Q 11-12-134JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
A $15\ \text{g}$ mass of nitrogen gas is enclosed in a vessel at a temperature $27^\circ\text{C}$. The amount of heat transferred to the gas, so that r.m.s. velocity of molecules is doubled, is about $\left[R = 8.3\ \text{J (K mol)}^{-1}\right]$
Answer: (B) $10\ \text{kJ}$
$v_{rms} \propto \sqrt{T}$, so doubling it needs $T = 4\times300 = 1200\ \text{K}$, i.e. $\Delta T = 900\ \text{K}$.
The vessel is closed, so heating is at constant volume. Nitrogen is diatomic, $C_V = \tfrac52R$, and $n = \dfrac{15}{28}$ mol.
$$Q = nC_V\Delta T = \frac{15}{28}\times\frac52\times8.3\times900 \approx 1.0\times10^4\ \text{J} = 10\ \text{kJ}$$
Solution by Sreeraj P, M.Sc Physics