Question

What is the correct equation that relates kinetic energy (\(E_{\text{ke}}\)) and pressure (\(P\)) of one mole of an ideal gas? - (V = Volume)

• A. \( P = \frac{2E_{\text{ke}}}{3V} \)
• B. \( P = \frac{3V}{2E_{\text{ke}}} \)
• C. \( P = \frac{3V^2}{2E_{\text{ke}}} \)
• D. \( P = \frac{2(E_{\text{ke}})^2}{3V} \)

Hint:

The kinetic energy of an ideal gas is related to pressure using: \[ P V = \frac{2}{3} E_{\text{ke}} \] This formula comes from the kinetic theory of gases, considering molecular motion and collisions.

Answer 1

The Correct Option is A

Step 1: Using the Kinetic Theory of Gases From kinetic theory, the pressure of an ideal gas is related to its kinetic energy by: \[ P V = \frac{2}{3} E_{\text{ke}} \] where: 
- \( P \) = Pressure 
- \( V \) = Volume 
- \( E_{\text{ke}} \) = Total kinetic energy of the gas 

Step 2: Expressing Pressure in Terms of Energy and Volume Rearranging the equation: \[ P = \frac{2E_{\text{ke}}}{3V} \] Thus, the correct answer is \( \mathbf{(1)} \ P = \frac{2E_{\text{ke}}}{3V} \).