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Exercises · 12.9

Q.At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at −20 ∘C-20\ ^\circ\text{C}? (atomic mass of Ar=39.9 u\text{Ar} = 39.9\ \text{u}, of He=4.0 u\text{He} = 4.0\ \text{u}).

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The rms speed depends only on temperature and molecular mass: vrms=3RT/Mv_{\text{rms}} = \sqrt{3RT/M}. Equating the rms speeds for argon and helium gives TAr=(MAr/MHe) THeT_{\text{Ar}} = (M_{\text{Ar}}/M_{\text{He}})\,T_{\text{He}}. With THe=253 KT_{\text{He}} = 253\ \text{K}, the required temperature is TAr≈2524 KT_{\text{Ar}} \approx 2524\ \text{K}.

The root mean square speed is a direct measure of the average kinetic energy of gas molecules. From the Kinetic Theory of Gases, the average translational kinetic energy per molecule is 32kT\frac{3}{2}kT, independent of the gas. This means that at the same temperature, lighter molecules move faster on average. But here we want the rms speed of heavy argon atoms to equal that of light helium atoms — so argon must be at a much higher temperature.

The formula for rms speed comes from equating kinetic energy:

12mvrms2=32kT\frac{1}{2} m v_{\text{rms}}^2 = \frac{3}{2} kT

where mm is the mass of one molecule. Multiplying by Avogadro's number NAN_A gives the molar form:

12Mvrms2=32RT\frac{1}{2} M v_{\text{rms}}^2 = \frac{3}{2} RT

where MM is the molar mass (in kg/mol) and RR is the universal gas constant. Solving:

vrms=3RTMv_{\text{rms}} = \sqrt{\frac{3RT}{M}}

This is the key relation. Notice that vrmsv_{\text{rms}} depends on the ratio T/MT/M.

  1. Convert the given temperature to Kelvin. Helium is at −20 ∘C-20\ ^\circ\text{C}.

THe=−20+273=253 KT_{\text{He}} = -20 + 273 = 253\ \text{K}

  1. Set the rms speeds equal. For argon at unknown TArT_{\text{Ar}} and helium at THeT_{\text{He}}:

3R TArMAr=3R THeMHe\sqrt{\frac{3R\,T_{\text{Ar}}}{M_{\text{Ar}}}} = \sqrt{\frac{3R\,T_{\text{He}}}{M_{\text{He}}}}

The 3R3R cancels. Square both sides:

TArMAr=THeMHe\frac{T_{\text{Ar}}}{M_{\text{Ar}}} = \frac{T_{\text{He}}}{M_{\text{He}}}

So:

TAr=MArMHe THeT_{\text{Ar}} = \frac{M_{\text{Ar}}}{M_{\text{He}}}\, T_{\text{He}}

This is a clean proportionality — the temperature must scale exactly with the molar mass.

  1. Plug in the atomic masses. The problem gives masses in atomic mass units (u). Since molar mass in g/mol has the same numerical value, the ratio is dimensionless:

MArMHe=39.94.0=9.975\frac{M_{\text{Ar}}}{M_{\text{He}}} = \frac{39.9}{4.0} = 9.975

Therefore: …

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