Question

derive relation dell H= dell U +dell nRT

Answer 1

♠ ANSWER :

❇ Relation between ∆ H = ∆ U + ∆ nRT

Consider a reaction,

A --------> B

At constant pressure, P. Let H ( A) be the enthalpy of the reactants and H ( B) be the enthalpy of the products, then the change in enthalpy, ∆ H will be

∆ H = H ( B) - H ( A),

Since H = U + pV

∆ H = [ U ( B) + pV ( B) ] - [ U (A) + p V (A) ]

∆ H = {U ( B ) - U ( A)} + p { V ( B) - V ( A )}

∆ H = ∆ U + p ∆ V.

ACCORDING TO IDEAL GAS EQUATION,

pV = nRT --- ( i )

For A,

pV ( A) = n (A) RT

For B,

pV (B ) = n (B) RT --- ( ii )

Now subtracting equation ( ii ) from ( i ),

{ p V ( B) - V ( A)} = n ( B) RT - n ( A ) RT

p ∆ V = { n ( B) - n ( A)} RT

❇ => p ∆ V = ∆ n RT.

Answer 2

Answer:

Explanation:

❇ Relation between ∆ H = ∆ U + ∆ nRT

Consider a reaction,

A --------> B

At constant pressure, P. Let H ( A) be the enthalpy of the reactants and H ( B) be the enthalpy of the products, then the change in enthalpy, ∆ H will be

∆ H = H ( B) - H ( A),

Since H = U + pV

∆ H = [ U ( B) + pV ( B) ] - [ U (A) + p V (A) ]

∆ H = {U ( B ) - U ( A)} + p { V ( B) - V ( A )}

∆ H = ∆ U + p ∆ V - - - - (i)

ACCORDING TO IDEAL GAS EQUATION,

pV = nRT --- ( ii)

For A,

pV ( A) = n (A) RT

For B,

pV (B ) = n (B) RT --- ( iii)

Now subtracting equation ( ii ) from ( i ),

{ p V ( B) - V ( A)} = n ( B) RT - n ( A ) RT

p ∆ V = { n ( B) - n ( A)} RT

❇ => p ∆ V = ∆ n RT.

Substitution it in---(i)

∆ H = ∆ U + ∆ nRT