The mean speed of the molecules of a hydrogen sample equal the mean speed of the molecules of helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample.

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Solution

Formula of mean speed:

$V_{a v g} = \sqrt{\frac{8 R T}{π M}}$

Now, according to question,

$\frac{8 R T_{1}}{π \times M_{H_{2}}} = \frac{8 R T_{2}}{π \times M_{H e}}$

$\frac{8 R T_{1}}{π \times 2} = \frac{8 R T_{2}}{π \times 4}$

$\frac{T_{1}}{T_{2}} = \frac{1}{2}$

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The mean speed of the molecules of a hydrogen sample equal the mean speed of the molecules of helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample.

Formula of mean speed:Vavg=√8RTπMNow, according to question,8RT1π×MH2=8RT2π×MHe 8RT1π×2=8RT2π×4 T1T2=12

Formula of mean speed:

$V_{a v g} = \sqrt{\frac{8 R T}{π M}}$

Now, according to question,

$\frac{8 R T_{1}}{π \times M_{H_{2}}} = \frac{8 R T_{2}}{π \times M_{H e}}$

$\frac{8 R T_{1}}{π \times 2} = \frac{8 R T_{2}}{π \times 4}$

$\frac{T_{1}}{T_{2}} = \frac{1}{2}$