1. Given relation

For the process, pressure increases linearly with temperature: $P \propto T$ So we can write $P = kT$ where $k$ is a constant during the process.

2. Use ideal gas equation

For an ideal gas, $PV = nRT$ Substitute $P = kT$: $kT\,V = nRT$ For $T \neq 0$, $kV = nR$ Since $n,R,k$ are constants, $V = \text{constant}$ Hence the process is isochoric.

So statement E is correct.

Also, since volume remains constant, statement C (volume increased) is false.

3. Work done

Work done by gas is $W = \int P\,dV$ But in an isochoric process, $dV = 0$ Therefore, $W=0$ So statement A is correct.

4. First law of thermodynamics

$Q = \Delta U + W$ Since $W=0$, $Q = \Delta U$ Thus heat added is not different from change in internal energy; it is equal to it.

So statement B is false.

5. Change in internal energy

For an ideal gas, internal energy depends only on temperature: $U \propto T$ The question says pressure increases linearly with temperature, and the gas goes from initial to final state along this relation. Since pressure increases with temperature, the final temperature is greater than the initial temperature.

Hence, $\Delta T > 0 \implies \Delta U > 0$ So statement D is correct.

6. Check all statements

• A: True
• B: False
• C: False
• D: True
• E: True

Therefore the correct set is: $A, D, E \text{ only}$

So the correct option is B.

7. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They match.