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Q. The work done during the expansion of a gas from a volume of $4\, dm^3$ to $6 \,dm^3$ against a constant external pressure of $3$ atm is :-

AIPMTAIPMT 2004Thermodynamics

Solution:

Work done (W) $=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})$
$=-3\times (6-4)=-6\,L.\,\,atm$ $=-6\times 101.32\,J$
$(\therefore \,1\,\,L\text{-atm=101}\text{.32}\,\text{J})$
$=-607.92\approx - 608\,J$