Saved Bookmarks
| 1. |
Derive the relation between DeltaH and DeltaU for an ideal gas. |
|
Answer» SOLUTION :1. When the system at constant pressure undergoes changes from an initial state with `H_1 U_1 V_1` and P parameters to a final state with `H_2 U_2 V_2` and P parameters, the CHANGE enthalpy `DeltaH`, is GIVEN by `DeltaH=U+PV` (2) At initial state `H_1=U_1+PV_1`......(1) At final state `H_2 =U_2+PV_2`.........(2) `(2)-(1)implies(H_2-H_1)=(U_2-U_1)+P(V_2-V_1)` ……(3) Considering `DeltaU=q+w, w=-PDeltaV` `DeltaH=q+w+PDeltaV` `DeltaH=q_(p)-PDeltaV+PDeltaV` `DeltaH=q_(p)` ……..(4) `q_p` is the HEAT absorbed at constant pressure and is considered as heat content. (3)Considered a closed system of glass which are chemically reacting to produce product gases at contant temperature and pressure with `V_1` and `V_f` as the total volumeof the reactant and product gases respectively and `n_1` and `n_f` are the number of mole of gaseous reactant and product. Then For reactant `PV_(i)= n_(i)RT` For product: `PV_(f)=n_(f)RT` Then considering reactant as initial state and product as final state. `P(V_(f)-V_(i))=(n_(f)-n_(i))RT` `PDeltaV=Deltan_(g)RT` We KNOW `DeltaH=DeltaU+PDeltaV` `:.DeltaH=DeltaU+Deltan_(g) RT` ………(5) |
|