During the power stroke in a four-stroke automobile engine, the piston is forced down as the mixture of combustion products and air undergoes an adiabatic expansion. Assume (1) the engine is running at 3400 cycles/min; (2) the gauge pressure immediately before the expansion is 20.0 atm; (3) the volumes of the mixture immediately before and after the expansion are 50.0 cm3 and 400 cm3, respectively; (4) the time interval for the expansion is one-fourth that of the total cycle; and (5) the mixture behaves like an ideal gas with specific heat ratio 1.40. Find the average power generated during the power stroke
W = integral of PdV
PV^(1.4) = constant
The Attempt at a Solution
So first I found out that it takes approximately .005s for the power stroke.
I found the constant to be approximately (using pascals and m^3) 1.9289. I then said P = 1.9289/V^1.4.
I put that into the integral so integral from .00005 to .0004 of 1.9289/V^1.4 dV. I get about 143 J. After doing the conversions to kw, I am wrong. I found (i know the answer) that I should be getting 150 J to turn into kW, so I don't understand what is wrong with my integral. The other solution that gives 150 J uses dE = nCv dT instead of PdV, but isnt' it the same thing as PdV? Thanks in advance!