The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm against constant external
pressure of 3 atm is :-
(2) +304 J
(3) -304 J
(
4-6 J
(1) -608 J
Answers
Answered by
10
Answer:
+304J
−304J
−6J
−608J
Answer :
D
Solution :
W=−pΔV,W=−3×(6−4)
W=−6×101.32(∴1 L atm=101.32 J)
W=-608 J
Explanation:
Similar questions