Chapter 6, Problem 6.23P

Interpretation Introduction

Interpretation:

The work done of a gas that compress from $0.88\text{ L}$ to $0.63\text{ L}$ by an external pressure of $2660\text{ mmHg}$ is to be calculated.

Concept introduction:

Work $\left(w\right)$ is the energy needed to move an object in the opposite direction of the force applied. The work done by the system is negative and the work done on the system is positive. The formula to calculate the work done against an external pressure is:

$w=-P\Delta V$

Here,

$w$ is the work done by the system or on the system.

$P$ is the external pressure.

$\Delta V$ is the change in the volume.

The conversion factor to convert $\text{mmHg}$$\text{torr}$ into $\text{atm}$ is:

$\text{1}\text{atm}=760\text{mmHg}$

The conversion factor to convert $\text{atm}\cdot \text{L}$ into $\text{J}$ is:

$\text{1}\text{atm}\cdot \text{L}=101.3\text{J}$

The expression to convert $\text{kJ}$ into $\text{J}$ is:

$\text{1}\text{kJ}=1000\text{J}$