Calculate the total pressure in a mixture of 10 g of dioxygen and 8g of diHydrogen confined in a vessel of 2 L at 27 degree celsius.
Answers
Answered by
4
Volume = V = 2 L = 0.002 m³
T = 273.15 + 27 = 300.15 °K
R = 8.314 J/ mole / °K
number of Moles of Oxygen = 10 gm / 32 gm/mol = 5/16
number of moles of Hydrogen: 8 gm / 2 gm/mol = 4
total number of moles in the gas mixture = 69/16
P V = n R T
P = 69/16 * 8.314 * 300.15 / 0.002 Pa
= 53.8 atm approximately or 5.38 * 10⁶ Pa
It seems a lot, for just 18 gms of a gas mixture in a 2 L container at room temperature, exerting 5 times atmospheric pressure.
T = 273.15 + 27 = 300.15 °K
R = 8.314 J/ mole / °K
number of Moles of Oxygen = 10 gm / 32 gm/mol = 5/16
number of moles of Hydrogen: 8 gm / 2 gm/mol = 4
total number of moles in the gas mixture = 69/16
P V = n R T
P = 69/16 * 8.314 * 300.15 / 0.002 Pa
= 53.8 atm approximately or 5.38 * 10⁶ Pa
It seems a lot, for just 18 gms of a gas mixture in a 2 L container at room temperature, exerting 5 times atmospheric pressure.
kvnmurty:
click on thanks button above;; select best answer
Similar questions