Question

6. Suppose you could suddenly increase the speed of every molecule in a gas by a factor of 2.

a. Would the RMS speed of the molecules increase by a factor of $2^{1/2},2,or{2}^2$? Explain.

b. Would the gas pressure increase by a factor of $2^{1/2},2$or $2^2$? Explain.

Short answer

a) RMS speed of the molecules increases by a factor of $2^{1/2},2$or $2^2$will result in an increase of two times in the RMS speed.

b) The gas pressure increase by a factor of $2^{1/2},2$or $2$will result in an increase of $2^2$in the gas pressure.

Step-by-step solution

Concept Introduction (Part a)

As temperatures increase, the molecules in a gas speed up, causing the root mean square molecular speed to increase as well. In other words, as a sample of gas's temperature increases, the molecules speed up and as a result the root mean square molecular speed increases as well.

Explanation (Part a)

By adding up the squares of all particle speeds and taking the root at the end, we can find the root mean square speed $v_{rms}$:

$v_{\mathrm{rms}}=\sqrt{\sum _i{v}_i^2}$

Change of numbers in every $v_i\to 2v_i$

$v_{\mathrm{rms}}^{\mathrm{\prime }}=\sqrt{\sum _i {\left(2v_i\right)}^2}=\sqrt{4\sum _i v_i^2}=2\sqrt{\sum _i v_i^2}=2v_{\mathrm{rms}}$

So, the speed of RMS will be double.

Final Answer (Part a)

Hence, RMS Speed will be double.

Concept Introduction (Part b) 

Pressure of the molecule is defined as the sum of all the forces applied to a wall divided by the wall's area. Therefore, a gas's pressure is the average linear momentum of its molecules.

Explanation (Part b)

While the RMS speed depends on the square root of the temperature, the pressure mainly relies on the temperature. So, $p\propto v_{\text{rms}}^2$.

We can see that the pressure quadruples when using the result from the previous part.

Final Answer (Part b)

Hence, the pressure of the gas will be increased by a factor$2^2.$