Introduction:
In this problem, we need to calculate the heat transfer (q) and work done (w) during an isoergic reversible change of an ideal monoatomic gas. Isoergic means that the change occurs at constant internal energy.

Given:
- Initial temperature (T1) = 25 degrees Celsius = 298 K
- Initial pressure (P1) = 1 atm
- Final volume (V2) = 2 times the original volume
- Monoatomic gas, so the number of moles (n) = 1

Solution:

Step 1: Converting Celsius to Kelvin:
The ideal gas law requires temperature to be in Kelvin. Thus, we convert the initial temperature from Celsius to Kelvin by adding 273 to the given value:
T1 = 298 K

Step 2: Calculating the final temperature:
Since the process is isoergic, the internal energy remains constant. Therefore, according to the ideal gas law, the final temperature (T2) can be calculated using the initial and final volumes and pressures:
(P1 * V1) / T1 = (P2 * V2) / T2

Since the number of moles (n) is constant, we can rewrite the equation as:
(P1 * V1) / T1 = (P2 * V2) / T2

Substituting the given values:
(1 atm * V1) / 298 K = (1 atm * 2V1) / T2

Simplifying the equation:
T2 = 2T1 = 2 * 298 K = 596 K

Step 3: Calculating q:
Since the process is isoergic, the change in internal energy (ΔU) is zero. Therefore, the heat transfer (q) is also zero.

Step 4: Calculating w:
The work done (w) during an isoergic process can be calculated using the equation:
w = -ΔU

Since ΔU is zero, the work done (w) is also zero.

Conclusion:
In this isoergic reversible change, the heat transfer (q) and work done (w) both are zero.