Problem 133 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following Column-I (a) Boyle's law (b) Charles' law (c) Graham's law (d) Ideal gas Column-II (p) \(\log \mathrm{P}=-\log \mathrm{V}+\) constant (q) \(\mathrm{r}=\frac{\mathrm{K} \cdot \mathrm{P}}{\sqrt{\mathrm{M}}}\) (r) \(\mathrm{d}=\frac{\mathrm{PM}}{\mathrm{RT}}\)

Step-by-Step Solution

Answer

Matches: (a) → (p), (b) → None, (c) → (q), (d) → (r).

Step 1: Understanding Boyle's Law

Boyle's Law describes the relationship between the pressure and volume of a gas at constant temperature. It is expressed mathematically as \( PV = \text{constant} \). This matches with expression (p) from Column-II, which simplifies to \( \log P = -\log V + \text{constant} \).

Step 2: Understanding Charles' Law

Charles' Law states that the volume of a gas is directly proportional to its temperature when pressure is constant. It is represented as \( V \propto T \). However, none of the expressions in Column-II explicitly match this relation, suggesting no direct assignment here.

Step 3: Understanding Graham's Law

Graham's Law involves the rates of effusion and is expressed as \( r = \frac{k}{\sqrt{M}} \), where \( r \) is the rate of effusion. Expression (q) in Column-II includes \( \frac{K \cdot P}{\sqrt{M}} \), closely resembling Graham's Law when considering pressure-related adjustments.

Step 4: Understanding Ideal Gas Law

The Ideal Gas Law combines several gas laws into \( PV = nRT \), and can lead to the expression for density \( d = \frac{PM}{RT} \). This matches expression (r) from Column-II, indicating the relationship between pressure, molar mass, and temperature.

Key Concepts

Boyle's Law

Boyle's Law focuses on how pressure and volume in a gas relate to each other when temperature is held constant. Imagine squishing a balloon. As you push down, the pressure increases while the volume decreases. This is the essence of Boyle's Law. Mathematically, it is expressed as \( PV = \text{constant} \). In simple terms, if you know the volume and pressure of a gas at a certain state and then change one, the other shifts to keep their product the same.

• If you double the pressure, the volume halves, and vice versa.
• This relationship helps explain why we need careful pressure adjustments when scuba diving to avoid the dangerous effects of pressure changes on our bodies.
• When linked to the phrase from Column-II: \( \log P = -\log V + \text{constant} \), it's a rearranged form illustrating the inverse relation between pressure and volume.

Charles' Law

Charles' Law tells us that the volume of a gas increases with temperature if pressure remains stable. Picture a hot air balloon rising. When the air inside is heated, it expands, causing the balloon to lift off.

• This law is captured by the formula \( V \propto T \), meaning volume \( V \) is proportional to temperature \( T \) in Kelvin.
• While there was no direct match, understanding this principle is crucial. It explains everyday phenomena like why car tires can expand in the heat.
• Charles' Law is a key part of understanding overall gas behavior under changing temperature conditions.

Graham's Law

Graham's Law focuses on effusion, the process whereby gas particles escape through a tiny hole. An important insight from this law is that lighter gases effuse faster than heavier ones.

• The mathematical representation is \( r = \frac{k}{\sqrt{M}} \), where \( r \) is the rate of effusion, and \( M \) is molar mass.
• Graham’s Law helps explain why helium balloons deflate faster than air-filled ones, as helium atoms are lighter.
• Matching to \( \frac{K \cdot P}{\sqrt{M}} \) from Column-II, this expression incorporates pressure, affecting effusion rates in real-world scenarios.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation encapsulating the behavior of gases by intertwining various simpler gas laws. This includes relationships of pressure, volume, temperature, and the amount of gas in moles.

• Expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature in Kelvin.
• It’s the grand formula that explains why varying any of these conditions affects the others.
• The relation \( d = \frac{PM}{RT} \) explained the density aspect in Column-II, showing how molar mass of a gas connects to its density given pressure and temperature conditions.
• Understanding this law gives a comprehensive view into how gases interact in different environments, helping in industries to optimize processes like ventilation and refrigeration.