Answer:
[tex]\large \boxed{\text{220 J}}[/tex]
Explanation:
1. Convert the pressure to atmospheres
[tex]p = \text{825 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \text{1.086 atm}[/tex]
2. Calculate the work done
[tex]w = -p\Delta V = -p(V_{f} - V_{i}) = -\text{(1.086 atm)}(\text{2.0 L - 4.0 L})\\\\ = -\text{1.086 atm} \times \text{(-2.0 L)} = \text{2.17 L$\cdot$atm}[/tex]
3. Convert litre atmospheres to joules
[tex]w = \text{2.17 L$\cdot$atm} \times \dfrac{\text{101.3 J}}{\text{1 L$\cdot$atm}} = \textbf{220 J}\\\\\text{The work done is $\large \boxed{\textbf{220 J}}$}[/tex]
Note that the volume is decreasing, so work is being done on the system.