Individual analysis of the provided processes is as follows:
In thermodynamics, work done by a gas is zero during an isochoric process, characterized by constant volume. If the question implies that pressure increases linearly with temperature, and option E is accurate, then the gas undergoes an isochoric process. Consequently, the work performed is indeed zero under these circumstances.
For an ideal gas, the change in internal energy (\(\(\Delta U\)\)) is directly proportional to the change in temperature (\(\(\Delta T\)\)). According to the first law of thermodynamics:
\(\Delta Q = \Delta U + \Delta W\)
Given that option E suggests an isochoric process, \(\Delta W = 0\). Therefore, \(\Delta Q = \Delta U\). This option is incorrect under an isochoric process.
If pressure increases linearly with temperature and volume remains constant, the volume does not increase. Hence, this option is also incorrect.
The internal energy of an ideal gas is a function of temperature. As temperature rises (indicated by the linear increase in pressure with temperature), the internal energy increases. Thus, option D is correct.
A linear relationship between pressure and temperature, as per the ideal gas law \(PV = nRT\), implies that the volume remains constant. A linear increase of pressure with temperature typically defines an isochoric process where volume is invariant.
Upon reviewing all options, only A, D, and E are deemed correct. The final answer is therefore:
A, D, E Only