Q 11-12-051JEE MainJEE Main 2024 (29 Jan, Shift 2)Easy
The temperature of a gas having $2.0\times10^{25}$ molecules per cubic metre at $1.38\ \text{atm}$ (Given, $k = 1.38\times10^{-23}\ \text{J K}^{-1}$) is:
Answer: (A) $500\ \text{K}$
$P = nkT$, where $n$ is the number of molecules per unit volume. With $1\ \text{atm} \approx 10^5\ \text{Pa}$:
$$T = \frac{P}{nk} = \frac{1.38\times10^5}{2\times10^{25}\times1.38\times10^{-23}} = \frac{10^5}{200} = 500\ \text{K}$$
Solution by Sreeraj P, M.Sc Physics