Answer: 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
