Question

The rms velocity of air at NTP will be…………………. if the density of air is $1.29{\text{ }}kg/{m^3}$.
A. $0.485m{s^{ - 1}}$
B. $0.485 \times {10^2}m{s^{ - 1}}$
C. $4.85 \times {10^2}m{s^{ - 1}}$
D. ${10^4}m{s^{ - 1}}$

Answer 1

${v_{rms}}$ is the rms velocity of the gas, which is in terms of gram molecular weight (M) and temperature (T). The formula must be changed to accommodate the density of air (ρ), as Gram molecular weight of air is irrelevant. NTP or STP conditions are given by 297K and 1 atm of pressure.
Formula used:
${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }}$

Complete answer:
Rms velocity or root mean square velocity for a gas is given by usually given by
${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }}$
Here,
Vrms is the rms velocity of the gas, in $m{s^{-1}}$
P is the Pressure, in atm
ρ is the density of the gas, in $kg.{m^{-3}}$
It is mentioned in the question about NTP conditions. The NTP condition is given by, Pressure of 1 atmosphere at room temperature i.e., ${27^o}C$.
So, now we have Pressure,
$P = 1\,atm = 101325Pa$
Here, we are changing the units of atmosphere from atm to Pascals for easier calculation) and density of air,
$\rho = 1.29kg/{m^3}$.
Substituting the above quantities in the formula, we have
${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }}$
$\Rightarrow {v_{rms}} = \sqrt {\dfrac{{3 \times 101325kg.{m^{ - 1}}.{s^{ - 2}}}}{{1.29kg.{m^{ - 3}}}}} = \sqrt {{\text{2,35,639}}{\text{.5}}{{\text{m}}^2}{s^{ - 2}}}$
$\Rightarrow {v_{rms}} = {\text{485}}{\text{.4}}m{s^{ - 1}} = 4.85 \times {10^2}m{s^{ - 1}}$
$\therefore {v_{rms}} = 4.85 \times {10^2}m{s^{ - 1}}$

So, the correct option is C.

Note:
One might ask how we arrived at the formula
${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }} \,\,$ from $\,\,{v_{rms}} = \sqrt {\dfrac{{3RT}}{\rho }}$
First, we need to change, RT using the Ideal Gas Equation, $PV = nRT$ and
Where
P is Pressure of the gas
V is Volume
n is no. of moles
R is Gas constant
T is temperature.
Using the ideal gas equation, we modify
$RT = \dfrac{{PV}}{n} = P \times \dfrac{V}{{Wt.}} \times M = P \times \dfrac{1}{\rho } \times M$.
Now substitute
$RT = P \times \dfrac{M}{\rho }$ in ${v_{rms}} = \sqrt {\dfrac{{3RT}}{\rho }}$
Thus, we have,
${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }}$.
In case, it’s hard to remember the formula you can easily derive it using the ideal gas equation, as stated above.
Also, remember that $1atm{\text{ }} = {\text{ }}101,325Pa$. You need to do this to avoid errors in the calculation part. As per the density is given in kg, we need to change the Pressure into a unit ‘Pa’. As, $1Pa = {\text{ }}1kg/m{s^2}$.