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

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: $$\left( 1.1 - 1 \right) \times 100 = 10\%$$