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")#

#= "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.