Question

An oxygen gas container has a volume of 20.0 L. How many grams of oxygen are in the container if the gas has a pressure of 883 mmHg at 24 ∘C? Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer : The mass of oxygen gas will be, 30.5 grams

Solution :

Using ideal gas equation,

$PV=nRT\\\\PV=\frac{w}{M}\times RT$

where,

n = number of moles of gas

w = mass of gas  = ?

P = pressure of the gas = $883mmHg=\frac{883}{760}atm$

conversion used : (1 atm = 760 mmHg)

T = temperature of the gas = $24^oC=273+24=297K$

M = molar mass of oxygen gas = 32 g/mole

R = gas constant = 0.0821 Latm/moleK

V = volume of gas = 20.0 L

Now put all the given values in the above equation, we get the mass of oxygen gas.

$(\frac{883}{760}atm)\times (20.0L)=\frac{w}{32g/mole}\times (0.0821Latm/moleK)\times (297K)$

$w=30.5g$

Therefore, the mass of oxygen gas will be, 30.5 grams