Determine the loss in the mass of 1 mole of an ideal monatomic gas held in a rigid container as it cools down by 10^ ^ ^ C . The gas constant is given as R = 8.3 J K^ -1 mol^ -1 .

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Correct answer: A. 1.38 × 10^ -15 kg

The change in internal energy of the gas during cooling at constant volume is given by

\Delta U = n C_v \Delta T

.

For a monatomic gas,

C_v = \dfrac{3}{2} R

.

Substituting the given values,

n = 1 \text{ mole}

,

R = 8.3 \text{ J K}^{-1} \text{mol}^{-1}

, and

\Delta T = 10 \text{ K}

:

\Delta U = 1 \times \dfrac{3}{2} \times 8.3 \times 10 = 124.5 \text{ J}

According to the mass-energy equivalence principle, the loss in mass is

\Delta m = \dfrac{\Delta U}{c^2}

.

Using

c = 3 \times 10^8 \text{ m s}^{-1}

:

\Delta m = \dfrac{124.5}{(3 \times 10^8)^2} = \dfrac{124.5}{9 \times 10^{16}} = 13.83 \times 10^{-16} \text{ kg}

\Delta m \approx 1.38 \times 10^{-15} \text{ kg}

Determine the loss in the mass of 1 mole of an ideal monatomic gas...

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