An ideal gas is expanding such that $$PT^3$$ = constant. The coefficient of volume expansion of the gas is:

We begin by recalling the ideal-gas equation

$$PV = nRT,$$

where $$P$$ is the pressure, $$V$$ is the volume, $$T$$ is the absolute temperature, $$n$$ is the number of moles and $$R$$ is the universal gas constant.

The problem further tells us that during the expansion the gas obeys the empirical relation

$$P\,T^{3} = \text{constant}.$$

For convenience, let us denote this constant by $$k$$, so that

$$P\,T^{3}=k \quad\Longrightarrow\quad P=\dfrac{k}{T^{3}}.$$

Now we insert this expression for $$P$$ into the ideal-gas equation. Substituting $$P=\dfrac{k}{T^{3}}$$ in $$PV=nRT$$, we have

$$\dfrac{k}{T^{3}}\;V = nRT.$$

Multiplying both sides by $$T^{3}$$ to clear the denominator,

$$kV = nR\,T^{4}.$$

Next we isolate $$V$$: dividing both sides by $$k$$,

$$V = \dfrac{nR}{k}\,T^{4}.$$

This shows explicitly that the volume varies as the fourth power of the temperature along the given path. To find the coefficient of volume expansion we recall its definition.

The (volumetric) coefficient of expansion $$\beta$$ is defined as

$$\beta = \dfrac{1}{V}\,\dfrac{dV}{dT}.$$

Using the expression $$V=\dfrac{nR}{k}\,T^{4}$$, we differentiate with respect to $$T$$:

$$\dfrac{dV}{dT} = \dfrac{nR}{k}\,\dfrac{d}{dT}\bigl(T^{4}\bigr) = \dfrac{nR}{k}\,(4T^{3}) = 4\,\dfrac{nR}{k}\,T^{3}.$$

But from the earlier result $$V=\dfrac{nR}{k}\,T^{4},$$ we observe that

$$\dfrac{nR}{k}\,T^{3}=\dfrac{V}{T}.$$

Substituting this in the derivative, we obtain

$$\dfrac{dV}{dT}=4\,\dfrac{V}{T}.$$

Now we plug this result into the definition of $$\beta$$:

$$\beta = \dfrac{1}{V}\,\dfrac{dV}{dT} = \dfrac{1}{V}\,\Bigl(4\,\dfrac{V}{T}\Bigr)=\dfrac{4}{T}.$$

Thus the coefficient of volume expansion for the gas undergoing the process $$PT^{3}=\text{constant}$$ is

$$\boxed{\dfrac{4}{T}}.$$

Hence, the correct answer is Option D.