If 64 grams of o2 gas occupy a volume of 56 liters at a temperature or 25°c, what is the pressure of the gas, in atmospheres (atm) ?

Using Ideal Gas Laws:

PV = nRT

Where,

P is Pressure = Not given

V is the Volume = 56L

n is number of moles of gas = Not given

R is the proportionality constant = 0.08206 L. atm/K. mol

T is the temperature = 25 + 273.15 = 298 K

Step 1:

Calculate n,

n = 2 x 15.999 = 31.998 g/mole

n (oxygen) = 64/31.998

= 2 moles of oxygen

Step 2:

Calculate Pressure,

P = nRT/V

Substitute the values,

P = 2 x 0.08206 x 298/56

P = 0.88 atm