27.5 JK^(-1) mol^(-1)`
After compression
`T_(2) = 250R, P_(2)=?`
`C_(p) = 27.5 + 8.314 JK^(-1) mol^(-1) = 35.814 JK^(-1) mol^(-1)`
`therefore C_(p)/C_(v) = (35.814)/(27.5) = 1.30 = gamma`
`therefore P_(1)^(1-gamma) .T_(1)^(gamma) = P_(2)^(1-gamma).T_(2)^(gamma)`
`therefore P_(2)^(1-gamma) = P_(1)^(1-gamma) (T_(1)/T_(2))^(gamma)`
`(P_(2))^(-0.3) = (2)^(-0.3) xx (200/250)^(1.30)`
`P_(2) = 5.2` atm
Let the final pressure be `P_(2)` and volume by `V_(2)`. `(therefore V = (nRT)/P = (3 xx 0.0821 xx 200)/2 = 24.63)` litre Now for volume `V_(2) : P_(1)V_(1)^(2) = P_(2)V_(2)^(2)`
`therefore [V_(2)/V_(1)]^(y) = P_(1)/P_(2)`
`therefore [V_(2)/(24.63)]^(1.3) = 2/(5.2)`
`therefore V_(2) = 11.9` litre
For adiabatic process q=0
Since whole of the work done is made on the cost of internal energy of the system.
Thus, `DeltaE=W = +4.157 kJ`
Also, `DeltaH = m xx C_(p) xx DeltaT = 3 xx 35.814 xx 50 = 5372.1 = 5.372` kJ
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