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The number of collisions made by a singl...

The number of collisions made by a single molecule with other molecules per `cm^(3)` per second. Is `Z_(1)`. At constant temperature by how much will `Z_(1)` change if the pressure is doubled in the vessel.

A

Increase `2` times

B

Decrease `2` times

C

Increase `0.5` times

D

Decrease `0.5` times

Text Solution

Verified by Experts

`Z_(1)=sqrt(2) pi sigma^(2)overline(c ) N^(*)` (`overline(c ) =` average velocity)
(`N^(*)=` Number of molecules per `cm^(3)`)
`Z_(1) prop P` (when `T` is constant)
`Z_(1) prop (1)/(sqrt(T))` (when `P` is constant)
`Z_(1) prop sqrt(T)` (when volume is constant)
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The number of bimolecular collisions per cm^(3) per second is Z_(11) . At constant temperature, by how much will Z_(11) change if the pressure is tripled in the vessel?

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Collision cross-section is an area of an imaginary sphere of radius sigma around the molecule within which the centre of another molecule cannot penetrate. The volume swept by a single molecule in unit time is V=(pisigma^(2))overline(u) where overline(u) is the average speed If N^(**) is the number of molecules per unit volume, then the number of molecules within the volume V is N=VN^(**)=(pisigma^(2)overline(u))N^(**) Hence, the number of collision made by a single molecule in unit time will be Z=N=(pi sigma^(2)overline(u))N^(**) In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is sqrt(2)overline(u) as shown below. Number of collision made by a single molecule with other molecule per unit time is given by Z_(1)=pisigma^(2)(overline(u)_("rel"))N^(**)=sqrt(2) pisigma^(2)overline(2)N^(**) The total number of bimolecular collisions Z_(1) per unit volume per unit time is given by Z_(1)=(1)/(2)(Z_(1)N^(**))or Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^(**))N^(**)=(1)/(sqrt(2))pisigma^(2)overline(u)N^(**2) If the collsion involve two unlike molecules then the number of collisions Z_(12) per unit volume per unit time is given as Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2) where N_(1) and N_(2) are the number of molecules per unit volume of the two types of molecules, sigma_(12) is the average diameter of the two molecules and mu is the reduced mass. The mean free path is the average distance travelled by a molecule between two successive collisions. We can express it as follows : lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1)) or" "lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^(**))implies(1)/(sqrt(2)pisigma^(2)overline(u)N^(**)) Three ideal gas samples in separate equal volume containers are taken and following data is given : {:(" ","Pressure","Temperature","Mean free paths","Mol.wt."),("Gas A",1atm,1600K,0.16nm,20),("Gas B",2atm,200K,0.16nm,40),("Gas C",4atm,400K,0.04nm,80):} Calculate number of collision by one molecule per sec (Z_(1)) .

Collision cross-section is an area of an imaginary sphere of radius sigma around the molecule within which the centre of another molecule cannot penetrate. The volume swept by a single molecule in unit time is V=(pisigma^(2))overline(u) where overline(u) is the average speed If N^(**) is the number of molecules per unit volume, then the number of molecules within the volume V is N=VN^(**)=(pisigma^(2)overline(u))N^(**) Hence, the number of collision made by a single molecule in unit time will be Z=N=(pi sigma^(2)overline(u))N^(**) In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is sqrt(2)overline(u) as shown below. Number of collision made by a single molecule with other molecule per unit time is given by Z_(1)=pisigma^(2)(overline(u)_("rel"))N^(**)=sqrt(2) pisigma^(2)overline(u)N^(**) The total number of bimolecular collisions Z_(11) per unit volume per unit time is given by Z_(11)=(1)/(2)(Z_(1)N^(**))"or" Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^(**))N^(**)=(1)/(sqrt(2))pisigma^(2)overline(u)N^(**2) If the collsion involve two unlike molecules then the number of collisions Z_(12) per unit volume per unit time is given as Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2) where N_(1) and N_(2) are the number of molecules per unit volume of the two types of molecules, sigma_(12) is the average diameter of the two molecules and mu is the reduced mass. The mean free path is the average distance travelled by a molecule between two successive collisions. We can express it as follows : lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1)) "or "lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^(**))implies(1)/(sqrt(2)pisigma^(2)overline(u)N^(**)) Three ideal gas samples in separate equal volume containers are taken and following data is given : {:(" ","Pressure","Temperature","Mean free paths","Mol.wt."),("Gas A",1atm,1600K,0.16nm,20),("Gas B",2atm,200K,0.16nm,40),("Gas C",4atm,400K,0.04nm,80):} Calculate ratio of collision frequencies (Z_(11)) (A:B:C) of following for the three gases.

