InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The diameter of an oxygen molecule is 3 `Ã…` The ratio of molecular volume to the actual volume occupied by the oxygen gas at STP isA. `2xx10^(-4)`B. `1xx10^(-4)`C. `1.5xx10^(-4)`D. `4xx10^(-4)` |
|
Answer» Correct Answer - D Here, diameter,d `=3Ã…` `thereforer=d/2=3/2xx10^(-10)m=3/2xx10^(-8)cm=1.5xx10^(-8)cm` Molecular volume of oxytgen gas, `V=4/3pir^(3)XN` N `to` Avogadros number Actual volume occupied by 1 mol of oxygen gas at STP=`22,400 cm^(3).` `thereforeV=4/3xx3.14xx(1.5xx10^(-8))^(3)xx6.023xx10^(23)=8.51cm^(3)` Therefore, ratio of the molecular volume to the actual volume of oxygen. `(V)/V_(1)=(8.51)/(22,400)=3.8xx10^(-4)=4xx10^(-4)` |
|
| 2. |
Molecular motion shows itself asA. temperatureB. internal energyC. frictionD. viscosity |
| Answer» Correct Answer - A | |
| 3. |
1 mole of a gas with `gamma=7//5` is mixed with 1 mole of a gas with `gamma=5//3`, then the value of `gamma` for the resulting mixture isA. `7/5`B. `2/5`C. `3/2`D. `(12)/(7)` |
| Answer» Correct Answer - C | |
| 4. |
The kinetic theory of gases gives the formula `PV=1/3Nmv^(2)` for the pressure P exerted by a gas enclosed in a volume V. The term Nm representsA. the mass of a mole of the gasB. the mass of the present in the volume VC. the averatge mass of one molecule of the gasD. the total number of molecules present in volume V |
| Answer» Correct Answer - B | |
| 5. |
Assertion: Each vibrational mode gives two degrees of freedom. Reason: By law of equipartition of energy, the energy for each degree of freedom in thermal equlibrium is `2k_(B)T.`A. If both assertion and reason are true and reason is the correct explanation os assertion.B. If both assertion and reason are true but reason is not be correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - C By law of equipartition of energy, the energy for each degree of freedom in thermal equilibrium is `1/2k_(B)T.` Each quadratic term from in the total energy rxpression of a molecules is to be conunted as a degree of freedom. Thus each vibriational mode gives 2 degrees of freedom i.e., kinetic and potential energy modes, corresponding to the energy `2((1)/(2)k_(BT))=k_(B)T.` |
|
| 6. |
According to the law of equipartition of energy, the energy associated with each degree of freedom is :A. `E=k_(B)T`B. `E=(1)/(2)k_(B)T`C. `E=3k_(B)T`D. `E=(3)/(2)k_(B)T` |
| Answer» Correct Answer - D | |
| 7. |
`0.014` kg of nitrogen is enclosed in a vessel at a temperature of `24^(@)C.` At which temperature the rms velocity of nitrogen gas is twice it’s the rms velocity at `27^(@)C`?A. 1200 KB. 600 KC. 300 KD. 150 K |
|
Answer» Correct Answer - A Using, `v_(rms)=sqrt((3RT)/(m))` or `((v_(rms))_(1))/((v_(rms))_(1))=sqrt((T_(1))/(T_(2)))" "(therefore"R and m are constant)"` According to question, `(v_(rms))_(2)=2(v_(rms))_(1)` `therefore(v_(rms))_(1)/(2(v_(rms))_(1))=sqrt((300)/(T_(2)))implies1/4=(300)/(T_(2))` `T_(2)=300xx4=1200K.` |
|
| 8. |
Which of the following diagrams (figure) depicts ideal gas behaviour?A. B. C. D. |
|
