A diatomic gas $$(\gamma = 1.4)$$ does 200 J of work when it is expanded isobarically. The heat given to the gas in the process is:

For an isobaric process (constant pressure), the work done by the gas is given by:

$$W = P \Delta V$$

Using the ideal gas law, $$PV = nRT$$, we can express the work as:

$$W = P \Delta V = nR \Delta T$$

Given that the work done by the gas is 200 J:

$$nR \Delta T = 200 \text{ J} \quad \text{(Equation 1)}$$

The heat supplied to the gas in an isobaric process is:

$$Q = n C_p \Delta T$$

where $$C_p$$ is the molar specific heat at constant pressure.

For a diatomic gas with $$\gamma = 1.4$$, the ratio of specific heats is $$\gamma = \frac{C_p}{C_v} = \frac{7}{5}$$. The molar specific heat at constant volume is $$C_v = \frac{5}{2}R$$ (since diatomic gases have 5 degrees of freedom). Therefore:

$$C_p = \gamma C_v = \frac{7}{5} \times \frac{5}{2}R = \frac{7}{2}R$$

Substituting into the expression for Q:

$$Q = n \left( \frac{7}{2}R \right) \Delta T = \frac{7}{2} (nR \Delta T)$$

Using Equation 1, where $$nR \Delta T = 200 \text{ J}$$:

$$Q = \frac{7}{2} \times 200 = 7 \times 100 = 700 \text{ J}$$

Thus, the heat given to the gas is 700 J.

Verification using the first law of thermodynamics:

The change in internal energy is:

$$\Delta U = n C_v \Delta T = n \left( \frac{5}{2}R \right) \Delta T = \frac{5}{2} (nR \Delta T) = \frac{5}{2} \times 200 = 500 \text{ J}$$

From the first law, $$\Delta U = Q - W$$:

$$500 = Q - 200 \implies Q = 700 \text{ J}$$

This confirms the result.

The correct option is D. 700 J.