JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 10)
The temperature of a gas having $$2.0 \times 10^{25}$$ molecules per cubic meter at $$1.38 \mathrm{~atm}$$ (Given, $$\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}$$) is :
500 K
300 K
200 K
100 K
Explanation
$$\begin{aligned}
& \mathrm{PV}=\mathrm{nRT} \\
& \mathrm{PV}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}} \mathrm{RT} \\
& \mathrm{N}=\text { Total no. of molecules } \\
& \mathrm{P}=\frac{\mathrm{N}}{\mathrm{V}} \mathrm{kT} \\
& 1.38 \times 1.01 \times 10^5=2 \times 10^{25} \times 1.38 \times 10^{-23} \times \mathrm{T} \\
& 1.01 \times 10^5=2 \times 10^2 \times \mathrm{T} \\
& \mathrm{T}=\frac{1.01 \times 10^3}{2} \approx 500 \mathrm{~K}
\end{aligned}$$
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