1. Given data

• Number of moles: $n = 5$
• Initial temperature: $T_1 = 20^\circ C = 293\,K$
• Adiabatic compression to $V_2 = \frac{V_1}{10}$
• Air is a diatomic ideal gas with rigid molecules

For a diatomic rigid gas, $C_V = \frac{5}{2}R, \qquad \gamma = \frac{C_P}{C_V} = \frac{7}{5}$

We need the change in internal energy: $\Delta U = n C_V (T_2 - T_1)$

2. Use adiabatic relation to find $T_2$

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

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

Hence, $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$

Since $\frac{V_1}{V_2} = 10, \qquad \gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5}$

Therefore, $\frac{T_2}{T_1} = 10^{2/5}$

Now, $10^{2/5} \approx 2.512$

Thus, $T_2 = 293 \times 2.512 \approx 736\,K$

So, $\Delta T = T_2 - T_1 = 736 - 293 = 443\,K$

3. Calculate change in internal energy

$\Delta U = n C_V \Delta T$

With $C_V = \frac{5}{2}R$

So, $\Delta U = 5 \times \frac{5}{2}R \times 443$

Using $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$,

$\Delta U = 12.5 \times 8.314 \times 443$

$\Delta U \approx 46031\,\text{J}$

$\Delta U \approx 46.0\,\text{kJ}$

4. Nearest integer

$X = 46$

5. Comparison with stored answer

Stored correct answer = $46$

This matches our derived value.