Given data

• Number of moles: $n=1$
• Gas is monatomic ideal gas
• Initial temperature: $T_1=100\,\text{K}$
• Final volume: $V_2=8V_1$
• Process: adiabatic
• Gas constant: $R=8\,\text{J mol}^{-1}\text{K}^{-1}$

Use adiabatic relation

For an ideal gas in an adiabatic process, $T V^{\gamma-1}=\text{constant}$

For a monatomic gas, $\gamma=\frac{C_P}{C_V}=\frac{5}{3}$ so $\gamma-1=\frac{2}{3}$

Therefore, $T_1 V_1^{2/3}=T_2 V_2^{2/3}$

Since $V_2=8V_1$, $T_2=T_1\left(\frac{V_1}{V_2}\right)^{2/3}$ $T_2=100\left(\frac{1}{8}\right)^{2/3}$

Now, $8^{2/3}=(2^3)^{2/3}=2^2=4$ so $\left(\frac{1}{8}\right)^{2/3}=\frac{1}{4}$

Hence, $T_2=100\times \frac{1}{4}=25\,\text{K}$

Find decrease in internal energy

For a monatomic ideal gas, $U=\frac{3}{2}nRT$

So decrease in internal energy is $\Delta U_{\text{decrease}}=\frac{3}{2}nR(T_1-T_2)$

Substitute values: $\Delta U_{\text{decrease}}=\frac{3}{2}\times 1\times 8\times (100-25)$ $=\frac{3}{2}\times 8\times 75$ $=12\times 75$ $=900\,\text{J}$

Final answer $\boxed{900}$

Comparison with stored answer

Stored correct answer = $900$

This matches the derived answer.