Question

A mixture of he and ne at a total pressure of 0.95 atm is found to contain 0.32 mol of he and 0.56 mol of ne. the partial pressure of ne is __________ atm

Answer 1

Answer : The partial pressure of neon gas is, 0.61 atm

Explanation : Given,

Moles of He = 0.32 mol

Moles of Ne = 0.56 mol

Total pressure = 0.95 atm

According to the Raoult's law,

`p_(Ne)=X_(Ne)* p_T`

where,

`p_(Ne)` = partial pressure of neon gas

`p_T` = total pressure of gas

`X_(Ne)` = mole fraction of neon gas

First we have to calculate the moles of fraction of neon gas.

`\text{Mole fraction of }Ne=\frac{\text{Moles of }Ne}{\text{Moles of }Ne+\text{Moles of }He}`

`\text{Mole fraction of }Ne=(0.56)/(0.56+0.32)=0.64`

Now we have to calculate the partial pressure of neon gas.

`p_(Ne)=X_(Ne)* p_T`

`p_(Ne)=0.64* 0.95atm`

`p_(Ne)=0.61atm`

Therefore, the partial pressure of neon gas is, 0.61 atm

Answer 2

The partial pressure of ne in atm is calculated as follows

moles of Ne /total moles x total pressure

total moles 0.32 + 0.56 = 0.88 mol

partial pressure = 0.56/0.88 x0.95= 0.6045 atm