Answer» We know that ideal gas equation is `pV =nRT` (a) When pressure p = constant From (i) volume `V prop` Temperature T (b) When T = constant From (i) pV = constant So, the graph is rectangular hyperbola. (C ) When V= Constant. From(i) `p prop T` So, the graph is straight line passes through the origin. (d) From (i) `p V prop T` `rArr " " (pV)/(T)=` constant `rArr` So, the graph hence through origin. So (d) is not correct. |
|
| 9. |
Two molecules of a gas have speeds of `9xx10^(6) ms^(-1)` and `1xx10^(6) ms^(-1)`, respectively. What is the root mean square speed of these molecules. |
|
Answer» For n- molecules we know that `V_("rms") =sqrt((V_(1)^(2)+V_(2)^(2)+V_(3)^(2)+.........+V_(n)^(2))/(n))" "[V_("rms")underset("square velocity")(= "root mean"]]` Where`V_(1),V_(2),V_(3) ..................V_(n)` are individual velocities of n-molecules of the gas. For two molecules, `V_("rms") =sqrt((V_(1)^(2)+V_(2)^(2))/(2)) " "[V_(1),V_(2),V_(3).............V_(n)"are individual velocity"]` `" Given " " "V_(1) 9xx 10^(6)m//s` `"and "" "V_(2) 1xx 10^(6) m//s` `:. " " V_("rms") =sqrt(((9xx10^(6))^(2)xx(1 xx 10^(6))^(2))/(2))` `=sqrt((81 xx 10^(12) + 1xx10^(12))/(2))` `=sqrt(((81 +1)xx10^(12))/(2))` `=sqrt((82 xx 10^(12))/(2))` `=sqrt(41)xx10^(6) m//s` |
|
| 10. |
The molecules of a given mass of a gas have root mean square speeds of `100 ms^(-1) "at " 27^(@)C` and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at `127^(@)C` and 2.0 atmospheric pressure?A. `(200)/(sqrt3)`B. `(100)/(sqrt3)`C. `(400)/(3)`D. `(200)/(3)` |
|
Answer» Correct Answer - A Here, `v_(rms)=100ms^(-1),T_(1)=27^(@)C=(27+273)K=300K` `P_(1)=1 atm,v_(rms2)=?,T_(2)=127^(@)C=(127+273)K=400 K` `P_(2) =2atm` From `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)),(V_(1))/(V_(2))=(P_(2))/(P_(1))2xx(300)/(400)=3/2` Again `P_(1)=1/3 (M)/(V_(1))v_(rms1)^(2)and P_(2)=1/3(M)/(V_(2))v_(rms_(2))^(2)` `therefore(v_(rms_(2))^(2))/(v_(rms_(1))^(2))xx(V_(1))/(V_(2))=(P_(2))/(P_(1))` `v_(rms_(2))^(2)=v_(rms_(1))^(2)xx(P_(2))/(P_(1))xx(V_(2))/(V_(1))=(100)^(2)xx2xx2/3` `v_(rms_(2))=(200)/(sqrt3)ms^(-1)` |
|
| 11. |
The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is v, at 480 K it becomesA. `4v_(rms)`B. `2v_(rms)`C. `(2v_(rms))/(2)`D. `(v_(rms))/(4)` |
|
Answer» Correct Answer - B `(v_(rms_(2)))/(v_(rms_(1)))=sqrt((T_(2))/(T_(1)))=sqrt((480)/(120))=2` `v_(rms_(2))=2(v_(rms))` |
|
| 12. |
Assertion : The ratio `C_(P)// C_(upsilon)` for a diatomic gas is more than that for a monoatomic gas. Reason : The moleculess of a monoatomic gas have more degrees of freedom than those of a diatomic gas.A. If both assertion and reason are true and reason is the correct explanation os assertion.B. If both assertion and reason are true but reason is not be correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - D For a monatomic gas, number of degree of freedom, n = 3, and for a diatomic gas, n= 5. As, `(C_(P))/(C_(V))=gamma=1+2/n,` For monatomic gas, `(C_(P))/(C_(V))=5/3=1.67and ` For diatomic gas, `(C_(P))/(C_(V))=7/5=1.4` `((C_(P))/(C_(V)))_("monatomic")gt((C_(P))/C_(V))_("diatomic")` |
|
| 13. |
