The temperature of a gas having 2.0 × 10²⁵ molecules per cubic meter at 1.38 atm (Given, k = 1.38 × 10⁻²³ J K⁻¹) is: (a) 500 K (b) 300 K (c) 200 K (d) 100 K

Answer: (a). P = nkT, where n is the number of molecules per m³. T = P/nk = 1.38 × 10⁵2.0 × 10²⁵ × 1.38 × 10⁻²³ (taking 1 atm ≈ 10⁵ Pa). T = (10⁵)/(200) = 500 K. So the answer is (a) 500 K.

The temperature of a gas having 2.0 × 10²⁵ molecules per cubic meter… | JEE Main 2024 (29 Jan, Shift 2)

Q36JEE Main 2024 · 29 JAN · SHIFT 2Medium

Gas Laws & Ideal Gas Equation

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

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Answer: (a)

$P = nkT$ , where $n$ is the number of molecules per m³.
$T = \frac{P}{nk} = \frac{1.38 \times 10^5}{2.0 \times 10^{25} \times 1.38 \times 10^{-23}}$ (taking 1 atm $\approx 10^5$ Pa).
$T = \frac{10^5}{200} = 500\text{ K}$ .
So the answer is (a) 500 K.

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