Question

The gas mixture inside one of the cylinders of an automobile engine expands against a constant external pressure of 0.98 atm, from an initial volume of 150 mL (at the end of the compression stroke) to a final volume of 800 mL. Calculatethe work done on the gas mixture during this process, and express it in joules.

Short answer

The work done during this process is -65 J.

Step-by-step solution

Work done

Work can possess a positive or negative sign based on several conventions. It is obtained by multiplying pressure with the volume, and the expression is:
$\mathbf{W}\mathbf{=}\mathbf{-}{\mathbf{P}}_{\mathbf{ext}}\mathbf{∆}\mathbf{V}$

Calculating the work done on the gas mixture during the process

The work done in a change of volume at a constant pressure can be given as:

${\text{W = -P}}_{\text{ext}}\text{∆V}$

${\text{P}}_{\text{ext}}$is the external pressure.

The change in volume, $\text{∆V}$can be given as:

$$\begin{gathered} {\text{∆V =V}}_{\text{2}}{\text{-V}}_{\text{1}} \\ \text{= 800 mL - 150 mL} \\ \text{= 650 mL} \\ \text{= 0.650 L} \end{gathered}$$

The external pressure is 0.98 atm.

The work done can be found as:

$$\begin{gathered} {\text{W = -P}}_{\text{ext}}\text{∆V} \\ \text{= - 0.98atm×0.650L} \\ \text{= - 0.64Latm} \end{gathered}$$

1 L atm is equal to 101.325 J.

0.64 L atm can be converted to joules as:

$$\begin{gathered} \text{W = - 0.64 L atm×}\frac{\text{101.325 J}}{\text{1 L atm}} \\ \text{= - 65 J} \end{gathered}$$

Hence, the work done during this process is -65 J.