Question

(3) 183 K
2 8 g of helium and 4 g of hydrogen gas are present
in a closed vessel at 127°C. If pressure exerted by
the gas mixture is 4.1 atm then the volume of the
container is
[NCERT Pg. 145]
(1) 32 L
(2) 16L
(3) 40 L
(4) 24 L
Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456​

Answer 1

Answer:

fxgjnbdeeruo

Explanation:

ghkiiiiuyuknkkiytre

Answer 2

Answer:

Number of moles of helium gas = mass of He/Molar mass of He = 8/4 = 2.

Number of moles of hydrogen gas = mass of H2/Molar mass of H2 = 4/2 = 2.

Total number of moles = 2 + 2 = 4.

The gas equation, PV = nRT is also applicable for mixture of gases, provided that they don't react with each other.

So, V = nRT/P = {4 × 0.082 × (273 + 127)}/4.1 = 32 L.

So, the volume of the container is 32 L.

Hope, this helps.

Explanation:

PLEASE MARK ME AS BRAINLEST ANSWER