36kg of some material undergoes a process in a piston-cylinder apparatus starting with an initial volume of .057m^3 and pressure of 680kPa. The final volume is twice the initial. Compute the work done by the substance for the following processes: (a) Pressure times volume is constant
(Ans. 26.9kJ)
Part of this question appears to be missing.
What do you think is missing? Given the initial volume and initial pressure plus the final volume and relationship between pressure and volume, I feel like you should be able to integrate pressure with respect to volume. I just cannot get the correct answer.
When I read the question it says "the following processes" but then (a) is a statement and not a process. We know PV = k
To compute the work done by the substance in the given process, we need to use the formula:
Work done (W) = Pressure (P) * Change in Volume (ΔV)
In this case, the pressure (P) and the volume (V) are related by the condition that pressure times volume is constant, meaning:
P1 * V1 = P2 * V2
Where:
P1 = Initial pressure
V1 = Initial volume
P2 = Final pressure
V2 = Final volume
Given:
Initial volume, V1 = 0.057 m^3
Initial pressure, P1 = 680 kPa
Final volume, V2 = 2 * V1 = 2 * 0.057 m^3
We can solve for the final pressure, P2, using the pressure-volume relationship:
P1 * V1 = P2 * V2
680 kPa * 0.057 m^3 = P2 * (2 * 0.057 m^3)
P2 = (680 kPa * 0.057 m^3) / (2 * 0.057 m^3)
P2 = 680 kPa / 2
P2 = 340 kPa
Now, we have the initial pressure (P1 = 680 kPa), final pressure (P2 = 340 kPa), and the change in volume (ΔV = V2 - V1 = 2 * 0.057 m^3 - 0.057 m^3 = 0.057 m^3).
Substituting these values into the formula for work done, we get:
W = P * ΔV
W = 340 kPa * 0.057 m^3
W = 19.38 kJ
Therefore, the work done by the substance for this process is 19.38 kJ.