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Estimate the number of collisions per se...

Estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP.The mean free path (average distance covered by a molecule between successive collisions)=1.38xx10^(-5)cm.`

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To estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP, we can follow these steps: ### Step 1: Convert Mean Free Path to Meters The mean free path (λ) is given as \(1.38 \times 10^{-5} \, \text{cm}\). We need to convert this into meters: \[ \lambda = 1.38 \times 10^{-5} \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 1.38 \times 10^{-7} \, \text{m} \] ...
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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(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^(**)) For a given gas the mean free path at a particular pressure is :

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.

The root mean square speed of an ideal gas is given by : u_("rms") = sqrt((3RT)/M) Thus we conclude that u_("rms") speed of the ideal gas molecules is proportional to square root of the temperature and inversely proportional to the square root of the molar mass. The translational kinetic energy per mole can also be given as 1/2Mu_(rms)^(2) . The mean free path ( lambda ) is the average of distances travelled by molecules in between two successive collisions whereas collision frequency (C.F.) is expressed as number of collisions taking place in unit time. The two terms lambda and C.F. are related by : C.F = (u_("rms")/lambda) If n represents number of moles, n0 is number of molecules per unit volume, k is Boltzmann constant, R is molar gas constant, T is absolute temperature and NA is Avogadro's number then which of the following relations is wrong ?

The root mean square speed of an ideal gas is given by : u_("rms") = sqrt((3RT)/M) Thus we conclude that u_("rms") speed of the ideal gas molecules is proportional to square root of the temperature and inversely proportional to the square root of the molar mass. The translational kinetic energy per mole can also be given as 1/2Mu_(rms)^(2) . The mean free path ( lambda ) is the average of distances travelled by molecules in between two successive collisions whereas collision frequency (C.F.) is expressed as number of collisions taking place in unit time. The two terms lambda and C.F. are related by : C.F = (u_("rms")/lambda) A jar contains He and H, in the molar ratio 1 : 5. The ratio of mean translational kinetic energy at the same temperature is

The root mean square speed of an ideal gas is given by : u_("rms") = sqrt((3RT)/M) Thus we conclude that u_("rms") speed of the ideal gas molecules is proportional to square root of the temperature and inversely proportional to the square root of the molar mass. The translational kinetic energy per mole can also be given as 1/2Mu_(rms)^(2) . The mean free path ( lambda ) is the average of distances travelled by molecules in between two successive collisions whereas collision frequency (C.F.) is expressed as number of collisions taking place in unit time. The two terms lambda and C.F. are related by : C.F = (u_("rms")/lambda) Which of the following relation is correct for an ideal gas regarding its pressure (P) and translational kinetic energy per unit volume (E) ?

Calculate the average volume available to a molecule in a sample of oxygen gas at S.T.P. Also calculate the average distance between neighbouring molecules if a oxygen molecule is assumed to be spherical.

The ratio of number of collisions per second at the walls of containers by He and O_(2) gas molecules kept at same volume and temperature is (assume normal incidence on walls)

The mean free path of collision of gas melecules varies with its diameter (d) of the molecules as

The constant motion and high velocities of gas particles lead to some important practical consquences. One such consquences is that gases mix rapidly when they come in contact. Take the stopper off a bottle of perfume, for instance, and the odour will spread rapidly through the room as perfume molecules mix with the molecules in the air. This mixing of different gases by random molecular motion and with frequent collision is called diffusion. A similar process in which gas molecules escape without collision through a tiny hole into a vacuum is called effusion. Both the processes follow Graham's law which is mathematically put as r prop sqrt(1//d) . The average distance travelled by molecules between successive collisions is called mean free path. Answer the following questions on the basis of the above information: Select the correct statement(s).

The constant motion and high velocities of gas particles lead to some important practical consquences. One such consquences is that gases mix rapidly when they come in contact. Take the stopper off a bottle of perfume, for instance, and the odour will spread rapidly through the room as perfume molecules mix with the molecules in the air. This mixing of different gases by random molecular motion and with frequent collision is called diffusion. A similar process in which gas molecules escape without collision through a tiny hole into a vacuum is called effusion. Both the processes follow Graham's law which is mathematically put as r prop sqrt(1//d) . The average distance travelled by molecules between successive collisions is called mean free path. Answer the following questions on the basis of the above information: The stop cocks of the bulbs X (containing NH_(3) ) and Y (containing HCl ), both under indentical conditions, are opened simultaneously. White fumes of NH_(4)Cl , are formed at point B . If AB=36.5 cm , then BC is approximately

HC VERMA ENGLISH-KINETIC THEORY OF GASES-All Questions
  1. Find the ratio of the mean speed of hydrogen molecules to the mean spe...

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  2. Shows a vessel partitioned by a fixed diathermoc separator. Different ...

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  3. Estimate the number of collisions per second suffered by a molecule in...

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  4. Hydrogen gas is contained in a closed vessel at 1atm(100kPa) and 300K....

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  5. Air is pumped into an automobile tyre's tube up to pressure pf 200kPa ...

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  6. Oxygen is filled in a closed metal jar of volume 1.0xx10^(-3)m^(3) at ...

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  7. An air bubble of radius 2.0mm is formed at the bottom of a 3.3m deep r...

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  8. 0.040g of He is kept in a closed container initially at 100.0^(@)C.The...

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  9. During an experiment, an ideal gas is found to obey an additional law ...

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  10. A vessel contains 1.60g of oxygen and 2.80g of nitrogen. The temperatu...

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  11. A uniform tube closed at aone end, contains a pellet of mercury 10cm l...

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  12. A glass tube sealed at both ends is 100cm long, It lies horizontally w...

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  13. A vessel of volume V(0) contains an ideal gas at pressure p(0)and temp...

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  14. One mole of an ideal gas undergoes a process p=(p(0))/(1+((V)/(V(0...

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  15. Show that the internal energy of the air (treated as an ideal gas) con...

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  16. Figure shows a cylindrical tube of radius 5cm and length 20cm. It is c...

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  17. shows a cylindrical tube of cross-sectional area A fitted with two fri...

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  18. Figure shows a large closed cylindrical tank containing water, Initial...

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  19. An ideal gas is kept in a long cylindrical vessel fitted with a fricti...

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  20. An ideal gas is kept in a long cylindrical vessel fitted with a fricti...

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