Question

The temperature at which the most probable speed of $C{{O}_{2}}$ molecules be twice as that of 50$^{0}C$ is
(A) 200$^{0}C$
(B) 1292 K
(C) 100$^{0}C$
(D) 646 K

Answer 1

Hint: Most probable speed(velocity) of any gaseous molecule is given by the formula $\sqrt{\dfrac{2RT}{M}}$based on Maxwell-Boltzmann distribution where this leads to the required answer .

Complete answer:
From the physical chemistry part in the previous classes, we have come across the term most probable speed in the kinetic theory of gases which provides a simplified explanation regarding many gaseous properties.
- The Maxwell-Boltzmann distribution is applicable to the particles having velocities in the three dimensions.
Now, let us consider the temperature of $C{{O}_{2}}$as ‘T’ whose velocity is doubled.
Also, most probable speed of gas will be directly proportional to the square root of temperature according to equation $\sqrt{\dfrac{2RT}{M}}$because the quantities R ( real gas constant) and M ( mass of one molecule of gas) remain constant.
Thus we can write the above equation as,
 ${{V}_{mps}}\propto \sqrt{T}$
For two components, we have $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\sqrt{\dfrac{{{T}_{1}}}{{{T}_{2}}}}$ ..…(1)
Given that ${{T}_{2}}$ = 50$^{0}C$ = 323 K and ${{T}_{1}}=T$ which is to be found.
When the most probable speed becomes twice, then equation number (1) becomes,
$2=\sqrt{\dfrac{T}{323}}$
\[\Rightarrow {{2}^{2}}=\dfrac{T}{323}\] $\Rightarrow T=4\times 323=1292K$

Therefore, the correct answer is option (B).

Additional information:
The most probable speed of the gas is dependent on the molar mass. When molar mass is increased, the most probable speed decreases, which means the highest point on the curve shifts to the left side.

Note:
There occurs confusion between the formulas for root mean square speed and most probable speed. Most probable speed is the one which informs us about the speed possessed by the maximum number of gas molecules and root mean square speed is square root of the average of square of velocity.