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)
What is the correct equation that relates kinetic energy (\(E_{\text{ke}}\)) and pressure (\(P\)) of one mole of an ideal gas?
- (V = Volume)
Show 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.
Updated On: Mar 3, 2026
- \( P = \frac{2E_{\text{ke}}}{3V} \)
- \( P = \frac{3V}{2E_{\text{ke}}} \)
- \( P = \frac{3V^2}{2E_{\text{ke}}} \)
- \( P = \frac{2(E_{\text{ke}})^2}{3V} \)
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The Correct Option is A
Solution and Explanation
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 \( \mathbf{(1)} \ P = \frac{2E_{\text{ke}}}{3V} \).
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