Name: Factors Affecting the Internal Energy
Uploaded: 2019-07-19
Duration: 3 min 47 s
Description: The internal energy of a thermodynamic system is the energy contained within it. Let's watch the video to learn more about the internal energy of a system.

Question

Derive the relationship between

∆ H

and

∆ U

for an ideal gas. Explain each term involved in the equation.

Accepted Answer

We know that, the heat absorbed at constant volume is equal to change in the internal energy i.e., $△ U = q_{v}$ . But most of chemical reactions are carried out not at constant volume, but in flasks or test tubes under constant atmospheric pressure $△ U = q_{p} — p △ V$ at constant pressure, where $q_{p}$ is heat absorbed by the system and $- p △ V$ represent expansion work done by the system.
Let us represent the initial state by subscript 1 and final state by 2.
We can rewrite the above equation as,

$U_{2} - U_{1} = p (V_{2} - V_{2})$

On rearranging, we get

$q_{p} = (U_{2} + {pV}_{2}) - (U_{1} + {pV}_{1}) . . . . . (1)$

Now, we can define another thermodynamic function, the enthalpy $H$ (Greek word enthalpies, to warm or heat content) as:

$H = U + pV . . . . . (2)$

So, equation (1) becomes

$q_{p} = H_{2} - H_{1} = △ H$

Although, $q$ is a path dependent function, $H$ is a state function because it depends on $U, p$ and $V$ , all of which are state functions. Hence, $q_{p}$ is independent of path.

For finite changes at constant pressure, we can write equation (2) as

$△ H = △ U + p △ V$

Since $p$ is constant, we can write:

$△ H = △ U + p △ V . . . . . . (3)$
Remember $△ H = q_{p}$ is heat absorbed by the system at constant pressure.
$△ H$ is negative for exothermic reactions which evolve heat during the reaction and $△ H$ is positive for endothermic reactions which absorb heat from the surroundings.

At constant volume $(△ V — 0), △ U = q_{p}$ , therefore, equation (3) becomes

$△ H = △ U = q_{r}$

Let us consider a reaction involving gases. If $V_{A}$ is the total volume of the gaseous reactants, $V_{B}$ is the total volume of the gaseous products, $n_{A}$ is the number of moles of gaseous reactants and $n_{B}$ is the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we can write

${pV}_{A} = n_{A} RT$

and ${pV}_{B} = n_{B} RT$

Thus, ${pV}_{B} - {pV}_{A} = n_{B} RT - n_{A} RT = (n_{B} - n_{A}) RT$

or $(V_{B} - V_{A}) p = (n_{B} - n_{A}) RT$

or $p △ V = n_{g} RT . . . . . . (4)$

Here, $△ n_{g}$ refers to the number of moles of gaseous products minus the number of moles of gaseous reactants.
Substituting the value of $p △ V$ from equation (4) in equation (3), we get

$△ H = △ U + △ n_{g} RT . . . . . . (5)$

NCERT Solutions for Chapter: Thermodynamics, Exercise 1: Long Answer Type

NCERT Chemistry Solutions for Exercise - NCERT Solutions for Chapter: Thermodynamics, Exercise 1: Long Answer Type

Questions from NCERT Solutions for Chapter: Thermodynamics, Exercise 1: Long Answer Type with Hints & Solutions

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