Question

calculate the temperature at which rms velocity of gas molecules is double its value at 27oC pressure of the gas remaining the same at what temp is the rms velocity of hydrogen molecules equal to that of an oxygen at 47oC?

Answer 1

$\huge \mathfrak {Answer:-}$

We know that,

Internal energy of an ideal gas ∝ Absolute temperature

Average K. E. of the gas molecules ∝ abs temperature

Now,

$\bold{P\:V\:=\:n\:R\:T}$

Average K. E. = $\frac{1}{2} mv^{2}$ $= \frac{3}{2}nRT$

For one molecule,

The translational K. E. with 3° of freedom of movement in 3d space.

Average K. E. =$\frac{1}{2} mv^{2} = \frac{3}{2} kT$

So,

rms velocity ∝ $\sqrt{t}$

$\frac{v2}{v1} = \sqrt{ \frac{T2}{T1} }$

$2 = \sqrt{ \frac{T2}{(273 + 27)} }$

$T2 = 1200°K \: or \: 927°C$

We assume the conditional of hydrogen and oxygen gas molecules are same.

So,

They are both diatomic.

$\frac{m1v1^{2} }{m2v2^{2} } = \frac{T1}{T2}$

$m1 = 2$ Hydrogen

$m2 = 32$ Oxygen

$v1 = v2$(Given)

$T2 = 320°K$

So,

$T1 = \frac{320° \times 2}{32} = 20°K$

$\huge {Be\:Brainly}$❤️

Answer 2

$<b>thanks for asking$

hope it helps