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Q12.2:
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, (0 °\text{C})). Show that it is 22.4 litres.

Solution :

Step 1: Establishing the Ideal Gas Equation

To determine the molar volume of an ideal gas at Standard Temperature and Pressure (STP), we utilize the Ideal Gas Law [Per the Ideal Gas Law equation $PV = nRT$]:

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

Step 2: Defining Given Constants and Variables

We are given the following conditions for STP (Standard Temperature and Pressure):

  • Number of moles ($n$): $1 \text{ mol}$
  • Pressure ($P$): $1 \text{ atm} = 1.01325 \times 10^5 \text{ Pa}$ (or $\text{N/m}^2$)
  • Temperature ($T$): $0^\circ\text{C} = 273.15 \text{ K}$
  • Universal Gas Constant ($R$): $8.314 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$

Step 3: Calculating Molar Volume ($V_m$)

Substituting the values into the rearranged Ideal Gas Law equation:

$V_m = \frac{(1 \text{ mol}) \times (8.314 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \times (273.15 \text{ K})}{1.01325 \times 10^5 \text{ Pa}}$

Performing the arithmetic multiplication in the numerator:

$Numerator = 8.314 \times 273.15 = 2270.9681 \text{ J}$

Dividing by the pressure:

$V_m = \frac{2270.9681}{101325} \text{ m}^3$

$V_m \approx 0.02241 \text{ m}^3$

Step 4: Unit Conversion to Litres

Given the conversion factor where $1 \text{ m}^3 = 1000 \text{ litres}$ [Per SI unit definition]:

$V_m = 0.02241 \text{ m}^3 \times 1000 \frac{\text{L}}{\text{m}^3}$

$V_m = 22.41 \text{ litres}$

Visual Representation of Molar Volume

1 mol of Ideal Gas Volume = 22.4 L Cube of side length ≈ 0.282 m

Final Solution: The molar volume of an ideal gas at STP is calculated to be approximately 22.4 litres.


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