Avogadro’s Law

1.0Statement of Avogadro’s Law

• According to this law ,at the same temperature and pressure equivalent volume of all gases contains equal number of molecules i.e. N1=N2 N_1 = N_2 if P,V and T are the same.
• The volume of a given gas is directly related to the amount of gas,when the temperature and pressure of the gas remain constant,equal volumes of gases under matching conditions contain the same number of entities.

2.0Formula For Avogadro’s Law

• If the temperature and pressure are kept constant, the volume of the gas will change in direct proportion to the number of moles of the gas.

$V\propto nV=kn$

• The constant k depends on the specific conditions (temperature and pressure) of the gas.
•  Increase the number of moles of gas, the volume increases proportionally, provided temperature and pressure are constant.

3.0Graphical Representation of Avogadro’s Law

4.0Derivation of Avogadro’s Law

• According to  ideal gas equation,

PV=nRT

$V=\frac{nRT}{P}$

Temperature and Pressure are constant

$V=\left(\frac{RT}{P}\right)n$

$V=kn\Rightarrow V\propto n$

$\frac{V}{n}=\text{Constant}$

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

At unchanged temperature and pressure, the volume of a gas rises as the number of gas molecules (or moles) increases.

5.0Real Life Applications of Avogadro’s Law

1. Molar Volume: At STP, one mole of an ideal gas occupies 22.4 liters, making it easy to calculate the gas volume from the moles or vice versa.
2. Gas Volume Measurement: Understanding the link between moles and volume ensures precise gas measurements in labs and industrial processes.
3. Gas Stoichiometry: Avogadro's Law helps predict gas volumes in chemical reactions, like determining the oxygen needed to burn a given volume of fuel in combustion reactions.

6.0Molar Volume of Gas

• The molar volume of a given gas is the volume contained by 1 mole of an ideal gas at (STP). The concept of molar volume is crucial in understanding how gases behave under different conditions and allows us to convert between moles and volumes of gases.

Standard Temperature and Pressure (STP)

Standard Temperature T=0℃ Or 273.156 K

Standard Pressure P=1 atmosphere atm or 1.013 kPa

At STP, the molar volume of an ideal gas is a constant value, which is approximately:

Molar Volume=22.4 L/mol

This indicates that 1 mole of any ideal gas occupies 22.4 liters at STP.

7.0Importance of Molar Volume

• The molar volume is helpful because it allows you to relate the amount of gas (in moles) to the volume it occupies, assuming the gas behaves ideally. For instance, knowing the molar volume at STP helps you calculate the volume of gas if you know the number of moles, or vice versa.

VolumeL=Number of molesn ✕  Molar Volume22.4 L/mol

8.0Solved Examples

1. At a particular temperature and pressure ,4 moles of Nitrogen occupy a volume of 30 litres.Then find volume occupied by 16 moles of oxygen at same temperature and pressure?

Solution:

From Avogadro’s Law

$V\propto n$

$\frac{V_1}{N_1}=\frac{V_2}{N_2}\Rightarrow \frac{30}{4}=\frac{V_2}{16}\Rightarrow V_2=120L$

2. How does the concept of Avogadro's number relate to Avogadro’s Law?

Solution:

Avogadro's number ($6.022✕10^{23}$ molecules/mol) represents the number of molecules or atoms in one mole of a substance. Avogadro’s Law states that the volume of a given gas is directly related to the number of moles, with 1 mole containing exactly $6.022✕10^{23}$ molecules. This connects the microscopic scale (individual molecules) to the macroscopic scale (volume), allowing us to relate the number of molecules in a gas to its volume.

3. A balloon with a volume of 34.5 L is filled with 3.2 moles of Helium gas.To what value will the balloon expand if another 8g of Helium is added.(Assume pressure and temperature does not change)

Solution:

Initial Volume V1=34.5 L

Initial amount of gas=3.2 mol

final amount of gas $n_2=n_1+n_{\text{He added}}$

Mass of Helium Added, $m_{\text{He added}}=4.00g/mol$

The pressure and temperature remains constant

$n_{\text{He added}}=m_{\text{He added}}\times \frac{1mol}{4.00g}=8g\times \frac{1mol}{4.00g}=2.0mol$

$n_2=n_1+n_{\text{He added}}=3.2+2.0=5.2mol$

$V_2=\frac{V_1{n}_2}{n_1}=\frac{34.5L\times 5.2mol}{3.2mol}=56L$

4. A party balloon has 2.50 mol of Helium gas in it at S.T.P.What is the volume of the balloon?

Solution:

$V=n\times \frac{22.4L}{1mol}$

$2.50mol\times \frac{22.4L}{1mol}=56L$

5. A sample of helium at STP has a mass of 32 g.What is the volume this mass of gas occupies?

Solution:

$n_{\text{He}}=m_{\text{He}}\times \frac{1mol}{4.00g}=32\times \frac{1mol}{4.00g}=8mol$

$V_{\text{He}}=n\times \frac{24.8L}{1mol}=8mol\times 24.8L/mol=198L$