Given data

• Number of moles: $n = 1$
• Gas: ideal monoatomic gas
• Temperature increase: $\Delta T = 50^\circ C = 50\,\text{K}$
• Process: constant pressure

Heat added at constant pressure

For an ideal gas, $E_1 = Q = n C_p \Delta T$

For a monoatomic ideal gas, $C_v = \frac{3R}{2}, \qquad C_p = C_v + R = \frac{5R}{2}$

Hence, $E_1 = 1 \cdot \frac{5R}{2} \cdot 50 = 125R$

Change in internal energy

For an ideal gas, $E_2 = \Delta U = n C_v \Delta T$

So, $E_2 = 1 \cdot \frac{3R}{2} \cdot 50 = 75R$

Find the ratio

$\frac{E_1}{E_2} = \frac{125R}{75R} = \frac{5}{3}$

Given, $\frac{E_1}{E_2} = \frac{x}{9}$

Therefore, $\frac{x}{9} = \frac{5}{3}$

$x = 9 \cdot \frac{5}{3} = 15$

Final answer $\boxed{15}$

Comparison with stored answer

Stored correct answer = $15$.

My derived answer matches the stored answer.