Respuesta :
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.
[tex]P_{O_2}=\frac{nRT}{V}=\frac{w_{O_2}RT}{M_{O_2}V}[/tex]
where,
[tex]P_{O_2}[/tex] = pressure of oxygen gas
n = number of moles of gas
[tex]w_{O_2}[/tex] = mass of oxygen gas = 48 g
[tex]M_{O_2}[/tex] = molar mass of oxygen = 32 g/mole
V = volume of gas = 22.4 L
T = temperature of gas = [tex]25^oC=273+25=298K[/tex]
R = gas constant = 0.0821 Latm/moleK
[tex]P_{O_2}=\frac{48g\times 0.0821Lam/moleK\times 298K}{32g/mole\times 22.4L}=1.638atm[/tex]
similarly, we have to calculate the pressure of water vapor.
[tex]P_{H_2O}=\frac{nRT}{V}=\frac{w_{H_2O}RT}{M_{H_2O}V}[/tex]
where,
[tex]w_{H_2O}[/tex] = mass of water vapor = 36 g
[tex]M_{H_2O}[/tex] = molar mass of water = 18 g/mole
[tex]P_{H_2O}=\frac{36g\times 0.0821Lam/moleK\times 298K}{18g/mole\times 22.4L}=2.184atm[/tex]
Now we have to calculate the total pressure of gas.
[tex]P=p_{O_2}+p_{H_2O}[/tex]
[tex]P=1.638+2.184=3.822atm=2904.72torr[/tex] (1 atm = 760 torr)
Therefore, the pressure of gas will be, 2904.72 torr
