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Question: Two containers are at the same temperature. The gas in the first container is at pressureand has molecules with massm1 and root-mean-square speedvrms1.The gas in the second is at pressure 2p1and has molecules with massm2and average speedvavg2=2vrms1. Find the ratio m1 / m2of the masses of their molecules.

Short Answer

Expert verified

Answer

The ratio of the masses of molecules in the two containers is 4.71 .

Step by step solution

01

Step 1: Given

  1. The first container is at pressure p1.
  2. The molecule with mass is m1.
  3. The rms speed is vrms1.
  4. The second container is at pressure p2 .
  5. The molecule with mass is m2.

The rms speed is vavg2=2vrms1.

02

Determining the concept

By using equations of average speed of the ideal gas Eq. (19-31) and therms speed of theideal gas Eq. (19-34), find the ratioof the masses (m1 /m2)of their molecule.

  1. The average speed of the ideal gas is,

vavg=8RTπM··············································1

  1. The rms speed of theideal gas is,

vrms=3RTM··············································2

where,γ=Cp/CV, Ris the gas constant, Tis the temperature, vis velocity and M is the molar mass.

03

Determining the ratio  m1 /m2)of the masses of their molecules          

Dividing Eq. (1) by Eq. (2),

vavg2vrms1=8RTπM23RTM1=8M13πM23

Forvavg2=2vrms1,we have-

m1m2=M1M2

From Eq. (3),

m1m2=M1M2m1m2=3π8vavg2vrms12m1m2=3π822m1m2=3π2m1m2=4.71

Hence, the ratio(m1 /m2) of the masses of their molecules is, 4.71 .

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