48 grams of oxygen gas and 36 of grams of water vapor are in a 22.4 l container. what would the pressure be in torr if the container is 25 c

Answer : The pressure of gas will be, 2904.72 torr

Solution : Given,

First we have to calculate the pressure of oxygen gas and water vapor by using ideal gas equation.

where,

= pressure of oxygen gas

n = number of moles of gas

= mass of oxygen gas = 48 g

= molar mass of oxygen = 32 g/mole

V = volume of gas = 22.4 L

T = temperature of gas =

R = gas constant = 0.0821 Latm/moleK

similarly, we have to calculate the pressure of water vapor.

where,

= mass of water vapor = 36 g

= molar mass of water = 18 g/mole

Now we have to calculate the total pressure of gas.

      (1 atm = 760 torr)

Therefore, the pressure of gas will be, 2904.72 torr

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