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Derive the relationship between `Delta H " and " Delta U` for an ideal gas. Explain each term involved in the equation. |
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Answer» From the first law of thermodyanmics, `q = Delta U + p Delta V` If the process carried out at constant volume, `Delta V = 0` Hence, `q_(v) = Delta U` [Here, `q_(v) =` Heat absorbed at constant volume, `Delta U =` change in internal energy] Similarly, `q_(p) = Delta H` Here, `q_(p) =` heat absorbed at constant pressure `Delta H =` enthalpy change of the system Enthalpy change of a system is equal to the heat absorbed or evolved by the system at constant pressure. As we know that at constant pressure, `Delta H = Delta U + p Delta V` Where, `Delta V` is the change in volume. This equation can be rewritten as `Delta H = Delta U + p(V_(f) - V_(i)) = Delta U + (pV_(f) - pV_(i))`....(i) where, `V_(i) =` initial volume of the system `V_(f) =` final volume of the system But for the ideal gases, `pV = nRT` So that `pV_(1) = n_(1) RT` and `pV_(2) = n_(2) RT` where, `n_(1) =` number of moles of the gaseous reactants `n_(2) =` number of moles of the gaseous products. Substituting these value in Eq. (i) we get `Delta H = Delta U + (n_(2) RT - n_(1) RT)` `Delta H = Delta U + (n_(2) - n_(1)) RT` or `Delta H = Delta U + Delta n_(g) RT` where, `Delta n_(g) = n_(2) - n_(1)` is the difference between the number of moles of the gaseous products and gaseous reactants. Putting the value of `Delta H " and " Delta U` we get `q_(p) = q_(v) + Delta n_(g) RT` Note conditions under which `q_(p) = q_(v) " or " Delta H = Delta U` (i) where reactions is carried out in a closed vessel so that volume remains constant i.e., `Delta V = 0` (ii) when reaction involves only solids or liquids or solutions but no gaseous reactant or product. This is because the volume change of the solids and liquids during a chemcial reaction are negligible. (iii) When reaction involves gaseous reactant and products but their number of moles are equal (i.e., `n_(p) = n_(r)`) e.g. `H_(2) (g) + Cl_(2) (g) rarr 2HCl (g)` `C(s) + O_(2) (g) rarr CO_(2) (g)` Since, `q_(p)` is different from `q_(v)` only in those reactions which involves gaseous reactants and products and `(n_(p))` gaseous `!= (n_(r))` gaseous. |
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