Depends on the process
This is a tricky one to answer, because it depends on what thermodynamic processes you are talking about.
For example, if his power allows him to compress the gas into a tiny ball, and then release it to go bang, there will not be any work (or bang) done. This is called adiabatic free expansion of a gas.
Lets make it isothermal
In order to do work and make a bang the way you are wanting, the process must be made isothermal. This is a reversible, constant temperature change. In order to keep temperature constant, your protagonist's power must include absorbing energy from the compressed gas to keep its temperature constant, then releasing that energy with the expansion.
In that case, the work done is
$$W_{1\rightarrow 2} = -nRT\log{\frac{V_2}{V_1}}.$$
A 2 foot radius sphere is 0.6096 meters radius in science units, and 0.949 m$^3$ in volume. A mol of ideal gas at standard pressure and temperature occupies 22.4 liters, so there are 42.3 mols of air in the compressed air. Standard temperature is 273 K. The ideal gas constant is 8.314 J / K / mol.
We can now do the calculation if we make some assumptions about how tightly the gas can be compressed. As you can see, the final compressed volume is a variable, and we can set it as we like. Either way you will get a lot of energy out thi way.
If we assume the gas is compressed into a 1 cm radius dot, we get a final volume of 4.19e-3 m$^3$, and energy of
$$-nRT\log\frac{V_2}{V_1} = 42.3 \text{mol} \cdot 8.314 \frac{\text{J}}{\text{K}\cdot\text{mol}}\cdot 273 \text{K} \log\frac{0.949 \text{ m}^3}{4.19\times10^{-6} \text{ m}^3} = 1.2 \text{MJ}$$
1.2 MJ is almost 100 times the energy of a .50 cal round. That's a lot of juice! Your only problem now is keeping your protagonist from being killed by his own blast.
