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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2024
  • Hard
  • MCQ

The temperature of a gas having \(2.0 \times 10^{25}\) molecules per cubic meter at \(1.38 \mathrm{~atm}\) (Given, \(\mathrm{k}=\) \(\left.1.38 \times 10^{-23} \mathrm{JK}^{-1}\right)\) is _______.

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Answer & Solution

Correct Answer

(A) \(500 \mathrm{~K}\)

Step-by-step Solution

Detailed explanation

\( \mathrm{PV}=\mathrm{nRT} \)

\( \mathrm{PV}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}} \mathrm{RT}\)

\(\mathrm{N}=\) Total no. of molecules

\(\mathrm{P}=\frac{\mathrm{N}}{\mathrm{V}} \mathrm{kT}\)

\(1.38 \times 1.01 \times 10^5=2 \times 10^{25} \times 1.38 \times 10^{-23} \times \mathrm{T}\)

\(1.01 \times 10^5=2 \times 10^2 \times \mathrm{T}\)

\(\mathrm{T}=\frac{1.01 \times 10^3}{2} \approx 500 \mathrm{~K}\)

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