An air bubble of volume `1.0 cm^(3)` rises from the bottom of a lake 40 m deep at a temperature of `12^(@) C`. To what volume does it grow when it reaches the surface, which is at a temperature of `35^(@) C`. ? Given `1 atm = 1.01 xx 10^(5) Pa`.A. `10.6xx10^(-6)m^(3)`B. `5.3xx10^(-6)m^(3)`C. `2.8xx10^(-6)m^(3)`D. `15.6xx10^(-6)m^(3)` |
|
Answer» Correct Answer - B Using, `P(1)=rho_(20+pgh` Here, `P_(2)=1.013xx10^(5)atm,h=40m` `rho=10^(3)kgm&(3)`(density of water). `g=9.8ms^(2)` `therefore P_(1)=1.013 xx10^(5)+10^(3)xx9.8xx40=493300 Pa` Now, `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` Here, `T_(1)=(12+273)=285K,T_(2)=(32+273)=328K,V_(1)=1xx10^(-6)m^(3)` `V_(2)` is the volume of the air bubble when it reaches the surface `thereforeV_(2)=(P_(1)V_(1)T_(2))/(T_(1)PP_(2))=((493300xx110^(-6)))/(285xx1.013xx10^(5))xx308` `=5.26xx10^(-6)m^(3)=5.3xx10^(-6)m^(3)` |
|
| 14. |
A gas mixture consists of 2 moles of oxygen and 4 of Argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system isA. 4 RTB. 9 RTC. 11 RTD. 15 RT |
|
Answer» Correct Answer - C For two moles of diatomic oxygen with no vibrational mode. `U_(1)=2xx5/2RT=5RT` For four moles of monoatomic Argon, `U_(2)=4xx3/2RT=6RT` `thereforeU=U_(1)+U_(2)=5RT+6RT=11RT` |
|
| 15. |
One kg of a diatomic gas is at pressure of `8xx10^4N//m^2`. The density of the gas is `4kg//m^3`. What is the energy of the gas due to its thermal motion?A. `3xx10^(4) J`B. `5xx10^(4) J`C. `6xx10^(4) J`D. `4xx10^(4) J` |
|
Answer» Correct Answer - B For a diatomic molecular ga, internal energy per molecule is `U=5/2NK_(B)T` `PV=NK_(B)T` From equations is (i) and (ii) `U=5/2PV` Here, `P=8xx10^(4)Nm^(-2),rho=4 kgm^(-3),m=1kg` `thereforeU=5/2xx8xx10^(4)xx1/4=5xx10^(4)J` |
|
| 16. |
Assertion: The ratio of rms speed and average speed of a gas molecules at a given temperture is `sqrt3:sqrt(8//pi)` Reason: `c_(rms)c_(av.)`A. If both assertion and reason are true and reason is the correct explanation os assertion.B. If both assertion and reason are true but reason is not be correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - B `c_(rms)=sqrt((3kT)/(m))and c_(av)=sqrt((8kT)/(pim))` `therefore(c_(rms))/(c_(av))=(sqrt3)/(sqrt((8//pi)))gt1` Thus, Assertion is true. Here Reason is also true because `c_(rms) gt c_(av)` but Reason is not the correct explanation of Assertion. |
|
| 17. |
An insulated container containing monoatomic gas of molar mass m is moving with a velocity `V_(0)`. If the container is suddenly stopped , find the change in temperature .A. `(mv_(0)^(2))/(2R)`B. `(mv_(0)^(2))/(3R)`C. `(3mv_(0)^(2))/(2R)`D. `(1)/(2)(mv_(0)^(2))/(R)` |
|
Answer» Correct Answer - B Here, M= mole mass of the gas `v_(o)` = initial speed Mass of gas = n m (n= number of moles in the gas) `therefore` Initial Kinetic energy of the gas `=1/2(nm)v^(2)` Final KE of gas = 0 Change in Kinetic energt, `DeltaK=1/2(nm)v^(2)` Let change in temperature of gas = `DeltaT` Change in internal energy of the gas `DeltaU=nC_(v)DeltaT=n((3)/(2)R)DeltaT` As, `DeltaU=DeltaK` `implies3/2nRDeltaT=1/2mnv_(o)^(2)` `impliesDeltaT=(mv_(0)^(2))/(3R)` |
|
| 18. |
An insulated container containing monoatomic gas of molar mass m is moving with a velocity `V_(0)`. If the container is suddenly stopped , find the change in temperature .A. `(mv_(0)^(2))/(2R)`B. `(mv(0)^(2))/(3R)`C. `(R )/(mv_(0)^(2))`D. `(3mv_(0)^(2))/(2R)` |
