Let's analyze the given processes individually:
In thermodynamics, work done by a gas is zero in an isochoric process, where volume remains constant. Since the question suggests the pressure increases linearly with temperature, the gas follows an isochoric process if option E is correct. Thus, the work done is indeed zero under these conditions.
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:
\(\Delta Q = \Delta U + \Delta W\)
Since option E suggests an isochoric process, \(\Delta W = 0\). Therefore, \(\Delta Q = \Delta U\). This option is false under an isochoric process.
If pressure increases linearly with temperature and volume is constant, the volume does not increase. Thus, this option is also false.
The internal energy of an ideal gas is a function of temperature. Since temperature increases (as pressure increases linearly with it), the internal energy increases. Thus, option D is true.
If pressure increases linearly with temperature, it means volume remains constant as per the ideal gas law \(PV = nRT\). A linear increase of pressure with temperature typically defines an isochoric process where volume does not change.
After evaluating all options, only A, D, and E are correct. Hence, the correct answer is:
A, D, E Only
Given: We are told that the pressure of the gas increases linearly with temperature. The relationship between pressure and temperature for an ideal gas is given by:
\[ P = aT + b, \] where \( a \) and \( b \) are constants. This suggests that the pressure increases linearly with temperature.
Since the pressure increases linearly with temperature, and for an ideal gas undergoing this process, the volume remains constant, there is no change in volume. The work done by the gas is given by: \[ W = P \Delta V. \] Since \( \Delta V = 0 \), the work done by the gas is zero. Thus, Option A is correct.
For an ideal gas undergoing any process, the heat added is related to both the work done and the change in internal energy. Since there is no change in volume in this case, the process is isochoric (constant volume), meaning all the heat added goes into increasing the internal energy. Therefore, Option B is incorrect because the heat added is equal to the change in internal energy.
As we established earlier, since this is an isochoric process (constant volume), the volume of the gas does not change. Thus, Option C is incorrect.
For an ideal gas, the internal energy depends only on the temperature. Since the temperature increases in this process, the internal energy of the gas increases. Thus, Option D is correct.
Given that no work is done by the gas and the pressure increases with temperature, this suggests that the volume remains constant. Hence, Option E is correct.
The correct answer is \( \boxed{A, D, E \text{ Only}} \).
Match List - I with List - II:

Choose the correct answer from the options given below:
In the following substitution reaction: