Question

A gas is made up of hydrogen and oxygen

molecules.
a) Which molecules moves faster?

b) Find the ratio of the velocities of

hydrogen and oxygen molecules.

Answer 1

a) Hydrogen because of its less density
b) The RMS velocity is inversely proportional to the square root of the molar mass of each gas, so the ratio requested is:
(1/√O) / (1/√H) = √H / √O = √(H/O) = √(1.007947 / 15.99943) =
0.2509959

Answer 2

Concept:

Root mean square velocity, $v_{rms}$ is given by,

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

where R = Gas constant

           T = Temperature

            m = mass

Given:

A gas is made up of hydrogen and oxygen molecules.

Mass of hydrogen molecule i.e., $H_2 (m_H) = 2X1 = 2 g/mole$

Mass of oxygen molecule i.e., $O_2 (m_O) = 2X16 = 32g/mol$

Find:

a) Which molecules move faster?

b) Find the ratio of the velocities of hydrogen and oxygen molecules.

Solution:

a) As we know,

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

So, according to the above equation, the molecule which has less mass will move faster.

Mass of hydrogen molecule is less than that of oxygen molecule, hence, Hydrogen molecules move faster.

b) For hydrogen molecule,

$v_{{rms}_H}$ = $\sqrt{\frac{3RT}{m_H} }$

$v_{{rms}_H} = \sqrt{\frac{3RT}{2} }$      -------------- (i)

For oxygen molecule,

$v_{{rms}_O} = \sqrt{\frac{3RT}{m_O} }\\v_{{rms}_O} = \sqrt{\frac{3RT}{32} }$             ------------- (ii)

Dividing equation (i) and (ii), we get

$\frac{v_{{rms}_H}}{v_{{rms}_O}} = \frac{\sqrt{\frac{3RT}{2} } }{\sqrt{\frac{3RT}{32} }}$

$\frac{v_{{rms}_H}}{v_{{rms}_O}} = \sqrt{\frac{32}{2} }$

$\frac{v_{{rms}_H}}{v_{{rms}_O}} = \sqrt{16 } = 4$

$v_{{rms}_H} : v_{{rms}_O} = 4:1$

Hence, the ratio of the velocities of hydrogen and oxygen molecules is 4:1.

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