# 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.)

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.