Verified Solution • 2025–26 Session moderate

QPages 34-37: Q27: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q27 Solution: 0.040 g of He is kept in a closed container initially at…

0.040 g of He is kept in a closed container initially at 100.0�C. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J. Use R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-Step Verified Solution

**Answer:** T2=196�CT_2 = 196�\text{C} **Step-by-Step Explanation:** m=0.040m = 0.040 g, M=4M = 4 g/mol, n=0.01n = 0.01 mol. Cv=32R=12.45C_v = \frac{3}{2}R = 12.45 J/mol-K. ΔU=nCvΔT=12⇒0.01×12.45×(T2−373)=12\Delta U = n C_v \Delta T = 12 \Rightarrow 0.01 \times 12.45 \times (T_2 - 373) = 12. T2−373=96.4⇒T2=469.4T_2 - 373 = 96.4 \Rightarrow T_2 = 469.4 K =196.4�C= 196.4�\text{C}.

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