Question

the ratio of the mass of 20L of hydrogen gas at 20 celeces and 10 atm and pressure is 20.then​

Answer 1

Answer:

Let there are n

1

moles of hydrogen and n

2

moles of helium in the given mixture. As Pv=nRT

Then the pressure of the mixture

P=

V

n

1

RT

+

V

n

2

RT

=(n

1

+n

2

)

V

RT

⇒2×101.3×10

3

=(n

1

+n

2

)×

20×10

−3

(8.3×300)

or, (n

1

+n

2

)=

(8.3)(300)

2×101.30×10

3

×20×10

−3

or, n

1

+n

2

=1.62....(1)

The mass of the mixture is (in grams)

n

1

×2+n

2

×4=5

⇒(n

1

+2n

2

)=2.5.....(2)

Solving the eqns. (1) and (2), we get

n

1

=0.74 and n

2

=0.88

Hence,

m

He

m

H

=

0.88×4

0.74×2

=

3.52

1.48

=

5

2

.

HOPE IT HELP YOU!