Root-mean-square speed of an O2 gas

[format1998]

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 (v_rms) 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

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

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

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

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



Please help! Thank you in advance. Any and all help is much appreciated!

 

[grzz]

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

 

[grzz]

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

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

 

[format1998]

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

 

[grzz]

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'.