Question

What will be the ratio of the root mean square speeds of the molecules of an ideal gas at 270k and 30k?

Answer 1

Answer: The ratio of the root mean square speeds of the molecules of an ideal gas is 3:1.

Explanation:

Given that,

Temperature $T_{1}= 270\ k$

Temperature $T_{2}= 30\ k$

We know that,

The root mean square speed of the molecule of an ideal gas is

$v_{rms}= \sqrt{\dfrac{3RT}{m}}$

The root mean square is directly proportional to temperature.

For temperature $T_{1}$,

$v_{rms}_{1}= \sqrt{\dfrac{3R\times 270}{m}}$

For temperature $T_{2}$,

$v_{rms}_{2}= \sqrt{\dfrac{3R\times 30}{m}}$

Divided equation (I) by equation (II)

$\dfrac{v_1}{v_{2}}=\sqrt{\dfrac{270}{30}}$

$\dfrac{v_{1}}{v_{2}}= \dfrac{3}{1}$

Hence, The ratio of the root mean square speeds of the molecules of an ideal gas is 3:1.

Answer 2

Given that,

Temperature, $T_1=270\ K$

Temperature, $T_2=30\ K$

To find,

The find the root mean square speeds of the molecules.

Solution,

The relation between root mean square speeds of the molecules and temperature is given by :

$v\propto \sqrt{T}$

So,

$\dfrac{v_1}{v_2}=\sqrt{\dfrac{T_1}{T_2}} \\\\\dfrac{v_1}{v_2}=\sqrt{\dfrac{270}{30}} \\\\\dfrac{v_1}{v_2}=3$

So, the ratio of the root mean square speeds of the molecules is 3:1.

Learn more,

RMS speed

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