Question

8. Molar mass of certain gas Ą is half that of B. If rms
speed of molecules of A at certain temperature is
200 ms. The rms speed of B at the temperature
half that of A will be​

Answer 1

let rms 0F A = VA=(3RT/M) = 200m/s & that of B as VB = 3RT'/M' WHERE T' = T/2 & M'= 2M

USING ABOVE 2 EQN WE GET VA/VB=(3RT/M) /{(3R(T/2)/2M} =( 2X2)1/2

VB= 200/2 =100m/s

Answer 2

Answer : The rms speed of B will be, 100 m/s

Explanation :

Formula used for root mean square speed :

$\nu_{rms}=\sqrt{\frac{3RT}{M}}$

where,

$\nu_{rms}$ = rms speed of the molecule

R = gas constant

T = temperature

M = molar mass of the gas

And the formula for two gases will be,

$\frac{\nu_A}{\nu_B}=\sqrt{\frac{T_{A}\times M_{B}}{T_{B}\times M_{A}}}$

Let the molar mass of gas B = 2M

Molar mass of gas A = 1M

Let temperature of gas A = 2T

Temperature of gas B = 1T

$\nu_A$ = rms speed of molecule A = 200 m/s

$\nu_B$ = rms speed of molecule B = ?

Now put all the given values in the above formula, we get

$\frac{200m/s}{\nu_{B}}=\sqrt{\frac{2T\times 2M}{1T\times 1M}}$

$\nu_B=100m/s$

Therefore, the rms speed of B will be, 100 m/s