- #1

- 26

- 0

## Homework Statement

0.280 mol of O

_{2}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?

O

_{2}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}=[itex]\sqrt{\frac{3kT}{m}}[/itex]

## The Attempt at a Solution

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

v

_{rms}=[itex]\sqrt{\frac{3kT}{m}}[/itex]=[itex]\sqrt{\frac{3(1.38E-23 J/K)(294K)}{0.00896 kg}}[/itex] = 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!