Question

The temperature at which RMS velocity of \(\mathrm{SO}_{2}\) molecules is half that of the He molecules at \(300 \mathrm{~K}\) is (1) \(150 \mathrm{~K}\) (2) \(600 \mathrm{~K}\) (3) \(900 \mathrm{~K}\) (4) \(1200 \mathrm{~K}\)

Short answer

1200 K

Step-by-step solution

Understand RMS Velocity Formula

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

Determine the RMS Velocity for He at 300 K

For helium (He) at 300 K, let the RMS velocity be \( v_{\text{He}} \). Then, \( v_{\text{He}} = \sqrt{\frac{3k \times 300}{m_{\text{He}}}} \).

Determine the RMS Velocity for SO\textsubscript{2}

To find the temperature where the RMS velocity of \( \mathrm{SO}_2 \) molecules is half that of \( \mathrm{He} \) molecules at 300 K, let the temperature be \( T_{\text{SO}_2} \) and name the RMS velocity as \( v_{\text{SO}_2} \). Then, \( v_{\text{SO}_2} = \frac{1}{2} v_{\text{He}} = \frac{1}{2} \sqrt{\frac{3k \times 300}{m_{\text{He}}}} \).

Substitute into RMS Velocity Equation

Use the RMS velocity formula for \( v_{\text{SO}_2} \): \( v_{\text{SO}_2} = \sqrt{\frac{3kT_{\text{SO}_2}}{m_{\text{SO}_2}}} \). Substitute the expression for \( v_{\text{SO}_2} \): \( \sqrt{\frac{3kT_{\text{SO}_2}}{m_{\text{SO}_2}}} = \frac{1}{2} \sqrt{\frac{3k \times 300}{m_{\text{He}}}} \).

Simplify and Solve for T

Square both sides to eliminate the square roots: \( \frac{3kT_{\text{SO}_2}}{m_{\text{SO}_2}} = \frac{1}{4} \times \frac{3k \times 300}{m_{\text{He}}} \). Cancel out the common factors: \( \frac{T_{\text{SO}_2}}{m_{\text{SO}_2}} = \frac{1}{4} \times \frac{300}{m_{\text{He}}} \) \( T_{\text{SO}_2} = \frac{300}{4} \times \frac{m_{\text{SO}_2}}{m_{\text{He}}} \).

Use Molar Masses

The molar mass of \( \mathrm{He} \) is approximately 4 g/mol, and the molar mass of \( \mathrm{SO}_2 \) is approximately 64 g/mol. Substitute these values: \( T_{\text{SO}_2} = \frac{300}{4} \times \frac{64}{4} \).

Calculate the Temperature

Simplify the expression to find the temperature: \( T_{\text{SO}_2} = 75 \times 16 = 1200 \text{ K} \).

Key concepts

Root Mean Square (RMS) velocity

The Root Mean Square (RMS) velocity is a critical concept in understanding the motion of gas molecules. It represents the square root of the average of the squares of the velocities of all molecules in a gas. The RMS velocity is given by the formula: \(\text{v}_{\text{rms}} = \textbackslash sqrt{\textbackslash frac{3kT}{m}}\). Here, \( k \) is the Boltzmann constant, \( T \) represents temperature, and \( m \) is the molar mass of the gas. The RMS velocity is essential in understanding kinetic theory, aiding in the calculations of kinetic energy, and revealing insights into how temperature and molecular mass influence molecule speeds.

Boltzmann constant

The Boltzmann constant ( \( k \) ) is a fundamental physical constant that plays a crucial role in the kinetic theory of gases. It bridges the macroscopic and microscopic worlds by relating the temperature of a gas to the kinetic energy of its molecules. The value of the Boltzmann constant is approximately \( 1.38 \times 10^{-23} \text{ J·K}^{-1} \). It appears in various equations, including the RMS velocity formula, where it helps determine how fast gas molecules move at a given temperature. The constant provides a quantitative measure to link thermodynamic properties with molecular statistics.

Molar mass of gases

The molar mass of a gas is the mass of one mole of its molecules, typically expressed in grams per mole (g/mol). It directly influences the RMS velocity of the gas. For example, helium (\(\text{He}\)) has a molar mass of approximately 4 g/mol, whereas sulfur dioxide (\(\text{SO}_2\)) has a molar mass of approximately 64 g/mol. In the RMS velocity equation \(\text{v}_{\text{rms}} = \textbackslash sqrt{\textbackslash frac{3kT}{m}}\), the molar mass (\