Question

Solve the following, show all work and units for your calculation: Let's say we have a mixture of hydrogen gas

(H2), and oxygen gas (O2). The mixture contains 6.7 mol hydrogen gas and 3.3 mol oxygen gas. The mixture is
in a 300 L container at 273 K and the total pressure of the gas mixture is 0.75 atm. What is the partial
pressure for each gas?​

Answer 1

Answer: The partial pressure of hydrogen gas is 0.503 atm and that of oxygen gas is 0.248 atm

Step-by-step explanation:

We are given:

Moles of hydrogen gas = 6.7 moles

Moles of oxygen gas = 3.3 moles

Mole fraction of a substance is given by:

`\chi_A=(n_A)/(n_A+n_B)`

Mole fraction of hydrogen gas,
`\chi_(H_2)=(6.7)/(6.7+3.3)=0.67`

Mole fraction of oxygen gas,
`\chi_(O_2)=(3.3)/(6.7+3.3)=0.33`

To calculate the total pressure of the container, we use the equation given by Raoult's law, which is:

`p_(A)=p_T* \chi_(A)`

where,

`p_A` = partial pressure of substance

`p_T` = total pressure

`\chi_A` = mole fraction of substance

• For hydrogen gas:

We are given:

`\chi_(H_2)=0.67\\p_T=0.75atm`

Putting values in above equation, we get:

`p_(H_2)=0.75atm* 0.67=0.503atm`

• For oxygen gas:

We are given:

`\chi_(O_2)=0.33\\p_T=0.75atm`

Putting values in above equation, we get:

`p_(O_2)=0.75atm* 0.33=0.248atm`

Hence, the partial pressure of hydrogen gas is 0.503 atm and that of oxygen gas is 0.248 atm