The work done during expansion of a gas from a volume of $4 \mathrm{dm}^3$ to $6 \mathrm{dm}^3$ against a…

The work done during expansion of a gas from a volume of $4 \mathrm{dm}^3$ to $6 \mathrm{dm}^3$ against a constant external pressure of $3 \mathrm{~atm}$ is $(1 \mathrm{~L} \mathrm{~atm}=101.32 \mathrm{~J})$ :
  1. $-6 \mathrm{~J}$
  2. $-608 \mathrm{~J}$
  3. $+304 \mathrm{~J}$
  4. $-304 \mathrm{~J}$

Solution

Work $=-P_{e x t} \times$ volume change $\begin{aligned} & =3 \times 101.32 \times(6-4) \\ & =6 \times 101.32 \\ & =-607.92 \mathrm{~J}=-608 \mathrm{~J} \end{aligned}$

Asked in: NEET 2004

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