Question

A piston has an external pressure of 14.0atm . How much work has been done if the cylinder goes from a volume of 0.180 liters to 0.530 liters. Express your answer in joules

Answer 1

Concepts and reason

The expansion or compression of gas takes place in a reversible process or an irreversible process based on the conditions used. An irreversible process may occur when internal pressure $\left( {{{\rm{P}}_{{\rm{int}}}}} \right)$ is greater than external pressure $\left( {{{\rm{P}}_{{\rm{ext}}}}} \right)$ because, the system does not attain equilibrium at every stage of the process.

Fundamentals

Pressure-volume work:

• Irreversible process: In this process, ${{\rm{P}}_{{\rm{in}}}}{\rm{ > }}{{\rm{P}}_{{\rm{ext}}}}$ and the expansion of gas takes place rapidly without maintaining equilibrium with the surroundings. Therefore, the pressure-volume $\left( {{\rm{PV}}} \right)$ work done against constant external pressure is given below:

${\rm{W = }} - {{\rm{P}}_{{\rm{ext}}}}\Delta {\rm{V}}$

$\begin{array}{l}\\{\rm{Where, 'W' is work done by gas,}}\\\\ & {\rm{'}}{{\rm{P}}_{{\rm{ext}}}}'{\rm{ external pressure,}}\\\\ & {\rm{'}}\Delta {\rm{V' is change in the volume}}{\rm{.}}\\\end{array}$

Given that,

$\begin{array}{c}\\{\rm{External pressure}}\left( {{{\rm{P}}_{{\rm{ext}}}}} \right){\rm{ = 14}}{\rm{.0atm}}\\\\{\rm{Initial volume}}\left( {{{\rm{V}}_{\rm{1}}}} \right) = 0.180{\rm{Liters}}\\\\{\rm{Final volume }}\left( {{{\rm{V}}_2}} \right) = 0.530{\rm{Liters}}\\\end{array}$

Work done ‘W’ is calculated below:

$\begin{array}{c}\\{\rm{W = }} - {{\rm{P}}_{{\rm{ext}}}}\Delta {\rm{V}}\\\\ = - \left( {{\rm{14}}{\rm{.0atm}}} \right)\left( {0.530 - 0.180} \right){\rm{Liters}}\\\\{\rm{ = }} - 4.9{\rm{Lit}} - {\rm{atm}}\\\end{array}$

Convert the units for work done from Lit-atm to joules.

$\begin{array}{l}\\{\bf{1}}{\rm{Lit - atm = 101}}{\rm{.3J}}\\\\{\rm{W}}\left( {{\rm{joules}}} \right)\\\\ - 4.9{\rm{Lit - atm = }} - {\rm{496}}{\rm{.37Joules}}\\\end{array}$

Ans:

Work done if the cylinder goes from a volume of 0.180 liters to 0.530 liters is $- 496.37{\rm{J}}{\rm{.}}$