JEE Advanced · 2016 · 22 May 2016 · Paper 1 · States of Matter

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The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases

x

times. The value of

x

Answer: 4

Solution

Mean free path

ℓ \propto T / P

and mean speed

\overset{c}{ˉ} \propto T

. Thus, diffusion coefficient

D \propto ℓ \times \overset{c}{ˉ} \propto (T / P) \times T = T^{3/2} / P

. Initially,

D_{1} \propto T_{1}^{3/2} / P_{1}

. After changes,

D_{2} \propto (4 T_{1})^{3/2} / (2 P_{1}) = 8 T_{1}^{3/2} / (2 P_{1}) = 4 T_{1}^{3/2} / P_{1} = 4 D_{1}

. Therefore,

x = 4

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Integer answer

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