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MCQEasyJEE 2024Molecular Speeds (rms, Average, Most Probable)

JEE Physics 2024 Question with Solution

At which temperature does the r.m.s. velocity of a hydrogen molecule equal that of an oxygen molecule at 47 °C47\,\text{°C}?

  • A

    80 K80\,\text{K}

  • B

    −73 K-73\,\text{K}

  • C

    4 K4\,\text{K}

  • D

    20 K20\,\text{K}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The r.m.s. velocity of a hydrogen molecule is equal to that of an oxygen molecule at 47 °C47\,\text{°C}.

Find: The temperature of hydrogen for which both r.m.s. velocities are equal.

Using the formula for r.m.s. velocity,

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

For equal r.m.s. velocities of hydrogen and oxygen,

3RTH2MH2=3RTO2MO2\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}

Squaring both sides,

3RTH2MH2=3RTO2MO2\frac{3RT_{H_2}}{M_{H_2}} = \frac{3RT_{O_2}}{M_{O_2}}

Canceling 3R3R,

TH2=TO2×MH2MO2T_{H_2} = T_{O_2} \times \frac{M_{H_2}}{M_{O_2}}

Now convert the oxygen temperature to Kelvin:

TO2=47+273=320 KT_{O_2} = 47 + 273 = 320\,\text{K}

Using MH2=2M_{H_2} = 2 and MO2=32M_{O_2} = 32,

TH2=320×232=20 KT_{H_2} = 320 \times \frac{2}{32} = 20\,\text{K}

Detailed Working

Given: Equal r.m.s. velocities for hydrogen and oxygen, with oxygen at 47 °C47\,\text{°C}.

Find: Temperature of hydrogen.

An equivalent form of the r.m.s. speed formula is

vrms=3kTmv_{\text{rms}} = \sqrt{\frac{3kT}{m}}

Since the speeds are equal,

3kTH2mH2=3kTO2mO2\sqrt{\frac{3kT_{H_2}}{m_{H_2}}} = \sqrt{\frac{3kT_{O_2}}{m_{O_2}}}

Therefore,

TH2mH2=TO2mO2\frac{T_{H_2}}{m_{H_2}} = \frac{T_{O_2}}{m_{O_2}}

So,

TH2=TO2×mH2mO2T_{H_2} = T_{O_2} \times \frac{m_{H_2}}{m_{O_2}}

Now,

mH2=2 u,mO2=32 um_{H_2} = 2\,\text{u}, \qquad m_{O_2} = 32\,\text{u}

and

TO2=47+273=320 KT_{O_2} = 47 + 273 = 320\,\text{K}

Substituting,

TH2=320×232=20 KT_{H_2} = 320 \times \frac{2}{32} = 20\,\text{K}

Therefore, the correct value is 20 K20\,\text{K}.

The solution concludes with 20 K20\,\text{K}, but the solution states "The Correct Option is C," while the listed options show 20 K20\,\text{K} as option D. Following the solution working, the defensible correct option is D.

Common mistakes

  • Using 4747 directly instead of converting 47 °C47\,\text{°C} to Kelvin is incorrect because gas-kinetic formulas require absolute temperature. First convert to 320 K320\,\text{K}, then substitute.

  • Inverting the mass ratio and using 322\frac{32}{2} instead of 232\frac{2}{32} is wrong because for equal r.m.s. speed, temperature is directly proportional to molecular mass in the relation Tm\frac{T}{m} being equal. Use TH2=TO2mH2mO2T_{H_2} = T_{O_2}\frac{m_{H_2}}{m_{O_2}}.

  • Confusing molecular mass with atomic mass leads to taking oxygen as 1616 instead of 3232. The question is about an oxygen molecule, so use O2=32 uO_2 = 32\,\text{u}, not 16 u16\,\text{u}.

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