A piston has an external pressure of 15.0 atmatm. How much work has been done in joules if the cylinder goes from a volume of 0.120 liters to 0.450 liters

Work done is  - 501.56 J

Explanation:

Given that;

External pressure P = 15.0 atm

Volume V1 = 0.120 liters

Volume V2 = 0.450 liters

Work done = ?

we know that; Work = -Pdv

where P is pressure and dv is change in volume

so we substitute our values into the equation

Work = -15.0 × ( 0.450 - 0.120)

= -15 × 0.33

= - 4.95 atm/L

we know that;

1 atm.L = 101.325 J

so

- 4.95 atm/L = 101.325 J × -4.95 atm/L ÷

= - 501.56 J

Therefore Work done is  - 501.56 J

i believe the answer is a

hey , the answer is c