Question on velocity of gas molecules?

Question

root mean square velocity
v=[(3*R*T/M)]^1/2; watz? this M.

calculate the root mean square velocity of oxygen molecules at 25 degree celcius.
watz da value of R used for this problem and why.
On reffering to my txt book
the answer is
v=(3*8.314*298/0.0319998)^1/2. How did they get these values.
Is the answer given in the book correct. I am confused?

Answer 1

R = 8.3145; it's a special constant, and the units on it vary depending on the situation.
(http://www.chem.queensu.ca/courses/06/chem112/Handout/Chemistry112MidyearExaminationInformationSheet.pdf to see some units for R)

T = temperature in kelvin (25 celsius = 287 K)

M = molar mass in kg (not grams!) 0.0319998kg/mol is the molar mass for O2

Answer 2

R = 8.314 J/K*mol...this is a constant, often used in equations.

T = 298 K or 298 Kelvin, a measure of temperature. To get Kelvin, just add your Celcius degree, in this case 23 C + 273 = 298 K.

M is the molecular weight. Oxygen is O=O and one oxygen molecule is 16 g/mol so O=O is 32 g/mol...
so we have to do the conversion to km...
32 g/mol x (1kg/1000g) = 0.032 kg/mol ...0.0319998 is a more accurate molecular weight...

So the answer given is correct.

Answer 3

i think of meters in step with 2d (m/s) makes the main experience. I in simple terms did a rapid calculation making use of gasoline "kinetic concept." An oxygen molecule at room temperature could have a root mean sq. speed of around 500 m/s. in case you had to, you need to transform to different gadgets (like kilometers in step with 2d or perhaps kilometers in step with hour), yet "meters in step with 2d" are mechanically utilized in correct equations and the values are genuine looking.