Question

What is the volume of 6.9 mol oxygen (O2) gas at 233 K and a pressure of 4.0 atm? (The universal gas constant is 0.0821 L•atm/mol•K.)

Answer 1

Volume of Oxygen O2 would be 33 L.

Solution : Given information - Mole (n) = 6.9 mol

Pressure (P) = 4.0 atm

Temperature (T) = 233 K

Universal gas constant (R) = 0.0821 L.atm/mol.K

Volume(V) is unknown , which we need to find.

Volume can be calculated by using the Ideal gas equation which is :

PV = nRT  (Ideal gas equation)

On rearranging the above formula we will get :

$V = \frac{nRT}{P}$

Where , P = pressure , V = volume , n = mole , R = universal gas constant , T = temperature.

We will plugin the values of P , n , R and T in the ideal gas equation and then calculate Volume (V)

$V = \frac{nRT}{P}$

$V = \frac{(6.9mol\times 0.0821\frac{L.atm}{mol.K}\times 233K)}{4.0atm}$

$V = \frac{131.99217 L}{4.0}$

V = 32.998 L or 33 L