Question

The rms velocity of hydrogen is \(\sqrt{7}\) times the rms velocity of nitrogen. If \(T\) 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) \(\mathrm{T}\left(\mathrm{H}_{2}\right)=\sqrt{7} \mathrm{~T}\left(\mathrm{~N}_{2}\right)\)

Short answer

(c) \( \mathrm{T}(\mathrm{H}_2) < \mathrm{T}(\mathrm{N}_2) \).

Step-by-step solution

Understanding RMS Velocity

The Root Mean Square (RMS) velocity is given by the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \) where \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the molar mass of the gas.

Formula for RMS Velocity Ratio

According to the problem, the RMS velocity of hydrogen is \( \sqrt{7} \) times the RMS velocity of nitrogen. Therefore, we can write: \( v_{rms}(H_2) = \sqrt{7} \cdot v_{rms}(N_2) \).

Express Each RMS Velocity in Terms of Temperature and Mass

For hydrogen: \( v_{rms}(H_2) = \sqrt{\frac{3kT(H_2)}{m(H_2)}} \). For nitrogen: \( v_{rms}(N_2) = \sqrt{\frac{3kT(N_2)}{m(N_2)}} \).

Set Up the Equation

Plug the expressions for \( v_{rms}(H_2) \) and \( v_{rms}(N_2) \) into the equation: \( \sqrt{\frac{3kT(H_2)}{m(H_2)}} = \sqrt{7} \cdot \sqrt{\frac{3kT(N_2)}{m(N_2)}} \).

Simplify the Equation

Remove the square roots by squaring both sides: \( \frac{3kT(H_2)}{m(H_2)} = 7 \cdot \frac{3kT(N_2)}{m(N_2)} \). This simplifies to \( \frac{T(H_2)}{T(N_2)} = 7 \cdot \frac{m(H_2)}{m(N_2)} \).

Compare Molar Masses

Hydrogen's molar mass \( m(H_2)=2 \) and nitrogen's molar mass \( m(N_2)=28 \). Substitute these values: \( \frac{T(H_2)}{T(N_2)} = 7 \cdot \frac{2}{28} \).

Solve for the Temperature Ratio

Simplify the fraction: \( \frac{T(H_2)}{T(N_2)} = \frac{7 \times 2}{28} = \frac{14}{28} = \frac{1}{2} \). Hence, \( T(H_2) = \frac{1}{2} T(N_2) \).

Determine the Correct Option

The ratio \( T(H_2) < T(N_2) \) matches option (c).

Key concepts

RMS Velocity

The Root Mean Square (RMS) velocity is a useful concept in kinetic theory. It helps us understand how fast molecules in a gas are moving, on average. Imagine it as the average speed of particles, but taking into account the direction as well. It's calculated using the formula:

• \( v_{rms} = \sqrt{\frac{3kT}{m}} \)

Here, \(k\) is the Boltzmann constant, \(T\) is the temperature in Kelvin, and \(m\) is the molar mass of the gas.
RMS velocity not only depends on how heavy the gas molecules are but also on the temperature of the gas. Higher temperatures increase the RMS velocity because particles move faster when they're warmer.
In the exercise, we see that the RMS velocity of hydrogen given is \(\sqrt{7}\) times that of nitrogen. This means hydrogen particles are moving considerably faster than nitrogen.

Temperature Dependence

Temperature plays a crucial role in determining the speed at which gas molecules move. In the RMS velocity formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), temperature \(T\) is directly proportional to the square of the velocity.
As the temperature increases, the kinetic energy of the gas particles increases as well. This means particles move faster, and consequently, the RMS velocity increases.

• Higher temperature = faster moving particles
• Lower temperature = slower moving particles

In the given exercise, to find the temperature relationship between hydrogen \((H_2)\) and nitrogen \((N_2)\), we understand that the equation is dependent on the temperature difference, leading to the conclusion: \(T(H_2) = \frac{1}{2} T(N_2)\). This means that for hydrogen to maintain a higher RMS velocity than nitrogen, the temperature of hydrogen must be lower than that of nitrogen.

Molar Mass

Molar mass refers to the mass of one mole of a substance, commonly expressed in grams per mole. It is a significant factor in calculating RMS velocity as seen in the equation \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where the molar mass \(m\) is in the denominator.
The relationship is inversely proportional, meaning:

• Lighter molecules (small \(m\)) move faster - hence larger RMS velocity
• Heavier molecules (large \(m\)) move slower - resulting in smaller RMS velocity

In the specific problem, hydrogen's molar mass \(m(H_2)=2\) g/mol is significantly lower than nitrogen's molar mass \(m(N_2)=28\) g/mol.
This large difference means hydrogen molecules are much faster than nitrogen molecules at the same temperature, which aligns with why hydrogen shows a greater RMS velocity compared to nitrogen in the given problem.