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A sample of argon gas at 1atm pressure a...

A sample of argon 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)`.

Text Solution

Verified by Experts

The correct Answer is:
114.51
`DeltaH = n xx C_(p) xx DeltaT` and `C_(p) = C_(v) + R = 12.48 + 8.314 = 20.794 JK^(-1) mol^(-1)`
For a given sample of argon gas, mole (n)
`n=(PV)/(RT) = (1 xx 1.25)/(0.0821 xx 300) = 0.05` mole
Also,for reversible adiabatic change `TV^(gamma-1)` = constant
`T_(2)V_(2)^(gamma-1) = T_(1)V_(1)^(gamma-1)`
or `T_(2)-T_(1) xx (V_(1)/V_(2))^(gamma-1) = 300 xx (1.25)/(2.50)^(1.66-1) = 300 xx (1/2)^(0.66)`
`T_(2) = 189.86` K
`DeltaT = T_(2)-T_(1) = 189.86 - 300 = -110.14` K
`DeltaH = 0.05 xx 20.794 xx (-110.14) = -114.51` J
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