Using the ideal gas equation, calculate the pressure of oxygen gas in a cylinder with a volume of 25.00 l. the oxygen masses 4.362 kg and room temperature is at 22.5 o c. how many moles of oxygen are there and what is the pressure of oxygen in atmospheres in the cylinder according to the ideal gas law?

Answer : The number of moles present in the oxygen gas is, 272.625 moles and the pressure of oxygen gas in atmosphere in the cylinder is, 264.56 atm

Solution : Given,

Mass of oxygen gas = 4.362 Kg = 4362 g      (1 Kg = 1000 g)

Molar mass of oxygen gas = 16 g/mole

First we have to calculate the moles of oxygen gas.

Now we have to calculate the pressure of oxygen gas by using ideal gas equation.

where,

P = pressure of gas

T = temperature of gas =

V = volume of gas = 25 L

n = number of moles of gas = 272.625 moles

R = gas constant = 0.0821 Latm/moleK

Now put all the given values in the ideal gas equation, we get the pressure of gas.

Therefore, the number of moles present in the oxygen gas is, 272.625 moles and the pressure of oxygen gas in atmosphere in the cylinder is, 264.56 atm

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