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  • An ensemble of gas particles consists of 1100 entities. Their speed distribution is given as follows: • 1000 particles with a speed of 100 m/s each, • 2000 particles

An ensemble of gas particles consists of 1100 entities. Their speed distribution
is given as follows:
• 1000 particles with a speed of 100 m/s each,
• 2000 particles with a speed of 200 m/s each,
• 4000 particles with a speed of 300 m/s each,
• 3000 particles with a speed of 400 m/s each,

• 1000 particles with a speed of 500 m/s each.
Determine the average speed and the root mean square (rms) speed of the
gas ensemble.

Option: 1

1656 m/s
 


Option: 2

17.4 m/s
 


Option: 3

27.50 m/s
 


Option: 4

2175.34 m/s


Answers (1)

best_answer

Let n_i represent the number of particles with speed v_i, where i indicates the speed category. The total number of particles, N, is 1100.
The average speed  \bar{v} 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}

The root mean square (rms) speed v_{rms} can be determined using the formula:

v_{\mathrm{rms}}=\sqrt{\frac{1}{N} \sum_i n_i v_i^2} ..
Substituting the given values, we find:

\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}Consequently, 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 4.

Posted by

Shailly goel

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