Collision cross-section is an area of an imaginary sphere of radius sigma around the molecule within which the centre of another molecule cannot penetrate. The volume swept by a single molecule in unit time is V=(pisigma^(2))overline(u) where overline(u) is the average speed If N^(**) is the number of molecules per unit volume, then the number of molecules within the volume V is N=VN^(**)=(pisigma^(2)overline(u))N^(**) Hence, the number of collision made by a single molecule in unit time will be Z=N=(pi sigma^(2)overline(u))N^(**) In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is sqrt(2)overline(u) as shown below. Number of collision made by a single molecule with other molecule per unit time is given by Z_(1)=pisigma^(2)(overline(u)_("rel"))N^(**)=sqrt(2) pisigma^(2)overline(u)N^(**) The total number of bimolecular collisions Z_(11) per unit volume per unit time is given by Z_(11)=(1)/(2)(Z_(1)N^(**))"or" Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^(**))N^(**)=(1)/(sqrt(2))pisigma^(2)overline(u)N^(**2) If the collsion involve two unlike molecules then the number of collisions Z_(12) per unit volume per unit time is given as Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2) where N_(1) and N_(2) are the number of molecules per unit volume of the two types of molecules, sigma_(12) is the average diameter of the two molecules and mu is the reduced mass. The mean free path is the average distance travelled by a molecule between two successive collisions. We can express it as follows : lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1)) "or "lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^(**))implies(1)/(sqrt(2)pisigma^(2)overline(u)N^(**)) Three ideal gas samples in separate equal volume containers are taken and following data is given : {:(" ","Pressure","Temperature","Mean free paths","Mol.wt."),("Gas A",1atm,1600K,0.16nm,20),("Gas B",2atm,200K,0.16nm,40),("Gas C",4atm,400K,0.04nm,80):} Calculate number of collision by one molecule per sec (Z_(1)) .

The ratio of number of collision per second at the walls of containers by H_(2) and Ne gas molecules kept at same volume and temperature

In a certain region of space there are only 5 molecules per cm^(3) of an average The temperature is 3K What is the average pressure of this gas ? .

CENGAGE CHEMISTRY-STATES OF MATTER-Exercises (Ture False)
  1. The number of collisions made by a single molecule with other molecule...

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  2. In the van der Waals equation (P + (n^(2)a)/(V^(2)))(V - nb) = nRT ...

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  3. Kinetic energy of a molecule is zero at 0^(@)C

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  4. A gas in a closed container will exert much higher pressure due to gra...

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  5. The graph between PV vs P at constant temperature is linear parallel t...

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  6. Real gases show deviation from ideal behaviour at low temperature and ...

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  7. In the microscopic model of the gas, all the moleculer are supposed to...

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  8. For real gases, at high temperature Z = 0 small value of a means gas...

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  9. Small value of a means, gas can be easily liqueifed.

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  10. Rate of diffusion is directly proportional to the square root of molec...

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  11. For ideal gases, Z = 1 at all temperature and pressure.

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  12. According to charles's law,

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  13. The pressure of moist gas is higher than pressure of dry gas.

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  14. Gases do not occupy volume and do not have force of attraction.

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  15. The van der Waal equation of gas is (P + (n^(2)a)/(V^(2))) (V - nb)...

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  16. Surface tension and surface energy have different dimensions.

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  17. The plot of PV vs P at particular temperature is called isovbar.

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  18. Equal volume of all gases always contains equal number of moles.

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  19. A gas with a = 0 cannot be liquified.

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  20. The van der waals constants have same values for all the gases.

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  21. All the molecules in a given sample of gas move with same speed.

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