# Root-mean-square speed of an O2 gas

• format1998
In summary, the problem involves finding the rms speed (vrms) of O2 gas with a quantity of 0.280 mol, a pressure of 3.50 atm, and a volume of 1.93 L. Using the ideal gas law, the temperature is calculated to be 294 K. Then, using the formula for vrms, the value is found to be 1.166 * 10-9 m/s. However, after checking the value of R and using the mass of one molecule of O2, the correct answer is 478 m/s. It is important to be careful with the use of 'm' as it can refer to the total mass of gas, the mass of one mole, or

## Homework Statement

0.280 mol of O2 gas is at a pressure of 3.50 atm and has a volume of 1.93 L.
What is the rms speed (vrms) of the gas molecules?

O2 gas
n = 0.280 mol @ 32 g/mol m = 0.00896 kg
P = 3.50 atm
V = 1.93 L

## Homework Equations

PV=nRT -> T = PV/nR

vrms=$\sqrt{\frac{3kT}{m}}$

## The Attempt at a Solution

T=$\frac{PV}{nR}$=$\frac{(3.5 atm * 1.93 L)}{.28 mol (0.0821 \frac{L*atm}{mol*K})}$ = 294 K

vrms=$\sqrt{\frac{3kT}{m}}$=$\sqrt{\frac{3(1.38E-23 J/K)(294K)}{0.00896 kg}}$ = 1.166 * 10-9 m/s

According to the answer key, the answer is 478 m/s. What am I doing wrong?

Check the value of R.
I think that R = 0.0831using your units for R.

format1998 said:
vrms=$\sqrt{\frac{3kT}{m}}$

Also I think that the 'm' in the above formula is the mass of ONE MOLECULE.

R = 0.0821 $\frac{L*atm}{mol*K}$
is the value that is on the book and other tables I found on the net

Using the mass of one molecule of O2 gave me 479 m/s. One digit off but I'll take it or maybe I'm still doing something wrong??

Thank you

Sorry. Your value of R is Ok in the units you are using. So i think that your mistake was in m.

One has to be extra careful in this topic because 'm' may stand for 'total mass of gas' or 'mass of one mole' or 'mass of one molecule'.