Answer pls thermo

Solution
(a) Irreversible Work done, W_irrev = −P_ext(V₂−V₁)
P_ext = 200 torr = 0.263 atm
V₁ = 12.7 L
V₂ = 12.7 + 3.3 = 16 L
Putting the given values, we get
W_irrev = −0.263(16−12.7)
W_irrev = −0.8679 L atm = −0.8679×101.325 = −87.94 J

(b) Reversible work done, $W_{rev}=-2.303nRT\log \frac{V_2}{V_1}$
T = 310 K
$n=\frac{given mass}{molar mass}=\frac{4.5}{16}$
Putting the given values, we get
$$\begin{gathered} W_{rev}=-2.303\times \frac{4.5}{16}\times 0.0821\times 310 \log \frac{16}{12.7} \\ {W}_{rev}=-1.65 \mathrm{L} \mathrm{atm}=-167.5 \mathrm{J} \end{gathered}$$