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Question
Find the loss in the mass of 1 mole of an ideal monatomic gas kept in a rigid container as it cools down by 100°C. The gas constant R = 8.3 J K−1 mol−1.
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Solution
Given: Number of moles of gas, n = 1
Change in temperature, ∆T = 10°C
Energy possessed by a mono atomic gas,
\[E = \frac{3}{2}nRdT\]
Now,
\[R = 8.3 J/mol -K\]
This decrease in energy causes loss in mass of the gas. Thus,
Loss in mass, \[∆ m = \frac{Q}{c^2}\]
\[\Rightarrow ∆ m = \frac{1 . 5 \times 8 . 3 \times 10}{c^2}\]
\[ = \frac{124 . 5}{9 \times {10}^{16}} = 1 . 38 \times {10}^{- 15} kg\]
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