QJEEM24Phbff9: JEE Main 2024 Physics - Class 11 JEEM24Phbff9 Solution: The temperature of a gas having 2.0 × 10 25 molecules per cubic …

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QJEEM24Phbff9: JEE Main 2024 Physics - Class 11 JEEM24Phbff9 Solution: The temperature of a gas having 2.0 × 10 25 molecules per cubic …

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

Multiple Choice Options

A 500 K
B 300 K
C 200 K
D 100 K

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Accepted Answer

$\begin{aligned} PV = nRT \\ PV = \frac{N}{N_{A}} RT \\ N = Total no. of molecules \\ P = \frac{N}{V} kT \\ 1.38 \times 1.01 \times 10^{5} = 2 \times 10^{25} \times 1.38 \times 10^{- 23} \times T \\ 1.01 \times 10^{5} = 2 \times 10^{2} \times T \\ T = \frac{1.01 \times 10^{3}}{2} \approx 500 K \end{aligned}$

QJEEM24Phbff9: JEE Main 2024 Physics - Class 11 JEEM24Phbff9 Solution: The temperature of a gas having 2.0 × 10 25 molecules per cubic …

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