Question

A gas cylinder contains 2.0 mol of gas X and 6.0 mol of gas Y at a total pressure of 2.1 atm. What is the partial pressure of gas Y? Use StartFraction P subscript A over P subscript T EndFraction equals StartFraction n subscript a over n subscript T EndFraction..

Answer 1

1.6 is the answer

Step-by-step explanation:

Answer 2

Answer : The partial pressure of X and Y gases are, 0.525 and 1.575 atm respectively.

Explanation : Given,

Moles of X = 2.0 mole

Moles of Y = 6.0 mole

Total pressure = 2.1 atm

Now we have to calculate the mole fraction of X and Y.

`\text{Mole fraction of }X=\frac{\text{Moles of }X}{\text{Moles of }X+\text{Moles of }Y}`

`\text{Mole fraction of }X=(2.0)/(2.0+6.0)=0.25`

and,

`\text{Mole fraction of }Y=\frac{\text{Moles of }Y}{\text{Moles of }X+\text{Moles of }Y}`

`\text{Mole fraction of }Y=(6.0)/(2.0+6.0)=0.75`

Now we have to calculate the partial pressure of X and Y.

According to the Raoult's law,

`p_i=X_i* p_T`

where,

`p_i` = partial pressure of gas

`p_T` = total pressure of gas = 2.1 atm

`X_i` = mole fraction of gas

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

`p_(X)=0.25* 2.1atm=0.525atm`

and,

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

`p_(Y)=0.75* 2.1atm=1.575atm`

Thus, the partial pressure of X and Y gases are, 0.525 and 1.575 atm respectively.