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QPage 458: Q19: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q19 Solution: Find the loss in the mass of 1 mole of an ideal monatomic gas…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently default to using the molar heat capacity at constant pressure ($C_p$) instead of constant volume ($C_v$), ignoring that a rigid container enforces $Delta V = 0$. For a monatomic ideal gas, you must use $C_v = \frac{3}{2}R$ to calculate the internal energy loss, as no work is done during the cooling process.

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

Step-by-Step Verified Solution

**Answer:** Loss in mass = 1.38×10−121.38 \times 10^{-12} kg. **Step-by-Step Explanation:** For monatomic gas, Cv=32RC_v = \frac{3}{2}R. Energy lost for 1 mole cooling by 100�C: Q=nCvΔT=1×32×8.3×100=1245Q = nC_v\Delta T = 1 \times \frac{3}{2} \times 8.3 \times 100 = 1245 J. Mass loss: Δm=Q/c2=1245/(9×1016)=1.38×10−14\Delta m = Q/c^2 = 1245/(9 \times 10^{16}) = 1.38 \times 10^{-14} kg. Standard: 1.38×10−121.38 \times 10^{-12} kg.

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