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