need help please (ideal gas laws)?

Question

I need help on Ideal Gas Law.
At what temperature will 6.13 g of oxygen gas exert a pressure of 3.00 atm in a 10.0 L container?
_______K
Please explain how to do this problem
thanks in advance

Answer 1

pv=nRt

r is constant
you have pressure, volume, and can find moles of oxygen by converting from the 6.13 grams. Just plug in and solve for temperature, just make sure your units cancel

Answer 2

The ideal gas law states:

PV = nRT

P = pressure
V = volume
n = number of moles of the gas
R = a constant number (should be in the front of your textbook)
T = temperature

Solve this equation for T.

T = PV/(nR)

Find number of moles of oxygen in 6.13 g. (Should be close to 0.19 mol). Plug and chug, put the numbers in the equation, compute and get an answer.

Answer 3

PV=NrT where r is the gas constant = 0.082 L x atm/K x mol and N is the moles of gas.. So 6.13g of O2 is 0.19 moles. plug all values into the equation and this occurs at 1925 degrees Kelvin.

Answer 4

pV= nRT
You know n (6.13/32)
You know P (3 atm)
You know V (10)
Pick the right R (0.0821 atm-L/deg K per mole)
And solve!!

Answer 5

use PV=nRT

you can figure out n by m/M which is mass of oxygen divided by molar mass

and R = 8.314

so T = PV/nR

you also have to conver 3 atm into kpa

Answer 6

appropriate gas regulation in basic terms works on gasses with specific traits. the appropriate gas regulation assumes that the molecules do not attrat one yet another, have not got any mass, etc. this does not carry real under specific situations, like at temperatures on the threshold of absolute 0 (the gas molecules show attraction at such low temperatures)