BITSAT 2015 Paper

Let there are n1 moles of hydrogen and n2 moles of helium in the given mixture. As Pv = nRT
Then the pressure of the mixture
P =

n1RT

V

+

n2RT

V

= (n1+n2)

RT

V

⇒ 2 × 101.3 × 103 = (n1+n2) ×

(8.3×300)

20×10−3

or (n1+n2) =

2×101.3×103×20×10−3

(8.3)(300)

or n1+n2 = 1.62 ... (1)
The mass of the mixture is (in grams)
n1 × 2 + n2 × 4 = 5
⇒ (n1+2n2) = 2.5 ... (2)
Solving the eqns. (1) and (2), we get
n1 = 0.74 and n2 = 0.88
Hence

mH

mHe

=

0.74×2

0.88×2

=

1.48

3.52

=

2

5