Understanding the Kinetic Molecular Theory (KMT)
The Kinetic Molecular Theory (KMT) is a model used to explain the behavior of gases. It assumes that gases consist of a large number of tiny particles in constant, random motion.
Key Postulates of KMT:
- Particle Size: Gas particles are so small compared to the distances between them that their volume is considered negligible.
- Motion: Gas particles move in straight lines until they collide with each other or the container walls.
- Collisions: All collisions between particles are perfectly elastic (no kinetic energy is lost).
- Forces: There are no attractive or repulsive forces between gas particles.
- Kinetic Energy: The average kinetic energy of gas particles is directly proportional to the absolute temperature.
Why Gases Deviate from Ideal Behavior
Ideal gases follow the Ideal Gas Law ($PV=nRT$) strictly. However, real gases deviate under two conditions:
- High Pressure: At high pressure, particles are squeezed together, so the volume they occupy is no longer negligible compared to the total volume.
- Low Temperature: At low temperatures, gas particles move slowly. Attractive intermolecular forces (which are ignored in KMT) become significant, causing the gas to condense or exert lower pressure than expected.
Problem Solving: Graham's Law of Diffusion
Problem: A saturated hydrocarbon $C_nH_{2n+2}$ diffuses twice as fast as $SO_2$. Find the volume occupied by the hydrocarbon at $27^\circ C$ (300 K) and 2 atm pressure (Assuming standard moles $n=1$ for calculation).
Step 1: Identify Molar Masses
- $M_{SO_2} = 32 + (16 \times 2) = 64 \text{ g/mol}$
- Let the hydrocarbon be $X$. According to Graham's Law: $\frac{r_X}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_X}}$
- Given $\frac{r_X}{r_{SO_2}} = 2$, then $2 = \sqrt{\frac{64}{M_X}}$
- $4 = \frac{64}{M_X} \Rightarrow M_X = 16 \text{ g/mol}$
Step 2: Identify the Hydrocarbon
- $M(C_nH_{2n+2}) = 12n + (2n+2) = 14n + 2$
- $14n + 2 = 16 \Rightarrow 14n = 14 \Rightarrow n = 1$
- The hydrocarbon is Methane ($CH_4$).
Step 3: Calculate Volume
- Using $PV = nRT$, $V = \frac{nRT}{P}$
- $V = \frac{1 \times 0.0821 \times 300}{2} = 12.315 \text{ Liters}$.