Question

If rms speed is increased 20% and volume is kept constant, then increase in pressure is

Hint:

For small percentage changes (<10%), you can use approximation: if \( y \propto x^n \), then \( \%\Delta y \approx n \times \%\Delta x \). Here, \( P \propto v^2 \), so \( 20\% \) change in \( v \) roughly gives \( 2 \times 20\% = 40\% \) change in \( P \). For exact values, square the multiplier: \( 1.2^2 = 1.44 \Rightarrow +44\% \).

Answer 1

Step 1: Understanding the Concept:
The Kinetic Theory of Gases relates macroscopic properties like pressure (\(P\)) and volume (\(V\)) to microscopic properties like the root-mean-square (rms) speed (\(v_{rms}\)) of gas molecules.
Step 2: Key Formula or Approach:
The pressure exerted by an ideal gas is given by:
\[ P = \frac{1}{3} \rho v_{rms}^2 = \frac{1}{3} \left(\frac{M}{V}\right) v_{rms}^2 \]
Since mass \(M\) and volume \(V\) are constant, pressure is proportional to the square of rms speed:
\[ P \propto v_{rms}^2 \]
Step 3: Detailed Explanation:
Let initial rms speed be \(v_1\), then increased speed:
\[ v_2 = 1.20v_1 \]
\[ \frac{P_2}{P_1} = (1.20)^2 = 1.44 \]
So, \(P_2 = 1.44P_1\).
Percentage increase:
\[ 44\% \]
Step 4: Final Answer:
The increase in pressure is 44%.