If `u_(1)` and `u_(2)` are the units selected in two systems of measurement and `n_(1)` and `n_(2)` their numerical values, then
If `u_(1)` and `u_(2)` are the units selected in two systems of measurement and `n_(1)` and `n_(2)` their numerical values, then
A
`n_(1)u_(1) = n_(2)u_(2)`
B
`n_(1)u_(1)+n_(2)u_(2) =0`
C
`n_(1)n_(2) = u_(1)u_(2)`
D
`(n_(1)+u_(3))=(n_(2) + u_(2))`
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a
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If u_(1), u_(2) and v_(1), v_(2) are the initial and final velocities of two particles before and after collision respectively, then (v_(2)-v_(1))/(u_(1)-u_(2)) is called ___________.
A physical quantity is measured and the result is expressed as nu where u is the unit used and n is the numberical value. If the result is expressed in various units then
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If u_(1) and u_(2) are the selected in two system of measurement and n_(1) and n_(2) their nomerical values, then
If u_(1) and u_(2) are the selected in two system of measurement and n_(1) and n_(2) their nomerical values, then
A
`n_(1)u_(1)=n_(2)u_(2)`
B
`n_(1)u_(1)+n_(2)u_(2) = 0`
C
`n_(1)n_(2)=u_(1)u_(2)`
D
`(n_(1) + u_(1)) =(n_(2) + u_(2))`
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^(**)) For a given gas the mean free path at a particular pressure is :
A
independent to temperature
B
decreases with rise in temperature
C
increases with rise in temperature
D
directly proportional to `T^(2)`
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)) .
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)) .
A
`4:1:4`
B
`1:4:4`
C
`4:3:2`
D
`1:2:4`
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