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Suppose a sample of gas is composed of 100 molecules, with speeds given below. Speed (m/s) Number 200 23 400 42 600 26 800 8 1000 1 According to the kinetic molecular theory, if the absolute temperature of the gas is halved, what is a reasonable estimation of the new distribution of speeds of the molecules

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Answer:

The answer is:

[tex]speed(\frac{m}{s}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (number)\\\\ 100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 23 \\\\ 200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 42 \\\\ 300 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 26 \\\\ 400 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8 \\\\ 500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \\\\[/tex]

Explanation:

Its total kinetic energy of a gas colloquial atom colloction is equal to absolute temperature alone, according to the programs that directly gas Theory.

Average speed for 1st option:

[tex]= \frac{(200 \times 23)+(400 \times 42)+(600 \times 26)+(800 \times 8) +1000}{100} \\\\= 444\ \frac{m}{s} \ \ \ \ \ \ \ \ \ \text{average speed of gas}[/tex]

 Average speed for 2nd option:

 [tex]= \frac{(200 \times 11) +(400 \times 21)+ (600 \times 13)+ (800 \times 4)}{11+21+13+4} \\\\= 440 \ \frac{m}{s}[/tex]

 Average speed for 3rd option:

[tex]= \frac{(200 \times 43) +(400 \times 50)+ (600 \times 7)}{100} \\\\ =328 \ \frac{m}{s}[/tex]

 Average speed for 4th option:

 [tex]= \frac{(200 \times 50) +(400 \times 25)+ (600 \times 13)+(800\times 8)+(1000 \times 4)}{100}\\\\ =382\ \frac{m}{s}[/tex]

  Average speed for 5th option:

[tex]= \frac{(100 \times 23) +(200 \times 42)+ (300 \times 26)+(400\times 8)+500}{100}\\\\ =222 \ \frac{m}{s}[/tex]

The average speed also increased by half because the temperatures had kinetic energy.[tex]\to 444 \ \frac{m}{s}\ to \ 222 \ \frac{m}{s}[/tex]

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