Question

If the average speed of molecules of an ideal gas in a container is doubled and the volume of the container is halved, then the increase in the pressure of the gas is:

• A. \( 100\% \)
• B. \( 400\% \)
• C. \( 800\% \)
• D. \( \mathbf{700\%} \)

Hint:

- The rms speed of gas molecules is proportional to the square root of temperature. - Pressure is proportional to \( v_{\text{rms}}^2 \) and inversely proportional to volume. - Use the ideal gas law \( PV = nRT \) to analyze pressure changes when volume and speed change.

Answer 1

The Correct Option is D

Step 1: Using the Ideal Gas Law For an ideal gas, the pressure is given by: \[ P = \frac{1}{3} m n v_{\text{rms}}^2 \] where: - \( P \) is the pressure, - \( m \) is the molecular mass, - \( n \) is the number density of molecules, - \( v_{\text{rms}} \) is the root mean square velocity of gas molecules. 

Step 2: Understanding the Effects of Given Changes 1. The rms speed \( v_{\text{rms}} \) is proportional to the square root of temperature: \[ v_{\text{rms}} \propto \sqrt{T} \] Since the average speed is doubled, we get: \[ v_{\text{rms}}' = 2 v_{\text{rms}} \] Squaring both sides: \[ v_{\text{rms}}'^2 = 4 v_{\text{rms}}^2 \] Since \( P \propto v_{\text{rms}}^2 \), the pressure increases 4 times due to this effect. 2. The volume is halved, and from the ideal gas equation: \[ PV = nRT \] Since temperature increases (from the increase in speed), using \( P \propto \frac{T}{V} \), halving the volume leads to another doubling of pressure. 

Step 3: Calculating the Total Increase in Pressure The total increase in pressure due to both effects: \[ P' = 4P \times 2 = 8P \] Thus, the percentage increase is: \[ \frac{P' - P}{P} \times 100 = \frac{8P - P}{P} \times 100 = 700\% \] 

Step 4: Verifying the Correct Option Comparing with the given options, the correct answer is: \[ \mathbf{700\%} \]