Question bank/JEE Main/Physics

JEE MainPhysicsModerateKinetic Theory of Gases

The following graph shows two isotherms for a fixed mass of an
ideal gas. The ratio of r.m.s speed of the molecules at temperatures
T_1 and T_2 is

Show answer and solution
PV=13M(vrms)12=RTPV = \frac{1}{3}M\left( v_{rms} \right)_{1}^{2} = RT
On lower curve, we have V=1m^3, corresponding to P=1×10^5Pa = 1
atmosphere. At T_1, we have
de oØ tks P=1×10^5Pa = 1 okrkoj.k ds
vuq:i gSA T_1 ij] gekjs ikl
1×1=13M(vrms)12=RT11 \times 1 = \frac{1}{3}M\left( v_{rms} \right)_{1}^{2} = {RT}_{1}
or(vrms)12=3M\left( v_{rms} \right)_{1}^{2} = \frac{3}{M}
Similarly, for upper curve we have
blh rjg] oØ ds fy, gekjs ikl gS
V=1m^3, P = 2 atmosphere
∴2×2=13M(vans)22=RT2\therefore 2 \times 2 = \frac{1}{3}M\left( v_{ans} \right)_{2}^{2} = {RT}_{2}
∴(vmas)22=4×3M=12M\therefore\left( v_{mas} \right)_{2}^{2} = \frac{4 \times 3}{M} = \frac{12}{M}
∴ Required ratio =12/M3/M=2\therefore\ Required\ ratio\ = \sqrt{\frac{12/M}{3/M}} = 2

Source: Grass-G / Ph-25 Physics paper 7 Jan.docx

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