A gas mixture is made up of kr (21.7 g), o2 (7.18 g), and co2 (14.8 g). the mixture has a volume of 23.1 l at 59 °c. calculate the partial pressure of each gas in the mixture and the total pressure of the gas mixture.

Answer: - partial pressure of Kr = 0.306 atm, partial pressure of oxygen = 0.264 atm and partial pressure of carbon dioxide = 0.396 atm

Total pressure is 0.966 atm

Solution: - moles of Kr = 21.7 g x (1mol/83.8g) = 0.259 mol

moles of oxygen = 7.18 g x (1mol/32g) = 0.224 mol

moles of carbon dioxide = 14.8 g x (1mol/44g) = 0.336 mol

Volume of container = 23.1 L and the temperature is 59 + 273 = 332 K

From ideal gas law equation, P = nRT/V

partial pressure of Kr = (0.259 x 0.0821 x 332).23.1 = 0.306 atm

partial pressure of oxygen = (0.224 x 0.0821 x 332) / 23.1 = 0.264 atm

partial pressure of carbon dioxide = (0.336 x 0.0821 x 332) / 23.1 = 0.396 atm

Total pressure of the gas mixture = 0.306 atm + 0.264 atm + 0.396 atm = 0.966 atm