|
Answer» Correct Answer - B If the container is suddenly stopped loss in kinetic energy of gas `=1/2(mn)v_(0)^(2),` where n is number of moles of gas. Let `DeltaT` is the fall in temperature of gas, Then,`n((3)/(2)RDeltaT)=1/2mnv_(0)^(2)` `DeltaT=(mv_(0)^(2))/(3R)` |
|
| 19. |
When an ideal gas is compressed adiabatically , its temperature rises the molecules on the average have more energy than before. The kinetic energy increases,A. because of collisions with moving parts of the wall onlyB. because of collisions with the entire wallC. because the molecules gets accelerated in their motion inside the volumeD. because the redistribution of energy amongest the molecules |
| Answer» When the gas is compressed adiabatically, the total work done on the gas increases its internal energy which in turn increase the KE of gas molecules and hence, the collisions between molecules also increases. | |
| 20. |
Assertion: All molecules of an ideal gas more with the same speed. Reason: There is no attraction between the molecules in an ideal gas.A. If both assertion and reason are true and reason is the correct explanation os assertion.B. If both assertion and reason are true but reason is not be correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 21. |
Assertion: In a mixture of gases at a fixed temperatue, the heavier molecule has the lower average speed. Reason: Temperature of a gas is a measure of the average kinetic energy of a molecule.A. If both assertion and reason are true and reason is the correct explanation os assertion.B. If both assertion and reason are true but reason is not be correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - B | |
| 22. |
Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same velocity V. The mass of the gas in A is `m_A,` and that in B is `m_B`. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be `DeltaP and 1.5 DeltaP` respectively. ThenA. `4m_(A)=9m_(B)`B. `2m_(A)=3m_(B)`C. `3m_(A)=2m_(B)`D. `9m_(A)=4m_(B)` |
|
Answer» Correct Answer - C Process is isothermal. Therefore, T = constant. Volume is increasing, therefore, pressure will decrease `(Pprop(1)/(V)).` In chamber `A,Deltap=(pA)i-(pA)f=(n_(A)RT)/(V)-(n_(A)RT)/(2V).` `=(n_(A)RT)/(2V)" "...(i)` In chamber `B,1.5Deltap=(pa),-(p_(B))f=(n_(B)RT)/(V)-(n_(B)RT)/(2V)` `=(n_(B)RT)/(2V)" "...(ii)` From eqs. (i) and (ii), `(n_(A))/(n_(B))=(1)/(1.5)=2/3,(m_(A)//M)/(m_(B)//M)=2/3or (m_(A))/(m_(B))=2/3therefore3m_(A)=2m_(B)` |
|
| 23. |
Assetion : Specific heat of a gas at constant pressure is greater than its specific heat at constant volume. This is because at constant pressure, some heat is spent in expansion of the gas.A. If both assertion and reason are true and reason is the correct explanation os assertion.B. If both assertion and reason are true but reason is not be correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
|
