Question
Physics Question on work done thermodynamics
The work done during the expansion of a gas from a volume of 4dm3 to 6dm3 against a constant external pressure of 3 atm is :-
A
-608 J
B
#ERROR!
C
-304 J
D
-6 J
Answer
-608 J
Explanation
Solution
The correct option is(A): -608 J.
Work done (W) =−Pext(V2−V1)
=−3×(6−4)=−6L.atm =−6×101.32J
(∴1L-atm=101.32J)
=−607.92≈−608J