MHT CET · Physics · Kinetic Theory of Gases
An ideal gas in a container of volume 500 c.c. is at a pressure of \(2 \times 10^{+5} \mathrm{~N} / \mathrm{m}^2\). The average kinetic energy of each molecule is \(6 \times 10^{-21} \mathrm{~J}\). The number of gas molecules in the container is
- A \(5 \times 10^{25}\)
- B \(25 \times 10^{23}\)
- C \(5 \times 10^{23}\)
- D \(2.5 \times 10^{22}\)
Answer & Solution
Correct Answer
(D) \(2.5 \times 10^{22}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{PV}=\mathrm{Nk}_{\mathrm{B}} \mathrm{T} \\ & \text { and, K.E./molecule }=\frac{3}{2} \mathrm{k}_{\mathrm{B}} \mathrm{T}=\frac{3}{2} \frac{\mathrm{PV}}{\mathrm{N}}\end{aligned}\)
\(\begin{aligned} \therefore \quad \mathrm{N} & =\frac{3}{2} \times \frac{\mathrm{PV}}{(\mathrm{K} . \mathrm{E} . / \mathrm{molecule})} \\ & =\frac{3}{2} \times \frac{2 \times 10^5 \times 500 \times 10^{-6}}{6 \times 10^{-21}}=2.5 \times 10^{22}\end{aligned}\)
\(\begin{aligned} \therefore \quad \mathrm{N} & =\frac{3}{2} \times \frac{\mathrm{PV}}{(\mathrm{K} . \mathrm{E} . / \mathrm{molecule})} \\ & =\frac{3}{2} \times \frac{2 \times 10^5 \times 500 \times 10^{-6}}{6 \times 10^{-21}}=2.5 \times 10^{22}\end{aligned}\)
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