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 :-

Question

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 :- (A) -608 J (B) + 304 J (C) -304 J (D) -6 J

Tardigrade Admin · Accepted Answer

Work done (W) =-Pext(V2-V1) =-3× (6-4)=-6 L. atm =-6× 101.32 J (∴ 1 L text-atm=101 text.32 textJ) =-607.92≈ - 608 J

Q. The work done during the expansion of a gas from a volume of $4 d m^{3}$ to $6 d m^{3}$ against a constant external pressure of $3$ atm is :-

-608 J

+ 304 J

-304 J

-6 J

Work done (W) $= - P_{e x t} (V_{2} - V_{1})$
$= - 3 \times (6 - 4) = - 6 L . a t m$ $= - 6 \times 101.32 J$
$(∴ 1 L -atm=101 .32 J)$
$= - 607.92 \approx - 608 J$

Q. 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 :-