The total internal energy of two mole monoatomic ideal gas at temperature T = 300 K will be _____________ J. (Given R = 8.31 J/mol.K)

Solution

Physics sir · 4 steps

A monoatomic gas has only 3 degrees of freedom, so each mole carries 3/2RT of energy.

1. Count degrees of freedom. A monoatomic molecule can only move along x, y and z. It has no rotation worth counting, so f = 3.
2. Equipartition. Each degree of freedom gets 1/2RT per mole. An ideal gas has no intermolecular potential energy, so this kinetic energy is all of its internal energy.

   U = n f/2 RT = 2 × 3/2 × 8.31 × 300

3. Put in numbers. 2 × 3/2 = 3, and 3 × 8.31 = 24.93. Then 24.93 × 300 = 7479.
4. So. The total internal energy is U = 7479 J, so the answer is 7479.

The common slip: 3739.5. Forgetting to multiply by the 2 moles gives the energy of one mole only, 3739.5 J, not 7479 J.

The idea to keep. Internal energy of an ideal gas is U = nf/2RT. Fix f first: 3 for monoatomic, 5 for diatomic at ordinary temperatures. Then multiply by the number of moles.

Answer from the NTA final answer key. Solution by PrepWiser.