# 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.)

##### 1 Answer
Dec 21, 2017

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")

$= \text{0.770 mols" ne "1 mol}$

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")

$= \text{0.995 mols" ~~ "1 mol}$

It is clear that with these numbers (did you copy down the question correctly?), the mols of gas changed. So $\Delta \left(n R T\right) \ne n R \Delta T$.

Instead, we begin with the definition:

$H = U + P V$

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

For a change in state,

$\textcolor{b l u e}{\Delta H} = \Delta U + \Delta \left(P V\right)$

$= \Delta U + {P}_{2} {V}_{2} - {P}_{1} {V}_{1}$

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

$= \textcolor{b l u e}{\text{44.0 L"cdot"atm}}$

Had we chosen to use $\Delta \left(n R T\right)$, we would still get it, as long as we DO change the mols of gas:

$\textcolor{b l u e}{\Delta H} = \Delta U + \Delta \left(n R T\right)$

$= \Delta U + {n}_{2} R {T}_{2} - {n}_{1} R {T}_{1}$

= "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")

$= \textcolor{b l u e}{\text{44.0 L"cdot"atm}}$

By the way, note that

$\Delta \left(P V\right) \ne P \Delta V + V \Delta P$

Actually,

$\Delta \left(P V\right) = P \Delta V + V \Delta P + \Delta P \Delta V$

In this case the $\Delta P \Delta V$ accounts for 10% of the $\Delta H$ value.