An ideal gas maintained at 25°C expands against a constant external pressure of 1.0 atm from an initial volume of 1.0L to a final volume of 10.0L. (a). what's the work done by the system (b). Assuming that at the end of the process, the system is at equilibrium with it's surrounding, what was the initial gas pressure?
To find the work done by the system, we can use the formula:
W = -Pext * ∆V
Where:
W = work done by the system
Pext = external pressure
∆V = change in volume
Given:
Pext = 1.0 atm
∆V = 10.0 L - 1.0 L = 9.0 L
(a). W = -1.0 atm * 9.0 L = -9.0 L.atm
Therefore, the work done by the system is -9.0 L.atm.
(b). Since the process is at equilibrium, the work done by the system is equal to the work done on the system. Therefore, the work done by the system is equal to the area under the curve in a PV diagram. This area is the area of a trapezoid, with the initial pressure as the smaller base and the final pressure as the larger base. The height of the trapezoid is the change in volume.
Using the formula for the area of a trapezoid:
Area = 1/2 * (Pinitial + Pfinal) * ∆V
Given:
Area = W = -9.0 L.atm
Pfinal = 1.0 atm
∆V = 9.0 L
-9.0 L.atm = 1/2 * (Pinitial + 1.0 atm) * 9.0 L
-9.0 = 4.5Pinitial + 4.5
-13.5 = 4.5Pinitial
Pinitial = -3.0 atm
Therefore, the initial gas pressure was -3.0 atm.