Ncert Exercises & Solutions

Derive the relationship between $∆ H$ and $∆ U$ for an ideal gas. Explain each term involved in the equation.

From the first law of thermodynamics,

q = ∆ U + p ∆ V

If the process carried out at constant volume,

∆ V = 0

Hence,

q_{v} = ∆ U

[Here,

q_{v} =

Heat absorbed at constant volume,

∆ U =

change in internal energy]

Similarly,

q_{p} = ∆ H

Here,

q_{b} =

heat absorbed at constant pressure

∆ 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,

∆ H = ∆ U + p ∆ V

where,

∆ V

is the change in volume.

This equation can be rewritten as

∆ H = ∆ U + p (V_{f} - V_{i}) = ∆ 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=number of moles of the gaseous reactants

Substituting these values in Eq. (i), we get

∆ H = ∆ U + (n_{2} RT - n_{1} RT)

∆ H = ∆ U + (n_{2} - n_{1}) RT

or ∆ H = ∆ U + ∆ n_{g} RT

where,

∆ n_{g} = n_{2} - n_{1}

is the difference between the number of moles of the gaseous products and gaseous reactants.

Putting the values of

∆ H

and

∆ U

we get

q_{p} = q_{v} + ∆ n_{g} RT