Question

At 400K, the root mean square (rms) speed of gas X (molecular mass=40) is equal to most probable speed of gas Y at 60K. The molecular mass of the gas Y is:

Answer 1

Hint : The formula for root mean square speed is ${{V}_{rms}}=\sqrt{\dfrac{3RT}{M}}$
And, the formula for most probable speed is ${{V}_{mps}}=\sqrt{\dfrac{2RT}{M}}$
As the question says, we will equate both the speeds and calculate the molar mass of the gas Y.

Complete step by step solution :
Root mean square speed of a gas is defined as the square root of the squared-average velocity of the gas molecules at a particular temperature.
Most probable speed of a gas is the speed possessed by the maximum number of gas molecules.

The formula for root mean square speed is ${{V}_{rms}}=\sqrt{\dfrac{3RT}{M}}$
And, the formula for most probable speed is ${{V}_{mps}}=\sqrt{\dfrac{2RT}{M}}$
Where R= gas constant,
T= absolute temperature
M= molar mass of the gas
Now we take ${{M}_{X}}$, ${{T}_{X}}$ as the molar mass and temperature of gas X and ${{M}_{Y}}$, ${{T}_{Y}}$ as the molar mass and temperature of gas Y
We now put the values in the formula of both the speeds and equate them to get the answer
${{V}_{rms}}x={{V}_{mps}}y$
$\sqrt{\dfrac{3R{{T}_{X}}}{{{M}_{X}}}}=\sqrt{\dfrac{2R{{T}_{Y}}}{{{M}_{Y}}}}$
We have canceled the R and the removed the square root (by squaring both sides)
$\dfrac{2{{T}_{Y}}{{M}_{X}}}{3{{T}_{X}}}={{M}_{Y}}$ Put the values of ${{M}_{X}}$ =40, ${{T}_{X}}$=400, ${{T}_{Y}}$=60
${{M}_{y}}=\,\dfrac{2\times 60\times 40}{3\times 400}$
Ans= ${{M}_{Y}}$= 4
Molar mass of gas Y is 4

So, the answer for the given question is 4.

Note : The formulas for the root mean square speed and most probable speed is similar. You can get confused while applying formulas. Convert the temperature into kelvin. You must see if ‘molar mass’ is given or ‘given mass’.