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Question: Consider a gas with density \(\rho \) and \(C\) as the root mean square velocity of its molecules co...

Consider a gas with density ρ\rho and CC as the root mean square velocity of its molecules contained in a volume. If the system moves as a whole with velocity v, then the pressure exerted by the gas is:
A)13ρC2\dfrac{1}{3}\rho {{C}^{2}}
B) 13ρ(C+v)2\dfrac{1}{3}\rho {{\left( C+v \right)}^{2}}
C) 13ρ(Cv)2\dfrac{1}{3}\rho {{\left( C-v \right)}^{2}}
D) 13ρ(C2v2)\dfrac{1}{3}\rho \left( {{C}^{2}}-{{v}^{2}} \right)

Explanation

Solution

The answer to this question is based on the basic concepts of physical chemistry that includes calculation based on the formula for calculating root mean square velocity that is,vrms=3RTM{{v}_{rms}} = \sqrt{\dfrac{3RT}{M}} and modifying this equation.

Complete Solution :
In the lower classes of chemistry, we have come across the concepts that deal with the thermodynamics of the system.
Let us now see what root means square velocity and its calculation.
- Root mean square velocity vrms{{v}_{rms}} is the measure of the speed of particles in a gas and this quantity takes the account of both molecular weight and temperature which directly affects the kinetic energy of a material.
- Now, in the data it is given that we have to consider root mean square velocity as CC and density as ρ\rho

- Now, by the formula used to calculate the root mean square velocity, we have
vrms=3RTM{{v}_{rms}}=\sqrt{\dfrac{3RT}{M}} ………(1)
where, R is the real gas constant, T is the temperature, M is the molecular mass.
We know by the ideal gas equation,PV=nRTPV=nRT
We know that, n=massmol.wt.n=\dfrac{mass}{mol.wt.}
PV=mMRT\Rightarrow PV=\dfrac{m}{M}RT
P=mMVRT\Rightarrow P=\dfrac{m}{MV}RT

- We know that density is given byρ=massvolume\rho =\dfrac{mass}{volume} , substituting this in above equation,
P=ρMRT\Rightarrow P=\dfrac{\rho }{M}RT
RTM=Pρ\Rightarrow \dfrac{RT}{M}=\dfrac{P}{\rho } …..(2)
Substituting equation number (2) in (1) we get,
vrms=C=3RTM=3Pρ{{v}_{rms}}=C=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3P}{\rho }}
By rearranging the above equation for pressure we get,
P=13ρC2\Rightarrow P=\dfrac{1}{3}\rho {{C}^{2}}
So, the correct answer is “Option A”.

Note: Note that the molar mass and molecular mass are two different quantities. Molar mass is the mass of mole of a substance and molecular mass/weight is mass of a molecule of substance based on the atomic weight of the C-12 atom.