Verified Solution • 2025–26 Session moderate

QJEEM19Che994: JEE Main 2019 Chemistry - Class 11 JEEM19Che994 Solution: 0.5 moles of gas A and x moles of gas B exert a pressure of 200 …

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 −- 1mol −- 1, x is :

Multiple Choice Options

  • A 2R4+R{{2R} \over {4 + R}}
  • B 2R4−R{{2R} \over {4 - R}}
  • C 4+R2R{{4 + R} \over {2R}}
  • D 4−R2R{{4 - R} \over {2R}}

Step-by-Step Verified Solution

We know,

PV = nRT

Given,

P = 200 Pa

V = 10 m3

T = 1000 K

n = 0.5 + x

∴\therefore    200 ×\times 10 = (0.5 + x) R ×\times 1000

⇒\Rightarrow     0.5 + x = 2R{2 \over R}

⇒\Rightarrow     x = 2R−12{2 \over R} - {1 \over 2}

⇒\Rightarrow    x = 4−R2R{{4 - R} \over {2R}}

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