Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The raito of their respective mean free times is closest to :

Heat and Thermodynamics PYQ — JEE Main 2020 · 9 Jan · Shift 2 · Q56 with Solution

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Correct answer: NO OPTION MATCHES THE CORRECT RESULT. THE PHYSICALLY CORRECT RATIO IS CLOSEST TO $$1.09$$ FOR $$\TAU_{\MATHRM{AR}}/\TAU_{\MATHRM{XE}}$$.

Mean free time formula

For a gas,

\tau = \frac{\lambda}{\bar v}

where $\tau$ is mean free time, $\lambda$ is mean free path, and $\bar v$ is mean speed.

Also,

\lambda = \frac{1}{\sqrt{2}\,n\sigma}

where $n$ is number density and $\sigma$ is collision cross-section.

Since both gases have the same number density,

\tau \propto \frac{1}{\sigma \bar v}

Collision cross-section

For spherical atoms,

\sigma = \pi d^2 = \pi (2r)^2 = 4\pi r^2

So,

\sigma \propto r^2

Thus,

\tau \propto \frac{1}{r^2 \bar v}

Mean speed dependence

At the same temperature,

\bar v \propto \frac{1}{\sqrt{M}}

where $M$ is molecular/atomic mass.

Therefore,

\tau \propto \frac{\sqrt{M}}{r^2}

Ratio of mean free times

Let argon = Ar and xenon = Xe. Then

\frac{\tau_{\text{Ar}}}{\tau_{\text{Xe}}} = \frac{\sqrt{M_{\text{Ar}}}/r_{\text{Ar}}^2}{\sqrt{M_{\text{Xe}}}/r_{\text{Xe}}^2} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{Xe}}}}\cdot \frac{r_{\text{Xe}}^2}{r_{\text{Ar}}^2}

Substitute values:

M_{\text{Ar}}=40,\quad M_{\text{Xe}}=140,\quad r_{\text{Ar}}=0.07\text{ nm},\quad r_{\text{Xe}}=0.10\text{ nm}

So,

\frac\tau_{\text{Ar}}{\tau_{\text{Xe}}} = \sqrt{\frac{40}{140}}\cdot \left(\frac{0.10}{0.07}\right)^2

Now,

\sqrt{\frac{40}{140}} = \sqrt{\frac{2}{7}} \approx 0.535

and

\left(\frac{0.10}{0.07}\right)^2 = \left(\frac{10}{7}\right)^2 = \frac{100}{49} \approx 2.041

Hence,

\frac\tau_{\text{Ar}}{\tau_{\text{Xe}}} \approx 0.535 \times 2.041 \approx 1.09

So the inverse ratio is

\frac{\tau_{\text{Xe}}}{\tau_{\text{Ar}}} \approx \frac{1}{1.09} \approx 0.92

This does not match any option.

Check likely intended interpretation

Sometimes such questions incorrectly use

\tau \propto \frac{1}{r\bar v}

which would give

\frac{\tau_{\text{Ar}}}{\tau_{\text{Xe}}} = \sqrt{\frac{40}{140}}\cdot \frac{0.10}{0.07} \approx 0.535 \times 1.429 \approx 0.76

Again, no option.

If instead ratio is taken as

\frac{\tau_{\text{Xe}}}{\tau_{\text{Ar}}} = \sqrt{\frac{140}{40}}\cdot \left(\frac{0.07}{0.10}\right)^2 \approx 1.871 \times 0.49 \approx 0.92

Still no option.

Thus, using the standard kinetic theory formula, the ratio should be closest to $1.09$ , which is absent from the choices.

Comparison with options

Given options are $2.3, 1.83, 4.67, 3.67$ . None is close to the physically correct result $\approx 1.09$ .

Therefore, the stored answer $\text{B} = 1.83$ appears inconsistent with standard theory.

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View written solutionFree

Correct answer: NO OPTION MATCHES THE CORRECT RESULT. THE PHYSICALLY CORRECT RATIO IS CLOSEST TO $$1.09$$ FOR $$\TAU_{\MATHRM{AR}}/\TAU_{\MATHRM{XE}}$$.

Mean free time formula

For a gas,

\tau = \frac{\lambda}{\bar v}

where $\tau$ is mean free time, $\lambda$ is mean free path, and $\bar v$ is mean speed.

Also,

\lambda = \frac{1}{\sqrt{2}\,n\sigma}

where $n$ is number density and $\sigma$ is collision cross-section.

Since both gases have the same number density,

\tau \propto \frac{1}{\sigma \bar v}

Collision cross-section

For spherical atoms,

\sigma = \pi d^2 = \pi (2r)^2 = 4\pi r^2

So,

\sigma \propto r^2

Thus,

\tau \propto \frac{1}{r^2 \bar v}

Mean speed dependence

At the same temperature,

\bar v \propto \frac{1}{\sqrt{M}}

where $M$ is molecular/atomic mass.

Therefore,

\tau \propto \frac{\sqrt{M}}{r^2}

Ratio of mean free times

Let argon = Ar and xenon = Xe. Then

\frac{\tau_{\text{Ar}}}{\tau_{\text{Xe}}} = \frac{\sqrt{M_{\text{Ar}}}/r_{\text{Ar}}^2}{\sqrt{M_{\text{Xe}}}/r_{\text{Xe}}^2} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{Xe}}}}\cdot \frac{r_{\text{Xe}}^2}{r_{\text{Ar}}^2}

Substitute values:

M_{\text{Ar}}=40,\quad M_{\text{Xe}}=140,\quad r_{\text{Ar}}=0.07\text{ nm},\quad r_{\text{Xe}}=0.10\text{ nm}

So,

\frac\tau_{\text{Ar}}{\tau_{\text{Xe}}} = \sqrt{\frac{40}{140}}\cdot \left(\frac{0.10}{0.07}\right)^2

Now,

\sqrt{\frac{40}{140}} = \sqrt{\frac{2}{7}} \approx 0.535

and

\left(\frac{0.10}{0.07}\right)^2 = \left(\frac{10}{7}\right)^2 = \frac{100}{49} \approx 2.041

Hence,

\frac\tau_{\text{Ar}}{\tau_{\text{Xe}}} \approx 0.535 \times 2.041 \approx 1.09

So the inverse ratio is

\frac{\tau_{\text{Xe}}}{\tau_{\text{Ar}}} \approx \frac{1}{1.09} \approx 0.92

This does not match any option.

Check likely intended interpretation

Sometimes such questions incorrectly use

\tau \propto \frac{1}{r\bar v}

which would give

\frac{\tau_{\text{Ar}}}{\tau_{\text{Xe}}} = \sqrt{\frac{40}{140}}\cdot \frac{0.10}{0.07} \approx 0.535 \times 1.429 \approx 0.76

Again, no option.

If instead ratio is taken as

\frac{\tau_{\text{Xe}}}{\tau_{\text{Ar}}} = \sqrt{\frac{140}{40}}\cdot \left(\frac{0.07}{0.10}\right)^2 \approx 1.871 \times 0.49 \approx 0.92

Still no option.

Thus, using the standard kinetic theory formula, the ratio should be closest to $1.09$ , which is absent from the choices.

Comparison with options

Given options are $2.3, 1.83, 4.67, 3.67$ . None is close to the physically correct result $\approx 1.09$ .

Therefore, the stored answer $\text{B} = 1.83$ appears inconsistent with standard theory.

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