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
B
50 K
C
273oC
D
100oK

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 solutionPhysics

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