Chemistry- pressure-volume work question

Question

A chemical process generates -615.4J of energy. If the piston expands from 155.9 ml to 4.025 L, what is the external pressure?

formula: w= -P(delta V)

notes:( I know that the system of the chemical process is doing the work in the surroundings so the work is going to be negative and the piston is expanding which is a sign of negative work, against the surrounding and external pressure)

Answer 1

Hello,

Great job identifying the correct equation.

w = P_extΔV

We know that the work generated, w, is -615.4 J. We know that ΔV is the change in volume (V₂ - V₁). We are told that the volume expands from (V₁) 155.9 ml to (V₂) 4.1025 L.

The trickiest part of this problem is the units. We are given work in Joules. A Joule is equal to kg•m²/s², so we need to get the volume and pressure to be in terms of these units. I see meters in the unit for Joules, so let's convert our volumes to m³ instead of ml or L.

V₁ = (155.9 ml)(1 L / 1000 mL)(1 m³ / 1000 L) = 1.56 x 10^-4 m³

V₂ = (4.1025 L)(1 m³ / 1000 L) = 4.1 x 10^-3 m³

Now, let's plug these values into our equation:

w = P_extΔV = P_ext(V₁ - V₂)

-615.4 kg•m²/s² = P_ext (1.56 x 10^-4 m³ - 4.1 x 10^-3 m³)

-615.4 kg•m²/s² = P_ext (-3.94 x 10^-3 m³)

P_ext = -615.4 kg•m²/s² / (-3.94 x 10^-3 m³)

Now, pay attention to the units. We have (kg•m²)/(s²•m³). The m² cancels out and we are left with m¹ on the bottom. So the units of Pressure we are left with are kg/m•s² which is equal to a Pascal.

P_ext = 156192.89 Pa or 156.19 kPa