Answer» Correct Answer - A In cases of `C_(v),` volume of the gas is kept constant and heat is required only for raising the temperature of one gram mole of the gas through `1^(@)C` or 1 K. No heat, what so ever, is spent in expansion of the gas. In case of `C_(p),` as pressure of the gas is kept constant, the gas would expand on heating. Therefore, some heat is sent in expansion of the gas against external pressure. This in addition to the amount of heat energy required for raising the temperature of one gram mole of the gas through `1^(@)C` or 1 K. Hence specific heat at constant pressure is greater than the specific heat at constant volume. |
|
| 24. |
The temperature of an ideal gas is increased from `27^(@)C to 127^(@)C`, the percentage increase in `V_(rms)` is [2013]A. `37%`B. `11%`C. `33%`D. `15.5%` |
|
Answer» Correct Answer - D `v_(rms)=sqrt((3RT)/(M))` `%"increase in" v_(rms)(sqrt((3RT_(2))/(M))-sqrt((3RT_(1))/(M)))/sqrt((3RT_(1))/(M))xx100` `=(sqrt(T_(2))-sqrtT_(1))/(sqrtT_(1))xx100=(sqrt(400)-sqrt(300))/(sqrt300)xx100` `=(20-17.32)/(17.32)xx100=15.5%` |
|
| 25. |
A gas in a cylinder. Its temperature is increased by `20%` on kelvin sacle and volume is reduced to `90%` how much percentage of the gas has to leak for pressure to remain constant?A. `20%`B. `25%`C. `30%`D. `40%` |
|
Answer» Correct Answer - B `PV = n RT " "(i)` `P((90V)/(100))=n_(1)R(T+(20)/(100)T)` `or 9/10PV=6/5m+(1)RT" "(ii)` Dividing (ii) by (i),we get `6/5(n_(1))/(n)=9/10or (n_(1))/(n)=(9xx5)/(6xx10)=3/4or n_(1)=3/4n` % of gas leaked `=(n-n_(1))/(n)xx100=(1-3/4)xx100=25%` |
|
| 26. |
The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, wll beA. PV =(5/32) TRB. PV=5RTC. PV=(5/2) RTD. PV= (5/16) RT |
|
Answer» Correct Answer - A As PV=nRT `n=(m)/("molecular mass")=5/32impliesPV=((5)/(32))RT` |
|
| 27. |
Consider an ideal gas with following distribution of spedds. (a) Calculate `V_("rms")` and henceT. `(m=3.0xx 10^(-26) kg)` (b) If all the molecules with speed `1000m//s` escape from the system, calculate new `V_("rms")` and hence T. |
|
Answer» (a) We know that `V_("rms")^(2) = (underset(i)(Sigma)n_(i)V_(i)^(2))/(Sigman_(i))` This is the rms speed for all molecules collectively. Now , `V_("rms") = ((underset(i)(Sigma)n_(i)v_(i)^(2))/(Sigman_(i)))^(1/2)` `=sqrt((n_(1)v_(1)^(2) +n_(2)v_(2)^(2)+n_(3)v_(3)^(2)+.........+n_(n)v_(n)^(2))/(n_(1)+n_(2) +n_(3)+............+n_(n)))` `=sqrt((n_(1)v_(1)^(2)+n_(2)v_(2)^(2)+n_(3)v_(3)^(2)+n_(4)v_(4)^(2)+n_(5)v_(5)^(2))/(n_(1)+n_(2)+n_(3)+n_(4)+n_(5)))` `=sqrt((10xx(200)^(2) +20xx(400)^(2)+40xx(600)^(2)+20xx(800)^(2)+10xx(1000)^(2))/(100))` `=sqrt((1000xx(4+32+144 +128 +100)))` `=sqrt(408 xx 1000) ~~ 639 m//s.` Now, according to kinetic theory of gasses `1/2 mv_("rms")^(2) =3/2 kg T" "[underset( m="mass of gaseous molecules")(k_(B) ="Boltzmann constant"]]` `T =1/3(mv_("rms")^(2))/(K_(B)) =1/3 xx (30xx10^(-26) xx 4.08 xx 10^(5))/(1.38 xx 10^(-23))` `=2.96 xx 10^(2) K =296 K` (b) If all the molecules with speed 1000`m//s` escape ,then `V_("rms")^(2) =(10xx (200)^(2) +20 xx(400)^(2) +40xx(600)^(2)+20 xx(800)^(2))/(90)` `=(10 xx100^(2) xx (1xx4 +2xx16 +4xx36 +2 xx 64))/(90)` `=10000 xx (308)/(9) =342 xx 1000 m^(2)//s^(2)` `V_("rms") =584 m//s` `"Again"" "T=1/3 (mv_("rms")^(2))/(K)` `=1/3 xx (3xx10^(-26) xx 3.42 xx 10^(5))/(1.38 xx 10^(-23))` `=2.478 xx 10^(2)` `=247 .8 -248 K` Note After escaping of molecules with speed of 1000 `m//s` the temperature in part (b) is 248 K whereas in part (a) before escaping of molecules the temperature was 297 K. thus evaprations facilitates cooling. |
