__Avogadro Law__

- It states that equal volumes of all gases under the same conditions of temperature and pressure

contain equal number of molecules. Mathematically,

__Ideal Gas Equation__

- The gases following Boyle’s law, Charles’law and Avogadro law firmly are termed as anideal gas.
- From Boyle’s law we have V ∝ 1/p (Constant T, n)
- From Charles’ law we have V ∝ T (Constant p, n)
- From Avogadro’s law we have V ∝ n (Constant T, n)
- Combining them we get, V ∝ nT/p
- or V = nRT/p

**pV = nRT** ………………………… (IDEAL GAS EQUATION)

R =pV/nT

R= Gas Constant. It is same for all gases.

**Problem:**

Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure.

**Solution:**

Using Ideal gas equation, pV = nRT ……….. (1)

p → Pressure of gas

V → Volume of gas

n→ Number of moles of gas

R → Gas constant

T → Temperature of gas

From equation (1),

p = n RT/V

n = Mass of Gas (m) /Molar Mass of gas (M)

Putting value of n in the equation, we have

p = m RT/ MV ————(2)

Density (ρ) = m /V —————-(3)

Putting (3) in (2) we get

p = ρ RT / M

Or ρ = PM / RT

**Problem:**

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm^{3} at 27°C. R = 0.083 bar dm3 K^{–1} mol^{–1}.

**Solution:**

m (Oxygen) = 8 g,

M (Oxygen) = 32 g/mol

m ( Hydrogen) = 4 g,

M (Hydrogen) = 2 g/mol

n (Amount of oxygen) = 8/ 32 = 0.25 mol

n (Amount of hydrogen) = 4/2 = 2 mol

According to ideal gas equation,

PV = n RT,

P X 1 = (0.25 + 2) X 0.083 **X** 300 = 56.02 bar

Total pressure of the mixture is 56.02 bars.

**Problem:**

Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar L K–1 mol^{-1}.

**Solution:**

According to ideal gas equation, PV = nRT

But n = Mass of Gas (m)/ Molar mass of Gas (M)

PV = (m/M) RT

For CO_{2} , M = 44 g/mol

Putting the values

1 x V = (8.8 **/** 44) x 0.083 x 304.1

= 5.05 L

Volume occupied is 5.05 L.

__Density and Molar Mass of a Gaseous Substance__

- As per ideal gas equation,

pV = nRT

Or n/V = p/RT

- As we know n = Mass of Gas (m) / Molar Mass of Gas (M) = m/M

So, m/MV = p/RT

d/M= p/RT

- Or Molar mass = M = dRT/p

**Problem:**

At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

**Solution:**

Using the formula ρ = Mp/RT

or p = ρ RT/ M

According to the question,

For certain oxide gas

2 = ρ RT/ M ……………(1)

For nitrogen

5 = ρ RT/28 ………………….(2)

From equation (1) & (2), we get

5/2 = M/28

Or

M = 5 X 28/ 2 = 70g/mol

**Problem:**

Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP?

**Solution:**

According to the ideal gas

Density ρ = p x M/ RT

According to the question,

5.46 = 2 x M/ 300 x R ……… (1)

ρ_{STP} = 1 x M/ 273 x R ……….. (2)

From equation (1) & (2),

ρ_{STP = }(1 x M/ 273 x R) x (300 x R/ 2 x M) = 300** /** 273 X 2 = 3.00 g/dm^{3}

Density of the gas at STP will be 3 g dm^{–3}.

**Problem:**

Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

**Solution:**

m ( Nitrogen) = 1.4 g

M (Nitrogen) = 28 g mol^{–1}

n (Amount of nitrogen) = 1.4 / 28 = 0.05 mol

Hence number of nitrogen molecule in 0.05 mol = 6.023 x 10^{23} x 0.05 = 3.0115 x 10^{22}

Number of electrons in one molecule of nitrogen = 14

Total number of electrons in 1.4 g of nitrogen = 3.0115 x 10^{22} x 14 = 4.22 x 10^{23}