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 \( \boxed{A, D, E \text{ Only}} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: