CBSE-XI-Physics
47: The Special Theory of Relativity
- #19Find 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.Ans : 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\,\mathrm{\,J/\,}\,\mathrm{\,mol\,}-K``
This decrease in energy causes loss in mass of the gas. Thus,
`` \,\mathrm{\,Loss\,}\,\mathrm{\,in\,}\,\mathrm{\,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}\,\mathrm{\,kg\,}``
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