Given data:
- Pressure (p) = 1 * 10^5 Pa
- Initial volume (V1) = 10 liters
- Final volume (V2) = 20 liters

Approach:
To calculate the increase in internal energy, we need to use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added (Q) minus the work done (W) on the system.

Step 1: Calculate the work done (W):
In this case, the process is isobaric, which means the pressure remains constant. The work done in an isobaric process can be calculated using the formula:

W = p * (V2 - V1)

Where p is the constant pressure and (V2 - V1) is the change in volume.

Substituting the given values, we get:

W = (1 * 10^5 Pa) * (20 liters - 10 liters)

W = 1 * 10^5 Pa * 10 liters

W = 1 * 10^6 J

Step 2: Calculate the change in internal energy (ΔU):
Since the process is reversible, the change in internal energy can be calculated using the formula:

ΔU = Q - W

In an isobaric process, the heat added (Q) can be calculated using the formula:

Q = p * (V2 - V1)

Substituting the given values, we get:

Q = (1 * 10^5 Pa) * (20 liters - 10 liters)

Q = 1 * 10^5 Pa * 10 liters

Q = 1 * 10^6 J

Now, substituting the values of Q and W into the equation for ΔU, we get:

ΔU = (1 * 10^6 J) - (1 * 10^6 J)

ΔU = 0 J

Conclusion:
The increase in internal energy (ΔU) of the ideal monoatomic gas undergoing a reversible isobaric process with a pressure of 1 * 10^5 Pa and volume increasing from 10 liters to 20 liters is zero. This implies that there is no change in the internal energy of the gas during this process.