Question:
What is the kinetic energy of 1 g of \( O_2 \) at 47°C?
What is the kinetic energy of 1 g of \( O_2 \) at 47°C?
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The total kinetic energy for n moles of an ideal gas is \( \frac{3}{2}nRT \), independent of the gas type.
Updated On: May 2, 2026
- \( 2.17 \times 10^2 \, \text{J} \)
- \( 2.24 \times 10^2 \, \text{J} \)
- \( 1.24 \times 10^2 \, \text{J} \)
- None of these
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The Correct Option is C
Solution and Explanation
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).
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).
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