Question
One mole of a non-ideal gas undergoes a change of state ( $${2.0\,atm,\,3.0L,\,95\left( K \right)}$$ → ( $${4.0\,atm,\,5.0L,\,245K}$$ ) with a change in internal energy, $$\Delta U = 30.0\,L\,atm\,.$$ The change in enthalpy $$\left( {\Delta H} \right)$$ of the process in $$L$$ atm is
A.
40.0
B.
42.3
C.
44.0
D.
not defined, because pressure is not constant
Answer :
44.0
Solution :
TIPS/Formulae :
$$\eqalign{
& \Delta H = \Delta U + {P_2}{V_2} - {P_1}{V_1} \cr
& {\text{Given}},\,\,\Delta U = 30.0L\,{\text{atm}} \cr
& {P_1} = 2.0\,{\text{atm}},\,{V_1} = 3.0L,\,{T_1} = 95K \cr
& {P_2} = 4.0\,{\text{atm}},\,{V_2} = 5.0\,L,\,{T_2} = 245K \cr
& \Delta H = \Delta U + {P_2}{V_2} - {P_1}{V_1} \cr
& = 30 + \left( {4 \times 5} \right) - \left( {2 \times 3} \right) \cr
& = 30 + 20 - 6 \cr
& = 44L\,{\text{atm}}. \cr} $$