Question

At 240.55 K, for one mole of an ideal gas, a graph of \( P \) (on y-axis) and \( V^{-1} \) (on x-axis) gave a straight line passing through the origin. Its slope (m) is 2000 J mol\(^{-1}\). What is the kinetic energy (in J mol\(^{-1}\)) of the ideal gas?

• A. \( 2000 \)
• B. \( 3000 \)
• C. \( 6000 \)
• D. \( 1500 \)

Hint:

For an ideal gas, kinetic energy is directly proportional to temperature and follows \( KE = \frac{3}{2} RT \).

Answer 1

The Correct Option is B

Step 1: Understanding the Given Data The ideal gas equation states: \[ P V = nRT \] Since \( P \) vs. \( V^{-1} \) gives a straight line through the origin, its slope represents: \[ m = nRT \] For one mole of gas: \[ m = RT = 2000 \text{ J mol}^{-1} \]
Step 2: Finding Kinetic Energy The kinetic energy per mole of an ideal gas is: \[ KE = \frac{3}{2} RT \] \[ KE = \frac{3}{2} \times 2000 = 3000 \text{ J mol}^{-1} \]