|
| 28. |
Ten small planes are flying at a speed of `150 km//h` in total darkness in an air space that is `20 xx 20 xx 1.5 km^(3)` in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are, On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximately by a sphere of radius 10 m.A. 125 hB. 220 hC. 432 hD. 225 h |
|
Answer» Correct Answer - D Here, `v=150Kmh^(-1)N=10` `V=20xx20xx1.2Km^(3).` Dimeter of plane, `d=2R=2xx10` `= 20 m 20xx10^(-3)km` `n=N/V=(10)/(20xx20xx1.5)=0.067Km^(-3)` Mean free path of a plane `lamda =(1)/(sqrt2pid^(2)n)` Time elapse before collision of two planes randomaly, `t=(lamda)/(v)=(1)/(sqrt2pid^(2)nv)` `=(1)/(1.414xx3.14xx(20)^(2)xx10^(-6)xx(0.0167)xx(150))` `=(10^(6))/(4449.5)=224.74h~~225h` |
|
| 29. |
If three molecules have velocities `0.5kms^(-1),1kms^(-1)and 2kms^(-1),` the ratio of the rms speed and average speed isA. `2.15`B. `1.13`C. `0.53`D. `3.96` |
|
Answer» Correct Answer - B rms speed, `v_(rms)=sqrt((v_(1)^(2)+v_(2)^(2)+v_(3)^(2))/(3))` `=sqrt(((0.5)^(2)+(1)^(2)+(2)^(2))/(3))=2.32kms^(-1).` Average speed, `v_(av)=(v_(1)+v_(2)+v_(3))/(3)=(0.5+1+2)/(3)=1.17kms^(-1)` `=(v_(rms))/(v_(av))=(1.32)/(1.17)~~1.13` |
|
| 30. |
Which one of the following is not an assumption in the kinetic theory of gases?A. The volume occupied by the molecules of the gas is negligible.B. The force of attraction between the molecules is negligable.C. The collision between the molecules are elastic.D. All molecules have same speed. |
|
Answer» Correct Answer - D Molecules of an ideal gas move randomly with differernt speeds. |
|
| 31. |
What is the average distance between atoms (interatomic distance) in water ? |
| Answer» A given mass of water in vapour state has `1.6710^3` times the volume of the same mass of water in liquid state (Ex. 13.1). This is also the increase in the amount of volume available for each molecule of water. When volume increases by `10^3 `times the radius increases by `V^(1/3)` or 10 times, i.e., 10 2 Å = 20 Å. So the average distance is 2 20 = 40 Å. | |
| 32. |
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms : [Hint : Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å ]. |
|
Answer» Atomic mass of a substance = M Density of the substance = `rho` Avogadro’s number = `N = 6.023 × 10^23` Volume of each atom `=4/3pir^3` Volume of N number of molecules `=4/3pir^3` N … (i) Volume of one mole of a substance `= (M)/(rho)` `=4/3pir^3 N=(M)/(P)` ` therefore r=root3((3M)/(4pirhoN)` For carbon: `M = 12.01 × 10^(–3) kg,` `rho = 2.22 × 10^3 kg m^(–3)` `thereforer=((3xx12.01xx10^-3)/(4pixx2.22xx10^3xx6.023xx10^23))^(1/3)` = 1.29 Å Hence, the radius of a carbon atom is 1.29 Å. For gold: ` M=197.00XX10^(-3) kg` `rho=19.32xx10^(3) kg m^(-3)` `thereforer=((3xx197xx10^-3)/(4pixx19.32xx10^3xx6.023xx10^23))^(1/3)` =1.59 Å Hence ,the radius of a gold atom is 1.59 Å For liquid nitrogen: `M=14.01xx10^(-3) kg ` `rho=1.00xx10^3 kg m^(-3)` `thereforer=((3xx14.01xx10^-3)/(4pixx1.00xx10^3xx6.023xx10^23))^(1/3)` =1.77 Å Hence, the radius of a liquid nitrogen atom is 1.77 Å. For lithium: `M=6.94 xx 10^(-3) kg ` `rho = 0.53xx10^3 kg m^(-3)` `thereforer=((3xx6.94xx10^-3)/(4pixx0.53xx10^3xx6.023xx10^23))^(1/3)` For liquid flurorine `M = 19.00 xx 10^(–3)` kg `rho = 1.14 × 103 kg m^(–3)` `thereforer=((3xx19xx10^-3)/(4pixx1.14xx10^3xx6.023xx10^23))^(1/3)` = 1.88 ÅHence, the radius of a liquid fluorine atom is 1.88 Å |
