Question

If the pressure remains constant, the temperature at which root mean square speed of the molecules of a gas will be twice of its value at 27 degree celsius ​

Answer 1

Answer:

Hope this would help you.

Answer 2

To find:

The temperature at which root mean square speed of the molecules of a gas will be twice of its value at 27 °C ?

Calculation:

Expression for RMS velocity of gases is :

$\therefore \: v_{rms} = \sqrt{ \dfrac{3RT}{m} }$

Now, initially at 27°C

$\implies \: v_{rms} = \sqrt{ \dfrac{3R(27 + 273)}{m} }$

$\implies \: v_{rms} = \sqrt{ \dfrac{3R(300)}{m} }$

Now, at required temperature T, the velocity is [2 × v_(rms)]:

$\therefore \: 2(v_{rms}) = \sqrt{ \dfrac{3RT}{m} }$

Dividing the equations:

$\therefore \: 2 = \sqrt{ \dfrac{T}{300} }$

$\implies \: 4= \dfrac{T}{300}$

$\implies \: T = 1200 \: K$

$\implies \: T = 1200 - 273$

$\implies \: T = {927}^{ \circ} C$

So, required temperature is 927°C.