What is the volume occupied by 10.9g of argon gas at a pressure of 1.34atm and a temperature of 461K ?

You would use the ideal gas law, pv = nrt.

With a little rearranging to solve for volume, you have v = nrt/p

You also will need to convert grams of Argon into moles, with grams/molecular mass

10.9g Argon * (1 mole Argon / 39.95g Argon) = .2728 moles Argon

Now to put in all the information:

v = (.2728 moles)(0.0821 atm*L/moles * Kelvin)(461 Kelvin)/(1.34 atm)

v = 7.71 Liters

Hello aeagle stud : Apply the Ideal gas Law as follows :

------------------------------------------------------------------------

PV = nRT = ( m / M ) ( R ) ( T )

V = ( m ) ( R ) T ) / ( P ) ( M )

V = ( 10.9 g ) ( 0.08205 atm - L/gmol - K ) ( 461 K ) / ( 1.34 atm ) ( 39.95 g/gmol )

V = 7.702 L <--------------------------