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Question

Physics Question on work done thermodynamics

The work done during the expansion of a gas from a volume of 4dm34\, dm^3 to 6dm36 \,dm^3 against a constant external pressure of 33 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(V2V1)=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})
=3×(64)=6L.atm=-3\times (6-4)=-6\,L.\,\,atm =6×101.32J=-6\times 101.32\,J
(1L-atm=101.32J)(\therefore \,1\,\,L\text{-atm=101}\text{.32}\,\text{J})
=607.92608J=-607.92\approx - 608\,J