### Ideal gas law for gas molecules

The gases are characterized by the lack of definite volume or shape. In the gaseous state, the matter has the properties of filling completely any available space to a uniform density. Low density and high compressibility are also pronounced properties of gases.Boyle's, Charles, and Avogadro's law gives the birth of the ideal gas law for gas molecules. Ideal gas law for n moles ideal gas,

Ideal gas law |

The equation gives us a connection between the pressure, volume, and temperature of a gas. Thus this equation is the equation of state for a gas molecule.

Problem

Why an ideal gas can not be liquefied?

Solution

The molecules to come close to each other for the liquefaction of gas molecules. The intermolecular attraction needed for the liquefaction of gas.

An ideal gas has no intermolecular attraction and hence ideal gas can not be liquefied.

Further putting the condition of the critical state to the ideal gas law.

PV = RT

or, (dP/dV)

At the critical state (dP/dV)

∴ (- RTc/Vc²) = 0

or, Tc = 0⁰ K

Thus for liquefying an ideal gas necessary to bring the temperature below 0⁰ K. Naturally this temperature not possible thus the gas can not be liquefied.or, (dP/dV)

_{T}= - RT/V²At the critical state (dP/dV)

_{T}= 0∴ (- RTc/Vc²) = 0

or, Tc = 0⁰ K

#### How to measure gas pressure?

ProblemHow to measure pressure excreted on the walls of a 3 liter of the flask when 7 grams of nitrogen are introduced into the same at 27°C?

Solution

PV = nRT

or, PV = (g/M)RT

Molecular weight of nitrogen 28 gm mol⁻¹

∴ P = (7/28) × (0.082 × 300)/3

= 2.05 atm

or, PV = (g/M)RT

Molecular weight of nitrogen 28 gm mol⁻¹

∴ P = (7/28) × (0.082 × 300)/3

= 2.05 atm

#### How to measure the density of gas molecules?

Ideal gas law for n mole gas

PV = nRT.

or, P = (g/V) × (RT/M)

∴ P = dRT/M

where d = density.

ProblemPV = nRT.

or, P = (g/V) × (RT/M)

∴ P = dRT/M

where d = density.

How to measure the density of ammonia gas at 100⁰ C, when confined by the pressure of 1600 mm Hg?

Solution

P = dRT/M

or, d = PM/RT

Molecular weight of ammonia = 17 gm mol⁻¹

P = 1600 mm Hg = 1600/760 atm

R = 82.1 cc atm mol⁻¹ K⁻¹

T = (273 +100) K

∴ d = (17 × 1600)/(760 × 82.1 × 373)

Density of ammonia = 0.00117 gm/cc

#### The molecular weight of gas molecules

Ideal gas law

PV = nRT

or PV = (g/M) × RT

where g = weight of the gas

M = molecular weight.

∴ PV = gRT/M

PV = nRT

or PV = (g/M) × RT

where g = weight of the gas

M = molecular weight.

∴ PV = gRT/M

Problem

Calculate the molecular weight of a gas, 12.8 gms of which occupy 10 liters at a pressure of 750 mm and at 27° C?

Solution

31.91 gm mol⁻¹

#### How to calculate the kinetic energy of the gas molecules?

Problem

How to calculate the total kinetic energy of 0.5 mol of an ideal gas at 273 K?

Solution

Total kinetic energy

= n (3RT/2)

= 0.05 {(3/2) × 8.314 × 273}

= 1702 Joule

= n (3RT/2)

= 0.05 {(3/2) × 8.314 × 273}

= 1702 Joule