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PV Work

The work done...

Question

The work done during the expression of a gas from $4 d m^{3} t o 6 d m^{3}$ against a constant external pressure of 3 atm is : (1 L atm = 101.32J).

- 6 J

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- 608 J

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+ 304 J

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- 304 J

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Solution

The correct option is B - 608 J
Work in expansion = - $P Δ V$
W = - 3 X (6 - 4) = -6 L atm
1 L atm = 101.32 J
W = -6 X 101.32 = -607.92J or -608J

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