Question

a mixture of N2 and Ar gases in a cylinder contains 7g of N2 and 8g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar the partial pressure of N2 is​

Answer 1

Given:7g of N2 i.e 0.25 moles

8g of Ar i.e 0.2 moles

Total pressure=27bars

Explanation:

we know that moles=Given mass /molecular mass

Now we know that

• partial pressure is equal to product of moles fraction and total pressure in closed system
• therefore,
• mole fraction of Nitrogen gas=0.25/0.45
• Now partial pressure=(0.25/0.45)27

=15bars

Answer 2

Given:

A mixture of N2 and Ar gases in a cylinder contains 7g of N2 and 8g of Ar. The total pressure of the mixture of the gases in the cylinder is 27 bar.

To find:

Partial pressure of N2.

Calculation:

Moles of Nitrogen = 7/28 = 1/4

Moles of Argon = 8/40 = 1/5

So, mole fraction of Nitrogen

$(\chi)_{N_{2}} = \bigg \{\dfrac{ \dfrac{1}{4} }{ (\dfrac{1}{5} + \dfrac{1}{4}) } \bigg \}$

$= > (\chi)_{N_{2}} = \bigg \{\dfrac{ \dfrac{1}{4} }{ (\dfrac{5 + 4}{20} ) } \bigg \}$

$= > (\chi)_{N_{2}} = \bigg \{\dfrac{ \dfrac{1}{4} }{ (\dfrac{9}{20} ) } \bigg \}$

$= > (\chi)_{N_{2}} = \dfrac{5}{9}$

So, partial pressure of Nitrogen

$\therefore \: P_{N_{2}} = P \times ( \chi)_{N_{2}}$

$= > \: P_{N_{2}} = 27 \times \dfrac{5}{9}$

$= > \: P_{N_{2}} = 15 \: bar$

So , final answer is:

$\boxed{ \bf{\: P_{N_{2}} = 15 \: bar}}$