What is the work being done when the pressure is 1.5 atm when the volume decreases in a container 35 L to 25 L?
so I set this up as p2= (p1 v1)/v2
1.5x35 /25 = .21
Did I do this correctly?
- AshLv 72 months agoFavorite Answer
At constant pressure,
Work, W = PΔV
W = P(Vf - Vi)
W = (1.5 atm)(25L - 35L)
W = -15 L atm
W = -15 L atm x ( 101.325 J / 1 L atm)
W = -1520 J
Since the work is negative, the work is done on the system