What is the relation between change in enthalpy (dH ) and change in internal (dE) for combustion of methan ?

For the given reaction:

                        CH₄(g) + 2O₂(g) ---------> CO₂(g) + 2H₂O(l)

                                          dH = dE + dnRT

                            Dn = no. of mole of products - no. of moles of reactants = 1– 3 = –2

                                             DH = DE – 2RT

1 mole of a real gas is subjected to a process from (2 bar, 40 lit.,300K) to (4 bar, 30 lit., 400 K). If change in internal energy is 35 kJ then calculate enthalpy change for the process.

                       DH = DU + D(PV)

                        D(PV) = P₂V₂ – P₁V₁

                                          = 4 × 30 – 2 × 40

                                   = 40 liter -bar = 4 kJ

                  so      DH = 35 + 4 = 24 kJ