Question

For a given gas at $1atm$ pressure, rms speed of the molecule is $200m/s$ at ${{127}^{o}}C$. At $2atm$ pressure and at ${{227}^{o}}C$, the rms speed of the molecule will be:
$A.80m/s$
$B.100\sqrt{5}m/s$
$C.80\sqrt{5}m/s$
$D.100m/s$

Answer 1

We should solve this problem by applying the concept of kinetic interpretation of temperature of gases. Root mean square speed of a gas is defined as the square root of square of average velocity of the molecules in the gas. rms speed of a molecule depends on only two factors. The two factors are temperature and molecular weight.
Formula used:
We are using the following relation of rms velocity in terms of temperature and molecular mass of molecules of the gas:-
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$.

Complete answer:
We know that rms velocity of the molecules of the gas depends only on temperature and molecular mass of the gas. Mathematically, which can be written as follows:-
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$Where, $R$is universal gas constant, $M$is molecular mass of the gas and $T$is the temperature.
For a given gas at $1atm$pressure, we have temperature, ${{T}_{1}}={{127}^{o}}C$. Temperature should be written in Kelvin scale. We know that to convert Celsius into kelvin we have to add $273$to the given value.
Hence, ${{T}_{1}}={{127}^{o}}C+273=400K$.
For gas at $1atm$pressure we have rms velocity, ${{v}_{1}}=200m/s$.
${{v}_{1}}=\sqrt{\dfrac{3R{{T}_{1}}}{M}}$
$200=\sqrt{\dfrac{3\times 400R}{M}}$…………… $(i)$
Now, for the same gas at $2atm$ pressure, we have ${{T}_{2}}={{227}^{o}}C={{227}^{o}}C+273=500K$
$M$ will be the same. rms velocity, ${{v}_{2}}$will be calculated as follows:-
$\Rightarrow {{v}_{2}}=\sqrt{\dfrac{3R{{T}_{2}}}{M}}$
$\Rightarrow {{v}_{2}}=\sqrt{\dfrac{3\times 500R}{M}}$……………. $(ii)$
On dividing $(i)$ by $(ii)$ we get,
$\Rightarrow \dfrac{200}{{{v}_{2}}}=\dfrac{\sqrt{\dfrac{3\times 400R}{M}}}{\sqrt{\dfrac{3\times 500R}{M}}}$
On simplification we get,
$\Rightarrow \dfrac{200}{{{v}_{2}}}=\sqrt{\dfrac{400}{500}}$
$\Rightarrow {{v}_{2}}=200\sqrt{\dfrac{5}{4}}$
$\Rightarrow {{v}_{2}}=\dfrac{200}{2}\sqrt{5}$
$\Rightarrow {{v}_{2}}=100\sqrt{5}m/s$

So, the correct answer is “Option B”.

Note:
It is very important for us to understand the basic difference between rms velocity and average velocity of the molecules of the gas. It should be noted that rms velocity of the molecules depends on molar mass $(M)$ only when the gases are different during the process. When gas is the same during the process then molar mass $(M)$ is taken as constant.