For an ideal gas, the equation of state is given by

$$ pV = nRT. $$

If during a process the pressure increases linearly with temperature (i.e., $$p \propto T$$), then from the ideal gas law (with a fixed amount of gas, $$n$$, and with the gas constant, $$R$$), it follows that

$$ \frac{p}{T} = \frac{nR}{V}. $$

Since $$nR$$ is constant, a linear relationship between $$p$$ and $$T$$ implies that $$\frac{p}{T}$$ remains constant throughout the process. This can only be true if the volume $$V$$ is constant. Hence, the process is isochoric.

For an isochoric (constant volume) process:

The work done by the gas is given by

$$ W = \int p\, dV, $$

and since $$dV = 0$$, we have $$W = 0.$$ (Statement A is true.)

The internal energy change for an ideal gas depends only on temperature:

$$ \Delta U = nC_V\Delta T. $$

If the temperature increases, then $$\Delta U > 0.$$ (Statement D is true.)

For an isochoric process, no expansion work is done, so any heat added goes entirely into increasing the internal energy. This means the heat added is not different from the change in internal energy (in fact, $$Q = \Delta U$$). (Statement B is false.)

Since the process is isochoric, the volume does not change (and certainly is not increased). (Statement C is false, and Statement E stating it is isochoric is true.)

Thus, the correct true statements are A, D, and E.