Answer to Question #160221 in Molecular Physics | Thermodynamics for Rashinda

Question

An ideal gas is maintained at constant pressure. If the temperature of the gas is increased from 200K to 600K, what happens to the rms speed of the molecules?
(a) It increases by a factor of 3.
(b) It remains the same
(c) It is one-third the original speed
(d) It is 3^(1/2) times the original speed.

Answer 1

Let's write the rms speed of the molecules at temperature $T_1$:

$$v_{rms, 1}=\sqrt{\dfrac{3RT_1}{M}}.$$

Let's write the rms speed of the molecules at temperature $T_2$:

$$v_{rms, 2}=\sqrt{\dfrac{3RT_2}{M}}.$$

Let's divide the first expression by the second one:

$$\dfrac{v_{rms, 1}}{v_{rms, 2}}=\sqrt{\dfrac{T_1}{T_2}}=\sqrt{\dfrac{200\ K}{600\ K}}=\dfrac{1}{\sqrt{3}},$$ $$v_{rms, 2}=\sqrt{3}v_{rms, 1}.$$

As we can see from the calculations, if the temperature of the gas is increased from 200K to 600K the rms speed of the molecules is $\sqrt{3}$ times the original speed. Therefore, the correct answer is (d).

Answer:

(d) It is $\sqrt{3}$ times the original speed.