Question

Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 10.0 atm.?
Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 10.0 atm.

Calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm exerted by the surroundings to a final volume of 15.0 L.

Now calculate the work done if this process is carried out in two steps: Step 1: First let the gas expand against a constant external pressure of 5.00 atm to a volume of 3.00 L. Step 2: Now let the gas expand to 15.0 L against a constant external pressure of 1.00

Answer 1

\[W=P\cdot\Delta V\]

Answer 2

the work done is always:

force * distance * cos(theta)

theta is the angle between the direction of force and the direction of displacement of the piston.

the force is the pressure times the surface area of the piston.

the convention is to follow the work done by the force exerted by the piston. in the case expansion, the force by the piston points towards the gas and the displacement points away from the gas. so the work done is:

(external pressure) * area * distance * cos(pi)
= - (external pressure) * (final volume - initial volume)
= - P(ext) * delta V

so when a gas increases in volume, the work done is negative and it indicates that the gas loses energy. essentially, we set the convention so that the delta E is negative when the gas spends energy to move against the external pressure.

Plugging in the numbers:

work done of the single-step process
= -(1.00 atm) * (15.00 L - 1.00L)
= - 14.0 L atm

work done of the two-step process
= -5.00 atm * (3.00 L - 1.00 L) + -1.00 atm * (15.00 - 3.00 L)
= -22.0 L atm

To convert L atm to Joule, the conversion factor is approx.

1 L * 1 atm = 10^-3 m^3 * 1.013 10^5 Pa = 101.3 Joules

so for example, -14.0 L atm (101.3 Joules) / (1 L atm) = 1.42 * 10^3 Joules