1. Given data

• Number of moles: $n=2$
• Initial temperature: $T_1=27^\circ C=300\,K$
• Initial volume: $V_1=V$
• Final volume: $V_2=2V$
• Process: adiabatic
• Gas is monatomic ideal gas, so $\gamma=\frac{C_p}{C_v}=\frac{5}{3}, \qquad C_v=\frac{3}{2}R$

2. Use adiabatic relation

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

So, $T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$

Substitute values: $300\,(V)^{\gamma-1}=T_2\,(2V)^{\gamma-1}$

Since $\gamma-1=\frac{5}{3}-1=\frac{2}{3}$ we get $300\,V^{2/3}=T_2\,(2V)^{2/3}$

$300=T_2\,2^{2/3}$

Hence, $T_2=\frac{300}{2^{2/3}}$

Now, $2^{2/3}\approx 1.587$

So, $T_2\approx \frac{300}{1.587}\approx 189\,K$

Thus, $\boxed{T_2\approx 189\,K}$

3. Change in internal energy

For an ideal gas, $\Delta U=nC_v(T_2-T_1)$

Here, $n=2, \qquad C_v=\frac{3}{2}R$

Therefore, $\Delta U=2\left(\frac{3}{2}R\right)(189-300)$

$\Delta U=3R(-111)$

Using $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$, $\Delta U=3\times 8.314\times (-111)$

$\Delta U\approx -2768\,\text{J}$

$\Delta U\approx -2.7\,\text{kJ}$

Thus, $\boxed{\Delta U\approx -2.7\,\text{kJ}}$

4. Match with options

• Final temperature $\approx 189\,K$
• Change in internal energy $\approx -2.7\,kJ$

So the correct option is: $\boxed{\text{D}}$

5. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.