There is no temperature at which the mode (most probable), the root mean ssquare velocity (v_"rms") and the average velocity have the same value in m/s.
In fact the probability distribution of molecular velocities (Maxwell distribution) is asymmetric. Look at this one:![enter image source here]()
Your teacher should chose one. The easiest one is v_"rms", because, in that case, you could use the simple formula:
v_"rms" = sqrt((3k_BT)/m)
where v_"rms" is the root mean square velocity in m/s;
k_B is Botzmann's constant = 1.38064852 × 10^"-23" ""m^2· kg · s^"-2" ·K^"-1"
T is the unknown temperature in kelvin,
m is the mass of a dioxygen (O_2) molecule in kg.
By assuming 1500 m/s is just the root mean square speed, v_"rms", by using molar mass of dioxygen (M_(O_2) = 0.031998 "kg"/"mol"), and knowing that:
k_B = (R " perfect gas constant")/(N_A " Avogadro's constant")=
=(8.31446 " "m^2· kg · s^"-2" ·K^"-1" · mol^"-1")/(6.02214·10^"23"mol^"-1"),
we can find T with the following calculation:
T=(v_"rms"^2 · M_(O_2))/(3R)=((1500 m/s)^2 · 0.031998 "kg"/"mol")/(3 · 8.31446" "m^2· kg · s^"-2" · K^"-1" · mol^"-1")= 2886 K
In which I have used the fact that a single dioxygen molecule multiplied by the Avogadro's constant yields the molar mass of dioxigen substance, M_(O_2) = 0.031998 "kg"/"mol" or 31.998 g/"mol".
The most probable speed v_"mp", average speed, v_"ave" and the rms speed v_"rms" are related in this way:
v_"mp"=sqrt(3/2) v_"rms"; " "v_"ave"=2/(sqrt(pi))v_"mp"
I hope this will help you and your teacher.
