1.

Show that in an isothermal expansion of an ideal gas, a `DeltaU = 0` and b. `DeltaH = 0`.

Answer» a. For one mole of an ideal gas, `C_(v) = ((delU)/(delT))_(v)`
Hence,
`DeltaU = C_(v) dT`
For a finite change, `DeltaU = C_(v) DeltaT`
For an isothermal process, `T` is constant so that `DeltaT = 0`
Therefore, `DeltaU = 0`.
b. We know that, `DeltaH = DeltaU + Delta(PV)`
For an ideal gas, `pV = RT`
`:. DeltaH = DeltaU + Delta(RT) = DeltaU + RDeltaT`
Since `DeltaT = 0` and `DeltaU = 0`
Therefore, `DeltaH = 0`


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