Problem 103 - A Complete Resource Book in Chemistry for JEE Main

Question

The \(\mathrm{ms}\) velocity of hydrogen is \(\sqrt{7}\) times the \(\mathrm{rms}\) velocity of nitrogen. If \(\mathrm{T}\) is the temperature of the gas (a) \(\mathrm{T}\left(\mathrm{H}_{2}\right)=\mathrm{T}\left(\mathrm{N}_{2}\right)\) (b) \(\mathrm{T}\left(\mathrm{H}_{2}\right)>\mathrm{T}\left(\mathrm{N}_{2}\right)\) (c) \(\mathrm{T}\left(\mathrm{H}_{2}\right)<\mathrm{T}\left(\mathrm{N}_{2}\right)\) (d) \(\mathrm{T}\left(\mathrm{H}_{2}\right)=\sqrt{7} \mathrm{~T}\left(\mathrm{~N}_{2}\right)\)

Step-by-Step Solution

Answer

(c) \( T(H_2) < T(N_2) \)

Step 1: Understand RMS Velocity Formula

The root mean square (rms) velocity of a gas is calculated using the formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

Step 2: Express Given Condition Mathematically

According to the problem, the \( ms \) velocity of hydrogen \( (v_{ms, H_2}) \) is \( \sqrt{7} \) times the \( rms \) velocity of nitrogen \( (v_{rms, N_2}) \). Thus, \( v_{ms, H_2} = \sqrt{7} \times v_{rms, N_2} \).

Step 3: Apply RMS Velocity Formula to Both Gases

For hydrogen, \( v_{ms, H_2} = \sqrt{\frac{3RT(H_2)}{M(H_2)}} \) and for nitrogen, \( v_{rms, N_2} = \sqrt{\frac{3RT(N_2)}{M(N_2)}} \). The molar mass \( M(H_2) = 2 \) and \( M(N_2) = 28 \).

Step 4: Substitute and Simplify

Substitute the expressions from Step 3 into the equation from Step 2: \[ \sqrt{\frac{3RT(H_2)}{2}} = \sqrt{7} \times \sqrt{\frac{3RT(N_2)}{28}} \].

Step 5: Square Both Sides to Eliminate the Square Roots

By squaring both sides, we get \( \frac{3RT(H_2)}{2} = 7 \times \frac{3RT(N_2)}{28} \). Simplify to \( \frac{3RT(H_2)}{2} = \frac{3RT(N_2)}{4} \).

Step 6: Further Simplify and Solve for Temperature Relationship

Cancel \( 3R \) from both sides: \( \frac{T(H_2)}{2} = \frac{T(N_2)}{4} \). Solve for \( T(H_2) \) in terms of \( T(N_2) \): multiply both sides by 2, \( T(H_2) = \frac{2}{4} T(N_2) \). Therefore, \( T(H_2) = \frac{1}{2} T(N_2) \), which implies \( T(H_2) < T(N_2) \).

Key Concepts

rms velocity

The root mean square (rms) velocity is a vital concept when studying the behavior of gases. It helps us determine how fast molecules in a gas are moving on average. The formula used to calculate the rms velocity is given by: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where:

• \( R \): the universal gas constant (8.314 J/mol K)
• \( T \): temperature of the gas in Kelvin
• \( M \): molar mass of the gas in kilograms per mole

This formula shows how the rms velocity depends on temperature and molar mass. When temperature increases, so does the rms velocity, meaning the gas molecules move faster. Conversely, a higher molar mass results in a slower rms velocity, as heavier molecules travel more slowly.

temperature relationship

In the study of gases, understanding how temperature relates to gas properties like velocity is crucial. According to kinetic theory, gas temperature is directly proportional to the average kinetic energy of its molecules. Higher temperature means more energy and faster-moving particles. In this exercise, hydrogen's rms velocity being \( \sqrt{7} \) times that of nitrogen underlines this relationship. When hydrogen gas achieves this condition, its temperature (\( T(H_2) \)) is less than nitrogen's temperature (\( T(N_2) \)). From this, we understand that lighter molecules (like hydrogen) need less energy (or lower temperature) to reach a given rms velocity compared to later molecules (like nitrogen). This illustrates the inverse relationship between temperature and molecular mass in fixing the kinetic properties.

molar mass

Molar mass, an essential concept in chemistry, signifies the mass of one mole of a substance. It's usually expressed in grams per mole or kilograms per mole. This property dramatically influences a gas's rms velocity. The formula for rms velocity, \( \sqrt{\frac{3RT}{M}} \), indicates that higher molar mass results in lower velocity. For example, the molar mass of hydrogen (\( M(H_2) = 2 \) g/mol) is much lower than that of nitrogen (\( M(N_2) = 28 \) g/mol). This means, under the same conditions, hydrogen molecules move faster compared to nitrogen molecules. It's crucial to note the mass inversely affects speed: if you have lighter molecules, expect quicker speeds at a given temperature. This exercise clearly demonstrates this principle by comparing the velocities of hydrogen and nitrogen.

universal gas constant

The universal gas constant, denoted as \( R \), is a cornerstone in the study of gases and their behaviors. Its value is constant, \( 8.314 \) J/(mol·K), and is derived from the ideal gas law, which combines pressure, volume, and temperature into one equation: \( PV = nRT \). In the context of the rms velocity formula, \( R \) helps bridge the macroscopic world of temperature with the microscopic movements of molecules. By understanding \( R \), scientists can predict how gases will behave under different conditions. Every time you apply the rms velocity or make calculations, \( R \) ensures that your results are universally applicable as long as the gases behave ideally. Hence, this constant is indispensable when exploring kinetic theory.