$$ C_P = \frac{5}{2} R \qquad \text{and} \qquad C_V = \frac{3}{2} R $$

For a constant pressure process, the heat added is related to the temperature change by

$$ Q = n C_P \Delta T. $$

Given that

$$Q = 500 \, \text{J},$$

$$n = 0.5 \, \text{mol},$$

$$R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1},$$

and the initial temperature $$T_i = 298 \, \text{K},$$

we first calculate $$C_P$$:

$$ C_P = \frac{5}{2} \times 8.3 = 20.75 \, \text{J K}^{-1} \text{mol}^{-1}. $$

Next, the temperature change is

$$ \Delta T = \frac{Q}{n C_P} = \frac{500 \, \text{J}}{0.5 \, \text{mol} \times 20.75 \, \text{J K}^{-1} \text{mol}^{-1}} \approx 48.2 \, \text{K}. $$

So, the final temperature is

$$ T_f = T_i + \Delta T \approx 298 \, \text{K} + 48.2 \, \text{K} \approx 346.2 \, \text{K}. $$

Rounded to the nearest units provided in the options, this is approximately $$348 \, \text{K}.$$

The change in internal energy is determined by

$$ \Delta U = n C_V \Delta T, $$

where

$$ C_V = \frac{3}{2} R = \frac{3}{2} \times 8.3 = 12.45 \, \text{J K}^{-1} \text{mol}^{-1}. $$

Then,

$$ \Delta U = 0.5 \, \text{mol} \times 12.45 \, \text{J K}^{-1} \text{mol}^{-1} \times 48.2 \, \text{K} \approx 300 \, \text{J}. $$

Thus, the final answers are:

Final temperature: approximately $$348 \, \text{K}$$

Change in internal energy: approximately $$300 \, \text{J}$$

These results correspond to Option B.