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 = (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 = (nRT)/(P)`

`V = ((6.9mol* 0.0821(L.atm)/(mol.K)* 233K))/(4.0atm)`

`V = (131.99217 L)/(4.0)`

V = 32.998 L or 33 L