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JEE Main 202131 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2021 · 31 August · Evening Shift · Physics · Kinetic Theory of Gases

A mixture of hydrogen and oxygen has volume 500 cm ^ 3 ,...

A mixture of hydrogen and oxygen has volume 500cm3,500{cm}^{3}, temperature 300K,300K, pressure 400kPa400kPa and mass 0.76g.0 . 76g . The ratio of masses of oxygen to hydrogen will be:

Options

  1. A16:316: 3
  2. B3:83: 8
  3. C8:38: 3
  4. D3:163: 16

Correct answer

A. 16:316: 3

Step-by-step solution

A number of moles of, O2=m132{O}_{2} = \dfrac{{m}_{1}}{32}, the number of moles of, H2=m22{H}_{2} = \dfrac{{m}_{2}}{2}. Using the ideal gas equation, PV=[m132+m22]253×300PV = \left[\dfrac{{m}_{1}}{32} + \dfrac{{m}_{2}}{2}\right]\dfrac{25}{3} \times 300,
400×103×500×10−6=[m132+m22]25×100⇒2×102=[m1m2×132+12]m2×25×100400 \times {10}^{3} \times 500 \times {10}^{- 6} = \left[\dfrac{{m}_{1}}{32} + \dfrac{{m}_{2}}{2}\right]25 \times 100 ⇒ 2 \times {10}^{2} = \left[\dfrac{{m}_{1}}{{m}_{2}} \times \dfrac{1}{32} + \dfrac{1}{2}\right]{m}_{2} \times 25 \times 100
⇒225=[m1m2×132+12]m2...(1)⇒ \dfrac{2}{25} = \left[\dfrac{{m}_{1}}{{m}_{2}} \times \dfrac{1}{32} + \dfrac{1}{2}\right]{m}_{2} . . . ( 1 )
⇒m1+m2=0.76⇒ {m}_{1} + {m}_{2} = 0 . 76
⇒m2[1+m1m2]=76100...(2)⇒ {m}_{2}\left[1 + \dfrac{{m}_{1}}{{m}_{2}}\right] = \dfrac{76}{100} . . . ( 2 )
(1)( 1 ) divided by (2)( 2 ) ; [m1m2×132+12][m1m2+1]=225×10076\dfrac{\left[\dfrac{{m}_{1}}{{m}_{2}} \times \dfrac{1}{32} + \dfrac{1}{2}\right]}{\left[\dfrac{{m}_{1}}{{m}_{2}} + 1\right]} = \dfrac{2}{25} \times \dfrac{100}{76}
Let assume,
⇒m1m2=x⇒x+1632[x+1]=876⇒76x+76×16=8×32x+8×32⇒76x+1216=256x+256⇒ \dfrac{{m}_{1}}{{m}_{2}} = x ⇒ \dfrac{x + 16}{32\left[x + 1\right]} = \dfrac{8}{76} ⇒ 76x + 76 \times 16 = 8 \times 32x + 8 \times 32 ⇒ 76x + 1216 = 256x + 256
960=180x⇒x=9618=163960 = 180x ⇒ x = \dfrac{96}{18} = \dfrac{16}{3}
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