A 25 × 10–3 m3 volume cylinder is filled with 1 mol of O2 gas at room temperature (300K). The molecular diameter of O2, and its root mean square speed, are found to be 0.3 nm, and 200 m/s, respectively. What is the average collision rate (per second) for an O2 molecule ?

Heat and Thermodynamics PYQ — JEE Main 2019 · 10 Apr · Shift 1 · Q65 with Solution

View written solutionFree

Correct answer: D

Use the mean collision rate formula

For a gas molecule, the average collision rate is

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

where $\lambda$ is the mean free path.

For hard-sphere molecules,

\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}

So,

z = \bar v\,\sqrt{2}\,\pi d^2 n

Here:

molecular diameter: $d = 0.3\,\text{nm} = 0.3 \times 10^{-9}\,\text{m} = 3 \times 10^{-10}\,\text{m}$
volume: $V = 25 \times 10^{-3}\,\text{m}^3$
number of molecules in $1$ mol: $N = N_A = 6.022 \times 10^{23}$
number density:

n = \frac{N}{V} = \frac{6.022 \times 10^{23}}{25 \times 10^{-3}}

Compute number density

n = \frac{6.022 \times 10^{23}}{2.5 \times 10^{-2}} = 2.41 \times 10^{25}\,\text{m}^{-3}

Compute collision cross-sectional factor

d^2 = (3 \times 10^{-10})^2 = 9 \times 10^{-20}\,\text{m}^2

Then,

\pi d^2 n = \pi \times 9 \times 10^{-20} \times 2.41 \times 10^{25}

= \pi \times 21.69 \times 10^5 approx 6.81 \times 10^6

Now multiply by $\sqrt{2}$ :

\sqrt{2}\,\pi d^2 n \approx 1.414 \times 6.81 \times 10^6 \approx 9.63 \times 10^6

Collision rate using the given speed

The question gives rms speed $v_{\text{rms}} = 200\,\text{m/s}$ . Taking this as the characteristic speed,

z \approx v_{\text{rms}}\,\sqrt{2}\,\pi d^2 n

z \approx 200 \times 9.63 \times 10^6 = 1.93 \times 10^9\,\text{s}^{-1}

This is of order

\boxed{10^9 \text{ to } 10^{10}\,\text{s}^{-1}}

So the nearest option is D.

Check via mean free path

\lambda = \frac{1}{9.63 \times 10^6} \approx 1.04 \times 10^{-7}\,\text{m}

Then,

z = \frac{200}{1.04 \times 10^{-7}} \approx 1.92 \times 10^9\,\text{s}^{-1}

Same result.

Comparison with stored answer

Stored correct answer is B ( $\sim 10^{12}$ ), but the calculation from the given data gives a value near $10^{9}$ – $10^{10}$ , so the nearest option is D.

Hence, I do not agree with the stored answer.

A likely issue is that either:

the given molecular speed should have been much larger (realistic rms speed for $\mathrm{O_2}$ at $300\,\text{K}$ is about $\sim 500\,\text{m/s}$ , still giving only order $10^{10}$ ), or
some data/options were intended differently.

Using the given numbers consistently, the answer is D.

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View written solutionFree

Correct answer: D

Use the mean collision rate formula

For a gas molecule, the average collision rate is

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

where $\lambda$ is the mean free path.

For hard-sphere molecules,

\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}

So,

z = \bar v\,\sqrt{2}\,\pi d^2 n

Here:

molecular diameter: $d = 0.3\,\text{nm} = 0.3 \times 10^{-9}\,\text{m} = 3 \times 10^{-10}\,\text{m}$
volume: $V = 25 \times 10^{-3}\,\text{m}^3$
number of molecules in $1$ mol: $N = N_A = 6.022 \times 10^{23}$
number density:

n = \frac{N}{V} = \frac{6.022 \times 10^{23}}{25 \times 10^{-3}}

Compute number density

n = \frac{6.022 \times 10^{23}}{2.5 \times 10^{-2}} = 2.41 \times 10^{25}\,\text{m}^{-3}

Compute collision cross-sectional factor

d^2 = (3 \times 10^{-10})^2 = 9 \times 10^{-20}\,\text{m}^2

Then,

\pi d^2 n = \pi \times 9 \times 10^{-20} \times 2.41 \times 10^{25}

= \pi \times 21.69 \times 10^5 approx 6.81 \times 10^6

Now multiply by $\sqrt{2}$ :

\sqrt{2}\,\pi d^2 n \approx 1.414 \times 6.81 \times 10^6 \approx 9.63 \times 10^6

Collision rate using the given speed

The question gives rms speed $v_{\text{rms}} = 200\,\text{m/s}$ . Taking this as the characteristic speed,

z \approx v_{\text{rms}}\,\sqrt{2}\,\pi d^2 n

z \approx 200 \times 9.63 \times 10^6 = 1.93 \times 10^9\,\text{s}^{-1}

This is of order

\boxed{10^9 \text{ to } 10^{10}\,\text{s}^{-1}}

So the nearest option is D.

Check via mean free path

\lambda = \frac{1}{9.63 \times 10^6} \approx 1.04 \times 10^{-7}\,\text{m}

Then,

z = \frac{200}{1.04 \times 10^{-7}} \approx 1.92 \times 10^9\,\text{s}^{-1}

Same result.

Comparison with stored answer

Stored correct answer is B ( $\sim 10^{12}$ ), but the calculation from the given data gives a value near $10^{9}$ – $10^{10}$ , so the nearest option is D.

Hence, I do not agree with the stored answer.

A likely issue is that either:

the given molecular speed should have been much larger (realistic rms speed for $\mathrm{O_2}$ at $300\,\text{K}$ is about $\sim 500\,\text{m/s}$ , still giving only order $10^{10}$ ), or
some data/options were intended differently.

Using the given numbers consistently, the answer is D.

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