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Question

 The root mean square velocity of the molecules in a sample of helium is $$ \frac { 5 }{ 7 }  $$ th that molecules in a sample of hydrogen. If the temperature of hydrogen gas is $$ 0^o C $$ that of helium sample is near about :-


A
0oC
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B
50 K
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C
273oC
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D
100oK
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Solution

The correct option is A $$5^0\ K$$

We know that,

$${{v}_{rms}}=\sqrt{3RT/M}$$

Now, the root mean square velocity of helium

$${{v}_{rms}}\left( He \right)=\sqrt{\dfrac{3R\times T}{M}}....(I)$$

Now, the root mean square velocity of hydrogen

$${{v}_{rms}}(H)=\sqrt{\dfrac{3R\times 273}{M}}.....(II)$$

Now, from equation (I) and (II)

  $$ \dfrac{{{v}_{rms}}\left( He \right)}{{{v}_{rms}}\left( H \right)}=\dfrac{\sqrt{\dfrac{3R\times T}{M}}}{\sqrt{\dfrac{3R\times 273}{M}}} $$

 $$ \dfrac{5}{7}=\sqrt{\dfrac{T\times 2}{273\times 4}} $$

 $$ \dfrac{25}{49}=\dfrac{T\times 2}{273\times 4} $$

 $$ T=\dfrac{273\times 25\times 4}{49\times 2} $$

 $$ T={{278}^{0}}C $$

 $$ T=278-273 $$

 $$ T={{5}^{0}}K $$

Hence, this is the required solution


Physics

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