Question

0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m³ at 1000 K. Given R is the gas constant in JK⁻¹ mol⁻¹, x is
(A) 2R/(4 + R)
(B) 2R/(4 - R)
(C) (4 + R)/2R
(D) (4 - R)/2R
[JEE Main 2019]

Answer 1

Answer:

4+R / 2R is answer is the option c

Answer 2

the value of x is (D) (4 - R)/2R

- Pressure of B, P₂ = mole fraction of B × total pressure

P₂ = {x/(x + 0.5)}×P

- From the ideal gas equation ,

P₂ × V = n × R × T

P₂ = x × R × T / V                    ( n = x = no. of moles )

- From  the above equations,

{x/(x + 0.5)} × P = x × R × T/V

Putting the values,

(x/x + 0.5) × 200 = x × 8.314 × 1000/10

On solving this,

We get,

x = (4 - R/2R)

- so the correct answer is option (D)