How would you determine the volume of 6 moles of helium gas at 0.34 atm pressure and 33 degrees C?

Question

please help...i need this asap.....

Answer 1

PV = nRT

V = (nRT)/P

V = (6*.0821*306)/.34

V = 443.34 L

Answer 2

Use the Ideal gas law. PV = nRT

  P = pressure

  V = volume

  n = # of moles

  R = gas constant (.0821 L*atm/ mol*K)  {* means multiply}

  T = temp. (in kelvin)

find volume:

1st       (.34 atm) V = (6 mol)(.0821 L*atm/mol*K)(306K)

needed to convert Celsius to Kelvin ( add 273 to C)

2nd     V = (6mol) (.0821 L* atm/ mol*K) (306K)

                                 divided by (.34atm)

3rd      multiply 6 mol by .0821 then by 306 K, the moles and kelvin will cancel

4th        divide the product by .34atm and atms cancel and left with V

V  = 443.34 L