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NCERT Exemplar · Q24

Q.A ballon has 5.0 g mole of helium at 7°C. Calculate

(a) the number of atoms of helium in the balloon,
(b) the total internal energy of the system.
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Using the Kinetic Theory of Gases, the number of atoms is found from Avogadro’s number and the mole count, and the internal energy follows from the equipartition theorem for a monatomic gas. The answers are 3.01×10243.01 \times 10^{24} atoms and 1.75×1041.75 \times 10^4 J.

The Kinetic Theory of Gases gives us a direct molecular picture. For a monatomic gas like helium, each atom moves independently in three dimensions, and its only energy is translational kinetic energy. The total internal energy of the gas is simply the sum of the kinetic energies of all atoms. This is why we can compute it from the temperature and the number of atoms — no need to think about vibrations or rotations, because helium atoms don’t have those.

The problem gives us 5.0 gram-moles of helium at 7°C. A gram-mole is just a mole — the standard unit. So we have n=5.0n = 5.0 mol. The temperature must be in kelvin for all gas-law calculations: T=7+273=280T = 7 + 273 = 280 K.

  1. Number of atoms One mole of any substance contains Avogadro’s number of particles, NA=6.02×1023N_A = 6.02 \times 10^{23} mol−1^{-1}. So the number of helium atoms is

N=nNA=5.0×6.02×1023=3.01×1024.N = n N_A = 5.0 \times 6.02 \times 10^{23} = 3.01 \times 10^{24}.

That’s a huge number — typical for macroscopic amounts of gas.

  1. Total internal energy For a monatomic ideal gas, the equipartition theorem says each atom has 32kBT\frac{3}{2} k_B T of kinetic energy, where kBk_B is Boltzmann’s constant. The total internal energy is therefore

U=N⋅32kBT.U = N \cdot \frac{3}{2} k_B T.

But it’s often more convenient to use the molar form: U=32nRTU = \frac{3}{2} n R T, where R=8.314R = 8.314 J/(mol·K) is the universal gas constant. This works because R=NAkBR = N_A k_B.

Substituting:

U=32×5.0×8.314×280.U = \frac{3}{2} \times 5.0 \times 8.314 \times 280. …

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