Chapter 6.6, Problem 25E

To determine

To Show: The expression for the work done by the gas when it is expanding by means of a piston inside a cylinder is ${{\displaystyle \int }}_{V_1}^{V_2}PdV$ .

Explanation of Solution

Given Information: The pressure of the gas is $P=P\left(V\right)$ , a function of its volume.

Concept Used:

• The work done by a force $F$ in moving a particle from $x_1$ to $x_2$ is given by

  $W={{\displaystyle \int }}_{x_1}^{x_2}F\left(x\right)dx$

• The force $F$ on a surface is given by

  $F=P\cdot A$

where $P$ is the pressure and $A$ is the area.

If $P\left(V\right)$ is the pressure of the gas at volume $V$ , then the force on the piston having area $A$ is

  $F\left(x\right)=P\left(V\right)\cdot A$

If the piston moves from a position $x$ to $x+dx$ , then the work done by the gas is

  $dW=P\left(V\right)\cdot A\cdot dx$

But $A\cdot dx$ is in the form $area\times length$ , which in turn is. It means that if the piston moves from $x$ to $dx$ , then

  $A\left(x+dx\right)-Ax=\left(V+dV\right)=V$

  $\Rightarrow Adx=dV$

So, replace $Adx$ in the expression for $dW$ to write

  $dW=P\left(V\right)dV$

Now integrate the above equation from the desired limits $V_1$ to $V_2$ so that

  $W_{V_1\to V_2}={{\displaystyle \int }}_{V_1}^{V_2}P\left(V\right)dV$