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Question: Density of a gas at one atm pressure is \(1.43 \times {10^{ - 2}}g\;c{c^{ - 1}}\). Calculate the RMS...

Density of a gas at one atm pressure is 1.43×10−2g  cc−11.43 \times {10^{ - 2}}g\;c{c^{ - 1}}. Calculate the RMS velocity of the gas.

Explanation

Solution

As we know that due to collision of molecules the speeds and kinetic energies of individual molecules keeps on changing and to calculate these speeds like RMS i.e the root mean square velocity as the name implies we first take square of the individual speeds and then their mean.

Complete Step by step answer:

As we know that due to collision of molecules the speeds and kinetic energies of individual molecules keeps on changing. However, at a given temperature the average kinetic energy of a gas molecule remains constant.
Let us assume that at a given temperature we have to calculate these speeds like RMS i.e the root mean square velocity as the name implies we first take the square of the individual speeds which may be assumed as V1,V2..{V_1},{V_2}..and n1,n2..{n_1},{n_2}..be the number of molecules then we will take mean and finally the square root of mean. Mathematically it can be represented as:
Vrms=3RTM{V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} where R is the gas constant, T is the temperature and M is the molecular mass of the gas.
Using this formula we can identify the RMS velocity of the gas. But in the question we are given with density and pressure of the gas and no information regarding which gas so to calculate the RMS speed we need to change the above formula in terms of pressure and density.
As we know that, PV=nRTPV = nRT and density=massvolumedensity = \dfrac{{mass}}{{volume}} using these two,
PV=nRT ⇒PM=dRT ⇒M=dRTP  PV = nRT \\\ \Rightarrow PM = dRT \\\ \Rightarrow M = \dfrac{{dRT}}{P} \\\
Putting the value of M in the RMS speed formula we get,
Vrms=3RTPdRT ⇒Vrms=3Pd  {V_{rms}} = \sqrt {\dfrac{{3RTP}}{{dRT}}} \\\ \Rightarrow {V_{rms}} = \sqrt {\dfrac{{3P}}{d}} \\\
Now, we know d=1.43×10−2g  cc−1d = 1.43 \times {10^{ - 2}}g\;c{c^{ - 1}}and , to calculate the speed in ms−1m{s^{ - 1}}we will change d=1.43  Kgm−3d = 1.43\;Kg{m^{ - 3}} P=1atm=1.013×105Nm−2P = 1atm = 1.013 \times {10^5}N{m^{ - 2}}thus putting these values in the formula:
Vrms=3Pd ⇒Vrms=3×1.013×1051.43 ⇒Vrms=461ms−1  {V_{rms}} = \sqrt {\dfrac{{3P}}{d}} \\\ \Rightarrow {V_{rms}} = \sqrt {\dfrac{{3 \times 1.013 \times {{10}^5}}}{{1.43}}} \\\ \Rightarrow {V_{rms}} = 461m{s^{ - 1}} \\\

Therefore, the correct answer is, Vrms=461ms−1{V_{rms}} = 461m{s^{ - 1}}.

Note: Remember that the RMS speed is directly proportional to the square root of the temperature and inversely proportional to the square root of mass, thus when the temperature is increased its RMS speed will increase.