For a spacecraft or a molecule to leave the moon, it must reach the escape velocity (speed) of the moon, which is 2.37 km/s. The average daytime temperature of the moon's surface is 365 K. What is the rms speed (in m/s) of an oxygen molecule at this temperature? rms speed = m/s

Vrms = 291 m/s

Explanation:

The root mean square velocity or vrms is the square root of the average square velocity and is. vrms=√3RTM. Where M is equal to the molar mass of the molecule in kg/mol.

Temperature = 365 K

Root mean square velocity = ?

molar mass of oxygen = 16 g/mol.

But xygen gas (O2) is comprised of two oxygen atoms bonded together. Therefore:

molar mass of O2 = 2 x 16

molar mass of O2 = 32 g/mol

Convert this to kg/mol:

molar mass of O2 = 32 g/mol x 1 kg/1000 g

molar mass of O2 = 3.2 x 10-2 kg/mol

Molar mass of Oxygen = 3.2 x 10-2 kg/mol

Vrms = √[3 (8.3145 (kg·m2/sec2) / K·mol) (365 K) / 3.2 x 10-2 kg/mol]

Vrms = 291 m/s

Vrms = 533 m/s

Explanation:

The formula for the root mean square molecular speed is given as;

V_rms=√ (3RT/M)

Where;

M is the molar mass of the molecule

T is Temperature

R is gas constant which has a value of 8.31 J/mol. k

V_rms is root mean square speed

Now let's calculate the molar mass of the molecule whuch in this case is oxygen.

Oxygen has a formula of O_2

Now, molar mass of one atom is 16 g/mol

Thus, molar mass of the 2 atoms would be; 2 x 16 = 32 g/mol = 0.032 kg/mol

T = 365K

So, plugging in the relevant values, we obtain;

V_rms = √[ (3x8.31x365) / 0.032]

Vrms = 533 m/s