1. Identify the process

The gas is heated at constant pressure because the initial pressure is given as $1.00\,\text{atm}$ and the standard interpretation here is heating under constant atmospheric pressure.

Given:

• Heat supplied: $q = 500\,\text{J}$
• Moles of Ar: $n = 0.5$ mol
• Initial temperature: $T_1 = 298\,\text{K}$
• Argon is a monoatomic ideal gas

2. Use molar heat capacity at constant pressure

For a monoatomic ideal gas, $C_{V,m} = \frac{3}{2}R, \qquad C_{P,m} = C_{V,m}+R = \frac{5}{2}R$

So, $C_{P,m} = \frac{5}{2}(8.3) = 20.75\,\text{J mol}^{-1}\text{K}^{-1}$

For $0.5$ mol, $nC_P = 0.5 \times 20.75 = 10.375\,\text{J K}^{-1}$

3. Find temperature rise

At constant pressure, $q = nC_P\Delta T$

Thus, $\Delta T = \frac{q}{nC_P} = \frac{500}{10.375} \approx 48.2\,\text{K}$

Therefore, $T_2 = T_1 + \Delta T = 298 + 48.2 \approx 346.2\,\text{K}$

Closest option: $T_2 \approx 348\,\text{K}$

4. Find change in internal energy

For an ideal gas, $\Delta U = nC_V\Delta T$

Now, $C_{V,m} = \frac{3}{2}R = \frac{3}{2}(8.3) = 12.45\,\text{J mol}^{-1}\text{K}^{-1}$

So for $0.5$ mol, $nC_V = 0.5 \times 12.45 = 6.225\,\text{J K}^{-1}$

Hence, $\Delta U = 6.225 \times 48.2 \approx 300\,\text{J}$

Alternatively, using $\Delta U = q - P\Delta V$ with constant pressure heating of an ideal gas, $P\Delta V = nR\Delta T = 0.5 \times 8.3 \times 48.2 \approx 200\,\text{J}$ so $\Delta U = 500 - 200 = 300\,\text{J}$

5. Match with options

The final temperature and internal energy change are:

• Final temperature $\approx 348\,\text{K}$
• $\Delta U \approx 300\,\text{J}$

So the correct option is B.