Kinetic Theory

JEE Main level, single correct, medium

139 of 168

Question 139 of 168

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:

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Answer: (A) 16:316:3

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 \Rightarrow 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)\Rightarrow \dfrac{2}{25}=\left[\dfrac{{m}_{1}}{{m}_{2}}\times \dfrac{1}{32}+\dfrac{1}{2}\right]{m}_{2}...(1)
⇒m1+m2=0.76\Rightarrow {m}_{1}+{m}_{2}=0.76
⇒m2[1+m1m2]=76100...(2)\Rightarrow {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\Rightarrow \dfrac{{m}_{1}}{{m}_{2}}=x\Rightarrow \dfrac{x+16}{32\left[x+1\right]}=\dfrac{8}{76} \Rightarrow 76x+76\times 16=8\times 32x+8\times 32 \Rightarrow 76x+1216=256x+256
960=180x⇒x=9618=163960=180x\Rightarrow x=\dfrac{96}{18}=\dfrac{16}{3}

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