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QPages 34-37: Q3: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q3 Solution: Find the number of molecules in of an ideal gas at 0�C and at a …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently forget to convert the pressure from mm of Hg to Pascals (SI units) before applying the ideal gas law, leading to massive order-of-magnitude errors. Always ensure $P$ is in Pa, $V$ in $m^3$, and $T$ in Kelvin to match the units of $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$.

Find the number of molecules in 1 cm31 \text{ cm}^3 of an ideal gas at 0�C and at a pressure of 10−5 mm10^{-5} \text{ mm} of mercury. Use R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-Step Verified Solution

**Answer:** N=3.54×1011N = 3.54 \times 10^{11} molecules **Step-by-Step Explanation:** P=10−5P = 10^{-5} mm of Hg =10−5×13600×9.8=1.33×10−3= 10^{-5} \times 13600 \times 9.8 = 1.33 \times 10^{-3} Pa. V=1 cm3=10−6 m3V = 1 \text{ cm}^3 = 10^{-6} \text{ m}^3, T=273T = 273 K.
n=PVRT=1.33×10−3×10−68.314×273=5.87×10−13 moln = \frac{PV}{RT} = \frac{1.33 \times 10^{-3} \times 10^{-6}}{8.314 \times 273} = 5.87 \times 10^{-13} \text{ mol}
N=5.87×10−13×6.023×1023=3.54×1011N = 5.87 \times 10^{-13} \times 6.023 \times 10^{23} = 3.54 \times 10^{11}

All Questions in this Chapter

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