Step 1: Since the process is isothermal and we are looking for the minimum work, we use the formula for reversible isothermal compression:
\[
W_{{rev}} = nRT \ln \left( \frac{P_2}{P_1} \right)
\]
Given:
\[
n = 1 { kmol}, R = 8.314 { kJ/kmol·K}, T = 400 { K}
\]
\[
P_1 = 200 { kPa}, P_2 = 1000 { kPa}
\]
\[
W = 1 \cdot 8.314 \cdot 400 \cdot \ln\left( \frac{1000}{200} \right)
= 3325.6 \cdot \ln(5)
\]
\[
\ln(5) \approx 1.6094 \Rightarrow
W \approx 3325.6 \cdot 1.6094 \approx 5365.28 \, {kJ/kmol}
\]