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  • We have two vessels of equal volume, one filled with hydrogen and the other with equal mass of Helium. The common temperature is <img alt="27^{\circ} \mathrm{C}" src="https://entrancecorner.oncodec

We have two vessels of equal volume, one filled with hydrogen and the other with equal mass of Helium. The common temperature is 27^{\circ} \mathrm{C}

Question : If the temperature of Helium is kept at 27^{\circ} \mathrm{C} and that of hydrogen is changed, at what temperature will its pressure become equal to that of helium ? The molecular weights of hydrogen and helium are 2 and 4 respectively.

Option: 1

-123^{\circ} \mathrm{C}


Option: 2

-140^{\circ} \mathrm{C}


Option: 3

-160^{\circ} \mathrm{C}


Option: 4

-183^{\circ} \mathrm{C}


Answers (1)

best_answer

Again, we have

\mathrm{pV}=\mathrm{nkT}

\mathrm{H}_{2} and \mathrm{He} have equal volumes. For having equal pressure, we must have
\mathrm{\mathrm{n}_{\mathrm{H}} \mathrm{T}_{\mathrm{H}}=\mathrm{n}_{\mathrm{He}} \mathrm{T}_{\mathrm{He}}}

\mathrm{or \, \quad \frac{T_{\mathrm{He}}}{\mathrm{T}_{\mathrm{H}}}=\frac{\mathrm{n}_{\mathrm{H}}}{\mathrm{n}_{\mathrm{He}}}=2}

\mathrm{Here \, \mathrm{T}_{\mathrm{He}}=27+273=300 \mathrm{~K}}
\mathrm{\therefore \mathrm{T}_{H}=\frac{1}{2} \mathrm{~T}_{\mathrm{He}}=150 \mathrm{~K}}
\mathrm{=150-273=-123^{\circ} \mathrm{C}}
 

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