Question

which equation is setup correctly to determine the volume of a 1.5 mol sample of oxygen gas at 22 degrees Celsius and 100 kPa

Answer 1

Hello!

We have the following data:


v (volume) = ? (in L)
n (number of mols) = 1,5 mol
T (temperature) = 22 ºC 
First let's convert the temperature on the Kelvin scale, let's see:
TK = TºC + 273,15
TK = 22 + 273,15
TK = 295,15

P (pressure) = 100 kPa → P = 100000 Pa → P ≈ 0,987 atm
R (gas constant) = 0,082 atm.L / mol.K

We apply the data above to the Clapeyron equation (gas equation), let's see:

$P*V = n*R*T$

$0,987*V = 1,5*0,082*295,15$

$0,987V = 36,30345$

$V = \dfrac{36,30345}{0,987}$

$\boxed{\boxed{V \approx 36,78\:L}} \end{array}}\qquad\checkmark$

I hope this helps. =)