Question

When the temperature of a gas is raised from 27°C to 90°C, the increase in the rms velocity of the gas molecules is:

• A. \( 10\% \)
• B. \( 15\% \)
• C. \( 20\% \)
• D. \( 17.5\% \)

Hint:

To calculate the change in \( v_{\text{rms}} \), use the proportionality \( v_{\text{rms}} \propto \sqrt{T} \). Convert temperatures to Kelvin before substitution and simplify using the square root property.

Answer 1

The Correct Option is A

Root Mean Square (rms) Velocity and Temperature Relationship

The root mean square velocity of gas molecules at a given temperature \( T \) is: \[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \] where:

• \( k \): Boltzmann constant
• \( m \): mass of one molecule 

Alternatively, using the ideal gas law: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where:

• \( R \): universal gas constant
• \( M \): molar mass of the gas

This shows that: \[ v_{\text{rms}} \propto \sqrt{T} \]

Step 1: Initial and Final Temperatures

• Initial temperature \( T_1 = 27^\circ C = 300\,K \)
• Final temperature \( T_2 = 90^\circ C = 363\,K \)

Step 2: Calculate Change in rms Velocity

Let: \[ v_1 = \sqrt{T_1} = \sqrt{300}, \quad v_2 = \sqrt{T_2} = \sqrt{363} \] Ratio: \[ \frac{v_2}{v_1} = \frac{\sqrt{363}}{\sqrt{300}} = \sqrt{\frac{363}{300}} = \sqrt{1.21} \approx 1.1 \]

Step 3: Percentage Increase

\[ \text{Percentage Increase} = (1.1 - 1) \times 100\% = 10\% \]

Conclusion:

The increase in rms velocity due to heating is: \[ \boxed{10\%} \]

Answer 2

Step 1: Understanding the Root Mean Square (rms) Velocity Formula The root mean square (rms) velocity of gas molecules is given by: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where: - \( R \) is the universal gas constant, - \( T \) is the absolute temperature in Kelvin, - \( M \) is the molar mass of the gas. Since \( v_{\text{rms}} \) is proportional to the square root of temperature, we can write: \[ v_{\text{rms}} \propto \sqrt{T} \] Step 2: Converting Temperatures to Kelvin Given initial and final temperatures in Celsius: \[ T_1 = 27^\circ C = 27 + 273 = 300 \text{ K} \] \[ T_2 = 90^\circ C = 90 + 273 = 363 \text{ K} \] Step 3: Finding the Change in \( v_{\text{rms}} \) The ratio of rms velocities at the two temperatures is: \[ \frac{v_{\text{rms,2}}}{v_{\text{rms,1}}} = \sqrt{\frac{T_2}{T_1}} \] \[ \frac{v_{\text{rms,2}}}{v_{\text{rms,1}}} = \sqrt{\frac{363}{300}} \] \[ \frac{v_{\text{rms,2}}}{v_{\text{rms,1}}} = \sqrt{1.21} \approx 1.1 \] Thus, the percentage increase in \( v_{\text{rms}} \) is: \[ \left( 1.1 - 1 \right) \times 100 = 10\% \]