# Answer to Question #28809 in Molecular Physics | Thermodynamics for Abhishek

Question #28809

One mole of an ideal gas at standard temprature and pressure occupies 22.4L(molar volume).What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of Hydrogen molecule to be about 1 A ( one angstrom).

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Expert's answer

A mole of anything contains an Avogadro constant ofconstituent particles, which is about 6.022 * 10^23 particles/mole.

Thus, in one mole of Hydrogen gas there is this many H2 molecules.

If diameter of each is about d = 1 A, the volume of onemolecule is (assuming spherical shape)

V = (4/3)*pi*(d/2)^3 = (4/3)*pi*(0.5 * 10^(-10) m)^3 = 5.236* 10^(-31) m^3.

To get the atomic (more correctly, molecular) volume of amole of H2, we'll multiply this by the number of molecules in a mole:

V_a = 5.236 * 10^(-31) (m^3 / particle) * 6.022 * 10^23(particles/mole) = 3.153 * 10^(-7) m^3 = 3.153 * 10^(-4) L

The molar volume constant is V_m = 22.4 L.

The ratio is

V_m / V_a = (22.4 L) / ( 3.153 * 10^(-4) L) = 71043.

Answer: about 70`000.

Thus, in one mole of Hydrogen gas there is this many H2 molecules.

If diameter of each is about d = 1 A, the volume of onemolecule is (assuming spherical shape)

V = (4/3)*pi*(d/2)^3 = (4/3)*pi*(0.5 * 10^(-10) m)^3 = 5.236* 10^(-31) m^3.

To get the atomic (more correctly, molecular) volume of amole of H2, we'll multiply this by the number of molecules in a mole:

V_a = 5.236 * 10^(-31) (m^3 / particle) * 6.022 * 10^23(particles/mole) = 3.153 * 10^(-7) m^3 = 3.153 * 10^(-4) L

The molar volume constant is V_m = 22.4 L.

The ratio is

V_m / V_a = (22.4 L) / ( 3.153 * 10^(-4) L) = 71043.

Answer: about 70`000.

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