air having a pressure of 40 psig and a volume of 8 cu ft expands isotermally to a pressure of 10 psig find the external work performed during the expasion
Note: psig means gauge pressure, equal to pressure in excess of atmospheric pressure.
To find the external work performed during the expansion, you can use the formula:
W = P1 * V1 * ln(V2/V1)
Where:
W = External work done
P1 = Initial pressure
V1 = Initial volume
V2 = Final volume
Given:
P1 = 40 psig
V1 = 8 cu ft
P2 = 10 psig (as the pressure reduces during the expansion)
First, you need to convert the pressures from psig (pounds per square inch gauge) to absolute pressure in psi (pounds per square inch). Absolute pressure is the pressure above zero pressure.
To convert from psig to psi, you add the atmospheric pressure at sea level, which is approximately 14.7 psi.
P1 (absolute) = P1 (psig) + atmospheric pressure
P1 (absolute) = 40 psig + 14.7 psi
P1 (absolute) = 54.7 psi
P2 (absolute) = P2 (psig) + atmospheric pressure
P2 (absolute) = 10 psig + 14.7 psi
P2 (absolute) = 24.7 psi
Now we can calculate the external work:
W = P1 * V1 * ln(V2/V1)
W = 54.7 psi * 8 cu ft * ln(V2/8 cu ft)
Since the expansion is isothermal (constant temperature), the ideal gas equation can be used to find V2. The ideal gas equation is:
PV = nRT
Where:
P = Pressure (absolute)
V = Volume
n = Number of moles (which remains constant for an isothermal process)
R = Ideal gas constant
T = Temperature (remains constant for an isothermal process)
Since the number of moles and temperature are constant, we can rewrite the equation as:
P1 * V1 = P2 * V2
Substituting the values:
54.7 psi * 8 cu ft = 24.7 psi * V2
438.4 = 24.7 * V2
V2 = 438.4 / 24.7
V2 = 17.74 cu ft
Now we can calculate the external work:
W = 54.7 psi * 8 cu ft * ln(17.74 cu ft / 8 cu ft)
W = 54.7 * 8 * ln(2.2175)
W ≈ 336.08 ft-lbf (to two decimal places)
Therefore, the external work performed during the expansion is approximately 336.08 foot-pound force (ft-lbf).