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An ensemble of gas particles consists of 1100 entities. The distribution of their speed is as follows:

1000 \text { particles with a speed of } 100 \mathrm{~m} / \mathrm{s} \text { each, }

2000 \text { particles with a speed of } 200 \mathrm{~m} / \mathrm{s} \text { each, }

4000 \text { particles with a speed of } 300 \mathrm{~m} / \mathrm{s} \text { each, }

3000 \text { particles with a speed of } 400 \mathrm{~m} / \mathrm{s} \text { each, }

1000 \text { particles with a speed of } 500 \mathrm{~m} / \mathrm{s} \text { each, }

Calculate the average speed and root mean square error (rms) speed of the gas ensemble.

Option: 1

758.60m/s


Option: 2

2175.34m/s


Option: 3

256.10m/s


Option: 4

15.20m/s


Answers (1)

best_answer

\text { Let } n_i \text { represent the number of particles with speed } v_i \text {, where } i \text { indicates the }

speed category. The total number of particles, N, is 1100

\text { The average speed } \bar{v} \text { can be calculated using the formula: }\bar{v}=\frac{1}{N} \sum_i n_i v_i

Substituting the given values, we have

\begin{aligned} \bar{v} & =\frac{1}{1100}(1000 \cdot 100+2000 \cdot 200+4000 \cdot 300+3000 \cdot 400+1000 \cdot 500) \\ & =\frac{1}{1100} \cdot 2200000 \\ & \approx 2000 \mathrm{~m} / \mathrm{s} . \end{aligned}

\text { The root mean square (rms) speed } v_{\text {rms }} \text { can be computed using the formula: }v_{\mathrm{rms}}=\sqrt{\frac{1}{N} \sum_i n_i v_i^2}

Substituting the given values, we get

\begin{aligned} v_{\mathrm{rms}} & =\sqrt{\frac{1}{1100}\left(1000 \cdot 100^2+2000 \cdot 200^2+4000 \cdot 300^2+3000 \cdot 400^2+1000 \cdot 500^2\right)} \\ & =\sqrt{\frac{1}{1100} \cdot 5206000000} \\ & \approx \sqrt{4732727.27} \\ & \approx 2175.34 \mathrm{~m} / \mathrm{s} . \end{aligned}Hence, the average speed of the gas ensemble is approximately 2000 m/s, and the root mean square speed is approximately 2175.34 m/s. Therefore, the correct option is 2.

 

Posted by

HARSH KANKARIA

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