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 .

Step-by-step solution

The change in internal energy of the gas during cooling at constant volume is given by U = n C_v T . For a monatomic gas, C_v = 3 2 R . Substituting the given values, n = 1 mole , R = 8.3 J K ⁻¹ mol ⁻¹ , and T = 10 K : U = 1 3 2 8.3 10 = 124.5 J According to the mass-energy equivalence principle, the loss in mass is m = U c^2 . Using c = 3 10^8 m s ⁻¹ : m = 124.5 (3 10^8)^2 = 124.5 9 10¹⁶ = 13.83 10⁻¹⁶ kg m 1.38 10⁻¹⁵ kg