Question

The temperature at which average speed of helium is same as that of CH4 at 27°C is

28°C

8°C

-150°C

-198°C​

Answer 1

use the answer attached

Answer 2

The average speed is the arithmetic means of the speeds of different molecules of gases.

Explanation:

The formula for average speed is: -$\sqrt\frac{8RT}{\pi M}$

T= Temperature in K.

R= Universal gas constant()

M= Molecular weight of the compound.

Now according to the question, avg. speed of He= avg. speed of methane.

Given T of methane= 27°C = (273 +27 = 300) 300K

M of methane= 16 u

T of helium=?

M of helium= 4u

So,$\sqrt\frac{8R300}{\pi 16} =\sqrt\frac{8RT}{/pi 4} \\\sqrt\frac{300}{16} = \sqrt\frac{T}{4}$

Squaring both the side,

$\frac{300}{16} =\frac{T}{4}\\T= \frac{300*4}{16}\\T= \frac{300}{4} \\T= 75$

temperature = 75K (75-273= -198)

The answer is -198 K.