Question

What is the kinetic energy of 1 g of \( O_2 \) at 47°C?

• A. \( 2.17 \times 10^2 \, \text{J} \)
• B. \( 2.24 \times 10^2 \, \text{J} \)
• C. \( 1.24 \times 10^2 \, \text{J} \)
• D. None of these

Hint:

The total kinetic energy for n moles of an ideal gas is \( \frac{3}{2}nRT \), independent of the gas type.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The average kinetic energy for a gas is given by \( KE = \frac{3}{2}nRT \), where n is the number of moles. This is the total translational kinetic energy.
Step 2: Detailed Explanation:
Given: mass of \(O_2\) = 1 g, Molar mass of \(O_2\) = 32 g/mol. Number of moles, \(n = \frac{1}{32} = 0.03125\) mol. Temperature, \(T = 47^\circ C = 47 + 273 = 320\) K. \(R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}\). Total kinetic energy = \(\frac{3}{2} nRT = \frac{3}{2} \times 0.03125 \times 8.314 \times 320\) = \(1.5 \times 0.03125 \times 2660.48\) = \(1.5 \times 83.14 = 124.71\) J ≈ \(1.24 \times 10^2\) J.
Step 3: Final Answer:
The kinetic energy is \(1.24 \times 10^2\) J, option (C).