COMEDK · Physics · 14. Kinetic Theory of Gases
An electric bulb of volume \(300 \mathrm{~cm}^3\) was sealed off during manufacture at a pressure of \(1 \mathrm{~mm}\) of mercury at \(27{ }^{\circ} \mathrm{C}\). The number of air molecules contained in the bulb is,
\((\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) , \(N_A=6.02 \times 10^{23}\), and \(1 \text{ mm of Hg} = 133.3 \text{ N/m}^2\))
- A \(9.67 \times 10^{17}\)
- B \(9.65 \times 10^{18}\)
- C \(9.67 \times 10^{16}\)
- D \(9.65 \times 10^{15}\)
Answer & Solution
Correct Answer
(B) \(9.65 \times 10^{18}\)
Step-by-step Solution
Detailed explanation
\(V = 3 \times 10^{-4}\) m\(^3\), \(T = 300\) K, \(P = \dfrac{101325}{760} \approx 133.32\) Pa \(N = \dfrac{PV \cdot N_A}{RT} = \dfrac{133.32 \times 3 \times 10^{-4} \times 6.02 \times 10^{23}}{8.31 \times 300}\)…
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