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QPages 34-37: Q2: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q2 Solution: Find the number of molecules of an ideal gas in a volume of at…

⚠️ Common Exam Traps & Conceptual Pitfalls

When applying the ideal gas law with $R=8.314; ext{J mol}^{-1} ext{K}^{-1}$, the volume must be in cubic metres. A frequent mistake is treating $1; ext{cm}^3$ as $10^{-3}; ext{m}^3$ instead of $10^{-6}; ext{m}^3$, which inflates the calculated moles (and thus the number of molecules) by a factor of 1000. Always convert units consistently or use $R=0.0821; ext{L atm mol}^{-1} ext{K}^{-1}$ with volume in litres.

Find the number of molecules of an ideal gas in a volume of 1.000 cm31.000 \text{ cm}^3 at STP.

Step-by-Step Verified Solution

**Answer:** N=2.69×1019N = 2.69 \times 10^{19} molecules **Step-by-Step Explanation:** Volume V=1.000 cm3=10−6 m3V = 1.000 \text{ cm}^3 = 10^{-6} \text{ m}^3. At STP, P=1.013×105P = 1.013 \times 10^5 Pa, T=273T = 273 K.
n=PVRT=1.013×105×10−68.314×273=4.46×10−5 moln = \frac{PV}{RT} = \frac{1.013 \times 10^5 \times 10^{-6}}{8.314 \times 273} = 4.46 \times 10^{-5} \text{ mol}
N=n×NA=4.46×10−5×6.023×1023=2.69×1019N = n \times N_A = 4.46 \times 10^{-5} \times 6.023 \times 10^{23} = 2.69 \times 10^{19}

All Questions in this Chapter

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