Question

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 1

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.

`P_(O_2)=(nRT)/(V)=(w_(O_2)RT)/(M_(O_2)V)`

where,

`P_(O_2)` = pressure of oxygen gas

n = number of moles of gas

`w_(O_2)` = mass of oxygen gas = 48 g

`M_(O_2)` = molar mass of oxygen = 32 g/mole

V = volume of gas = 22.4 L

T = temperature of gas =
`25^oC=273+25=298K`

R = gas constant = 0.0821 Latm/moleK

`P_(O_2)=(48g* 0.0821Lam/moleK* 298K)/(32g/mole* 22.4L)=1.638atm`

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

`P_(H_2O)=(nRT)/(V)=(w_(H_2O)RT)/(M_(H_2O)V)`

where,

`w_(H_2O)` = mass of water vapor = 36 g

`M_(H_2O)` = molar mass of water = 18 g/mole

`P_(H_2O)=(36g* 0.0821Lam/moleK* 298K)/(18g/mole* 22.4L)=2.184atm`

Now we have to calculate the total pressure of gas.

`P=p_(O_2)+p_(H_2O)`

`P=1.638+2.184=3.822atm=2904.72torr` (1 atm = 760 torr)

Therefore, the pressure of gas will be, 2904.72 torr