Part a: three gases (8.00 g of methane, ch_4, 18.0g of ethane, c_2h_6, and an unknown amount of propane, c_3h_8) were added to the same 10.0- l container. at 23.0 degrees c, the total pressure in the container is 5.40 atm. calculate the partial pressure of each gas in the container. part b: a gaseous mixture of o_2 and n_2 contains 37.8 % nitrogen by mass. what is the partial pressure of oxygen in the mixture if the total pressure is 405 mmhg?

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

Moles of methane gas =

Moles of ethane gas  =

Moles of propane gas =

Mole fraction of methane =

Similarly, mole fraction of ethane and propane :

Partial pressure of each gas can be calculated by the help of Dalton's' law:

Partial pressure of methane gas:

Partial pressure of ethane gas:

Partial pressure of propane gas:

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas =

Moles of oxygen gas  =

Mole fraction of nitrogen=

Similarly, mole fraction of oxygen

Partial pressure of each gas can be calculated by the help of Dalton's' law:

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

Partial pressure of oxygen gas:

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.