Question

When is PV (pressure-volume) work negative?

Answer 1

Step 1: Pressure-volume work

Consider an ideal gas having a definite mass of n moles in a cylinder fitted with a movable piston. Let $\mathrm{A}$be the area of the cross-section of the cylinder. Let the gas expand from ${\mathrm{V}}_1$ to ${\mathrm{V}}_2$ against constant external pressure exerted on the piston. Due to this expansion, the piston moves upward by a distance $\mathrm{d}$. Hence, the work done in the system is negative.

Step 2: Derivation

$\mathrm{Work}\mathrm{Done}=\mathrm{Opposing}\mathrm{Force}\times \mathrm{Displacement}$

$\mathrm{W}=\mathrm{F}\times \mathrm{d}...\left(1\right)$

We know that pressure is the force per area, this can be written as

$\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$

Rearranging the equation, we get

$\mathrm{F}=\mathrm{P}\times \mathrm{A}...\left(2\right)$

Substitute equation (2) in equation (1).

$\mathrm{W}=\mathrm{P}\times \mathrm{A}\times \mathrm{dx}...\left(3\right)$

But, $\text{A × dx}$ is equal to $(\mathrm{\Delta V}={\mathrm{V}}_2–{\mathrm{V}}_1)$; therefore, the equation becomes

$\mathrm{W}=-\mathrm{P}\times \mathrm{\Delta V}$

Here,

$\mathrm{dx}$ is the change in displacement.

$∆\mathrm{V}$ is the change in volume.

${\mathrm{V}}_1$ is the initial volume.

${\mathrm{V}}_2$ is the final volume.

Step 3: Explanation

1. Whether work is defined as having a positive sign or a negative sign is a matter of convention. Heat flow is defined from a system to its surroundings as negative; using that same sign convention, we define work done by a system on its surroundings as having a negative sign because it results in a transfer of energy from a system to its surroundings. This is an arbitrary convention.
2. The work done by gas expanding against an external pressure is therefore negative, corresponding to work done by a system on its surroundings. Conversely, when a gas is compressed by external pressure, $\mathrm{\Delta V}$ is negative and that makes the work negative.

$$\begin{gathered} {\mathrm{Work}}_{\left(\mathrm{gas}\right)}=\mathrm{External}\mathrm{pressure}\times \mathrm{Change}\mathrm{in}\mathrm{volume} \\ \mathrm{W}=-\mathrm{P}∆\mathrm{V} \end{gathered}$$

Here, the negative ($-$) sign indicates that the system has lost energy.

Therefore, whenever a gas expanded against that external pressure, certain energy must be transferred to the environment. As a result, the negative work reduces the total energy of the gas.