Question

If the temperature of an ideal gas is increased from 200 K to 600 K, 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 of the original speed. (d) It is !3 times the original speed. (e) It increases by a factor of 6.

Answer 1

Answer: (d) It is !3 times the original speed.

Explanation: The rms speed of a gas is related to its temperature by the formulae below;

U(r.m.s) =√(3RT)/M

Where;

T represents the temperature.

R represents the gas constant.

M represents the molar mass of the gas.

Therefore, if the temperature increases from 200k to 600k

The temperature has then increased by a factor of 3,

However, we must note that temperature in the formulae is included in the square-root

Recall,

U(r.m.s) =√(3RT)/M

Consequently, temperature (T) can now be represented by (3T).

The inference drawn from this is that the root-mean-square speed would increase by a factor of √3

Therefore, option (d) is correct.

Answer 2

Complete question:

If the temperature of an ideal gas is increased from 200 K to 600 K, 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 of the original speed. (d) It is $\sqrt{3}$ times the original speed. (e) It increases by a factor of 6.

Answer:

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

Explanation:

Temperature is a macroscopic property of objects that is intrinsically connected with the microscopic world, because is related with the speed of the microscopic particles it is composed, more precisely is related with the average speed of the microscopic particle. On an ideal gas that relation is contained in the Root Mean Square speed or RMS speed of the molecule on the gas, it states:

$v_{rms}=\sqrt{\frac{3RT}{M}}$

With M the molar mass, R the ideal gas constant and T the temperature of the gas. For 200K temperature:

$v_{200}=\sqrt{\frac{3R200}{M}}$(1)

For 600K temperature:

$v_{600}=\sqrt{\frac{3R(600)}{M}}=\sqrt{\frac{3R(3*200)}{M}}=\sqrt{3} \sqrt{\frac{3R(200)}{M}}$(2)

Note that the term $\sqrt{\frac{3R(200)}{M}}$ is the same on (1) so we can write (2) as:

$v_{600}=\sqrt{3}v_{200}$

So, the new rms speed is $\sqrt{3}$ the original speed.