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# 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?

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- 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

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