Question

An ideal gas undergoes a change of state (2.0 atm. 3.0 L, 95 K ) to (4.0 atm. 5.0 L, 245 K) with a change in internal energy, `DeltaU` = 30.0 L atm. The change in enthalpy (`DeltaH`) of the process in L atm is (A) 44 (B) 42.3 (C) ?

Why don't we use the formula for work as nr(change in temp.) to get the work done by taking n=1.
When to use (change in pv) and when to use(change in temp.)

Answer 1

Well, every natural variable has changed, and so the mols changed as well. Apparently, the starting mols is not `1`!

`"1 mol gas" stackrel(?" ")(=) (P_1V_1)/(RT_1) = ("2.0 atm" cdot "3.0 L")/("0.082057 L"cdot"atm/mol"cdot"K"cdot"95 K")`

`= "0.770 mols" ne "1 mol"`

The final state also presents the same problem:

`"1 mol gas" stackrel(?" ")(=) (P_2V_2)/(RT_2) = ("4.0 atm" cdot "5.0 L")/("0.082057 L"cdot"atm/mol"cdot"K"cdot"245 K")`

`= "0.995 mols" ~~ "1 mol"`

It is clear that with these numbers (did you copy down the question correctly?), the mols of gas changed. So `Delta(nRT) ne nRDeltaT`.

Instead, we begin with the definition:

`H = U + PV`

where `H` is enthalpy, `U` is internal energy, and `P` and `V` are pressure and volume.

For a change in state,

`color(blue)(DeltaH) = DeltaU + Delta(PV)`

`= DeltaU + P_2V_2 - P_1V_1`

`= "30.0 L"cdot"atm" + ("4.0 atm" cdot "5.0 L" - "2.0 atm" cdot "3.0 L")`

`= color(blue)("44.0 L"cdot"atm")`

Had we chosen to use `Delta(nRT)`, we would still get it, as long as we DO change the mols of gas:

`color(blue)(DeltaH) = DeltaU + Delta(nRT)`

`= DeltaU + n_2RT_2 - n_1RT_1`

`= "30.0 L"cdot"atm" + ("0.995 mols" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "245 K" - "0.770 mols" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "95 K")`

`= color(blue)("44.0 L"cdot"atm")`

By the way, note that

`Delta(PV) ne PDeltaV + VDeltaP`

Actually,

`Delta(PV) = PDeltaV + VDeltaP + DeltaPDeltaV`

In this case the `DeltaPDeltaV` accounts for `10%` of the `DeltaH` value.