The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm is (1 L atm = 101.32 J)
- -6 J
- -608 J
- +304 J
- -304 J
Answer
(2) -608 J
Solution
For expansion against a constant external pressure (irreversible):
w = -pex V
V = 6 - 4 = 2 dm3 = 2 L
⇒ w = -3 2 = -6 L atm
= -6 101.32 J
= -607.9 -608 J
Answer: (2)