Q 11-12-006NEETNEET 2023Top questionMedium
The temperature of a gas is $-50°$ C. To what temperature the gas should be heated so that the rms speed is increased by $3$ times?
Answer: (B) $3295°$ C
$v_{rms} = \sqrt{\dfrac{3RT}{M}}$, so $v_{rms} \propto \sqrt{T}$.
"Increased by 3 times" means the speed increases by three times its value, so the new speed is $v + 3v = 4v$. Then
$$T_2 = 16T_1 = 16 \times (273 - 50) = 16 \times 223 = 3568\ \text{K} = 3295°\text{C}$$
(If the new speed were only $3v$, the temperature would be $9 \times 223 = 2007$ K, which is not among the options.)
Solution by Sreeraj P, M.Sc Physics