Question

An ideal gas expands against a constant pressure of $2.026 \times {10^5}\,{\text{N}}{{\text{m}}^{ - 2}}$ from $5\,{\text{d}}{{\text{m}}^3}$ to $1.5\,{\text{d}}{{\text{m}}^3}$. If the change in the internal energy is $418$J. calculate the change in the enthalpy.

Answer 1

Enthalpy change is determined as the sum of internal energy change and pressure-volume change product. Enthalpy change is greater than the change in internal energy for the expansion process. As the pressure volume works positively in the expansion process.

Formula used: ${{\Delta H}}\,{\text{ = }}\,{{\Delta U}}\,\,{{ + p\Delta V}}$

Complete answer
The relation between internal energy and enthalpy is as follows:
${{\Delta H}}\,{\text{ = }}\,{{\Delta U}}\,\,{{ + p\Delta V}}$
Where,
${{\Delta H}}$ is the change in enthalpy.
${{\Delta U}}$ is the change in internal energy during the thermodynamic process..
${\text{p}}$ is the pressure.
${{\Delta V}}$ is the change in volume during the thermodynamic process.
The change in volume is determined by subtracting the initial volume from final volume.
${{\Delta V = }}\,{{\text{V}}_{{\text{final}}}} - \,{{\text{V}}_{{\text{initial}}}}$
Substitute $5\,{\text{d}}{{\text{m}}^3}$for final volume and $1.5\,{\text{d}}{{\text{m}}^3}$ for initial volume.
$\Rightarrow {{\Delta V = }}\,5\,{\text{d}}{{\text{m}}^3} - \,1.5\,{\text{d}}{{\text{m}}^3}$
$\Rightarrow {{\Delta V = }}\,3.5\,{\text{d}}{{\text{m}}^3}$
Convert the volume change from dm to m as follows:
$\Rightarrow 1\,{\text{d}}{{\text{m}}^3} = \,{\left( {{\text{0}}{\text{.1}}} \right)^{\text{3}}}\,{{\text{m}}^{\text{3}}}$
$\Rightarrow 3.5\,{\text{d}}{{\text{m}}^3}{\text{ = }}\,3.5\, \times {10^{ - 3}}{{\text{m}}^3}$. If the change in the internal energy is
Substitute $418$J for the change in the internal energy, $2.026 \times {10^5}\,{\text{N}}{{\text{m}}^{ - 2}}$for the change in internal energy, and $3.5\, \times {10^{ - 3}}{{\text{m}}^3}$for volume change.
$\Rightarrow {{\Delta H}}\,{\text{ = }}\,{{\Delta U}}\,\,{{ + p\Delta V}}$
$\Rightarrow {{\Delta H}}\,{\text{ = }}\,418\,{\text{Nm}}\,{\text{ + }}\,2.026 \times {10^5}\,{\text{N}}{{\text{m}}^{ - 2}}\, \times 3.5\, \times {10^{ - 3}}{{\text{m}}^3}$
$\Rightarrow {{\Delta H}}\,{\text{ = }}\,418\,{\text{Nm}}\,{\text{ + }}\,709.1\,{\text{Nm}}$
$\Rightarrow {{\Delta H}}\,{\text{ = }}\,1127.1\,{\text{Nm}}$
So, If the change in the internal energy is $418$J. Calculate the change in the enthalpy is $1127.1\,{\text{Nm}}$.

Therefore, the change in enthalpy is $1127.1\,{\text{Nm}}$.

Note: The total number of carbon atoms and hydrogen atoms remains the same on both sides of the pyrolysis reaction. The carbon number in each product of pyrolysis will be less than the carbon number of the reactant. Pyrolysis gives lower hydrocarbons from higher hydrocarbons. Wurtz reaction gives higher hydrocarbons from lower hydrocarbons. If a single type of alkane is used as a reactant then, the product formed by the Wurtz reaction will have double carbon number than the carbon present in the reactant. Wurtz reaction is also known as Wurtz fittig reaction.