Introduction:
In this question, we are asked to calculate the temperature at which the root mean square velocity of O2 gas is equal to the most probable velocity of Ne (20 g mol-1).

Formula:
The root mean square velocity of a gas is given by the formula:
v_rms = √(3kT/m)
where k is the Boltzmann constant, T is the temperature in Kelvin, and m is the molar mass of the gas.

The most probable velocity of a gas is given by the formula:
v_mp = √(2kT/m)

Calculation:
Molar mass of O2 = 32 g mol-1
Molar mass of Ne = 20 g mol-1

Let's assume the temperature at which the root mean square velocity of O2 is equal to the most probable velocity of Ne to be T°C.

The velocity of O2 (v_rms) at T°C can be calculated as follows:
v_rms = √(3k(T+273)/m)
v_rms = √(3 x 1.38 x 10^-23 x (T+273)/(32/1000))
v_rms = 4.87 x 10^2 √(T+273)

The velocity of Ne (v_mp) at T°C can be calculated as follows:
v_mp = √(2k(T+273)/m)
v_mp = √(2 x 1.38 x 10^-23 x (T+273)/(20/1000))
v_mp = 5.37 x 10^2 √(T+273)

Now, equating these two velocities, we get:
4.87 x 10^2 √(T+273) = 5.37 x 10^2 √(T+273)

Squaring both sides, we get:
2.37 x 10^5(T+273) = 2.89 x 10^5(T+273)

Simplifying, we get:
T = 8°C

Conclusion:
Therefore, the temperature at which the root mean square velocity of O2 gas at 300 K is equal to the most probable velocity of Ne (20 g mol-1) is 8°C.