Question

An ideal gas occupying a volume of $2{\text{ d}}{{\text{m}}^3}$ and pressure of $5{\text{ bar}}$ undergoes isothermal and irreversible expansion against an external pressure of ${\text{1 bar}}$. Find the work done in the process.

Answer 1

An ideal having volume $2{\text{ d}}{{\text{m}}^3}$ and pressure of $5{\text{ bar}}$ first undergoes through isothermal expansion then undergoes with irreversible expansion. During this work is done by the system against the external pressure. We have to calculate the amount of work done by the system in doing so.
Formula Used:
${\text{PV = }}$Constant

Complete answer:
We are given ideal gas which expands isothermally and irreversibly. For isothermal expansion we know that the product of pressure and volume always remains constant. This can be shown by the given relation as,
${\text{PV = }}$Constant
Let us consider the initial pressure of gas be ${{\text{P}}_i}$ and volume of gas be${{\text{V}}_i}$. According to the question it expands isothermally to pressure ${{\text{P}}_f}$ and volume${{\text{V}}_f}$. By using the relation of isothermal expansion we can write as,
${{\text{P}}_i}{{\text{V}}_i}{\text{ = }}{{\text{P}}_f}{{\text{V}}_f}$
According to given question, ${{\text{P}}_i}{\text{ = 5 bar}}$, ${{\text{V}}_i}{\text{ = 2 d}}{{\text{m}}^3}$ and ${{\text{P}}_f}{\text{ = 1 bar}}$. On substituting the given values we get the ${{\text{V}}_f}$ as,
${{\text{V}}_f}{\text{ = }}\dfrac{{{{\text{P}}_i}{{\text{V}}_i}}}{{{{\text{P}}_f}}}$
${{\text{V}}_f}{\text{ = }}\dfrac{{{\text{5 }} \times {\text{ 2}}}}{1}{\text{ d}}{{\text{m}}^3}{\text{ = 10 d}}{{\text{m}}^3}$
Thus we get the final volume. Now for calculating the work done we can use the following relation as,
Work done against external pressure ${\text{ = - }}{{\text{P}}_{ext.}}\left( {{{\text{V}}_f}{\text{ - }}{{\text{V}}_i}} \right)$
The external pressure is equal to one bar. On substituting the values we get the work done as,
Work done ${\text{ = - }}{{\text{P}}_{ext.}}\left( {{{\text{V}}_f}{\text{ - }}{{\text{V}}_i}} \right)$
Work done ${\text{ = - 1}}\left( {{\text{10 - 2}}} \right)$
Work done ${\text{ = - 800 J}}$
Thus on converting the units we get the work done. Hence the work done is ${\text{ - 800 J}}$. Therefore we can say that the final volume of the gas is ${\text{ 10 d}}{{\text{m}}^3}$. The total work done in doing so against external pressure of one bar is ${\text{ - 800 J}}$.

Note:
By converting the given units into standard units we get the work done in joules. The product of pressure and volume always remains constant for isothermal expansion, since temperature is the same throughout the expansion. For converting $atm$ to joule we multiply it with $101.34$.