|
| 33. |
At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at `47^(@)C`?A. 10 KB. 20 KC. 30 KD. 40 K |
|
Answer» Correct Answer - B Now, rms velocity of `H_(2)` molecule = rms velocity of `O_(2)` molecule `sqrt((3RxxT)/(2))=sqrt((3Rxx(47+273))/(32))` `T=(2xx320)/(32)=20K` |
|
| 34. |
The heat capacity per mole of water is (R is universal gas constant)A. 9RB. `9/2R`C. 6RD. 5R |
|
Answer» Correct Answer - A We treat water like a solic. For each atom average energy is `3k_(B)T.` Water molecule has three atoms, two hydrogen and one oxygen. The total energy of one mole of water is `U=3xx3k_(B)TxxN_(A)=9RT" "(thereforeK_(B)=(R)/(N_(A)))` `therefore` heat capacity per mole of water is `C=(DeltaQ)/(DeltaT)=(DeltaU)/(DeltaT)=9R` |
|
| 35. |
The ratio of the molar heat capacities of a diatomic gas at constant pressure to that at constant volume isA. `7/5`B. `3/2`C. `3/5`D. `5/2` |
|
Answer» Correct Answer - A For a diatomic gas, Molar heat capacity at constantpressure, `C_(P)=7/2R` Molar heat capacity at constant volume , `C_(V)=5/2R` `(C_(P))/(C_(V))=(7//2R)/(5//2 R)=7//5` |
|
| 36. |
Three moles of oxygen ar mixed with two moles of helium. What will be the ratio of specific heats at constant pressure and constant volume for the mixture ?A. `2.5`B. `3.5`C. `1.5`D. 1 |
|
Answer» Correct Answer - C For a monatomic gas like helium `gamma_(He)=5/3` For a diatomic gas like oxygen `gamma_(O_(2))=7/5` `therefore gamma_(mix)=(3xxgamma_(O_(2))+2xxgamma_(He))/((3+2))` `=(3xx7/5+2xx5/3)/(5)=(21/5+10/3)/(5)=(113)/(15xx5)=1.5` |
|
| 37. |
If `C_(p) and C_(v)` denoted the specific heats of unit mass of nitrogen at constant pressure and volume respectively, thenA. `C_(p)=C_(v)=( R)/(28)`B. `C_(p)-C_(v)=(R )/(7)`C. `C_(p)-C_(v)=(R )/(14)`D. `C_(p)-C_(v)=R` |
|
Answer» Correct Answer - A using Mayers equation, for 1 mole of gas, `C_(P)-C_(V)=R` `therefore"for unit mass,"C_(P)-C_(V)=R/M=(R)/(28)` `" "(therefore"M for nitrogen=28)"` |
|
| 38. |
N molecules each of mass m of gas A and 2 N molecules each of mass 2m of gas B are contained in the same vessel which is maintined at a temperature T. The mean square of the velocity of the molecules of B type is denoted by `v^(2)` and the mean square of the x-component of the velocity of a tye is denoted by `omega^(2)`. What is the ratio of `omega^(2)//v^(2) = ?`A. `3:2`B. `1:3`C. `2:3`D. `1:1` |
|
Answer» Correct Answer - C The mean sequare velocity of gas molecules is giben by `v^(2)=(3kT)/(m).` For gas A, `v_(A)^(2)=(3kT)/(m)" "...(i)` For a gas molecule. `v^(2)-v_(x)^(2)+v_(y)^(2)+v_(z)^(2)=3v_(x)^(2)" "(thereforev_(x)^(2)=v_(y)^(2)=v_(z)^(2))` or `v_(x)^(2)=(v^(2))/(3)` From eqn. (i), we get `w^(2)=v_(x)^(2)=[(3kT)/(m/(3))]=(kT)/(m)" "...(ii)` For gas B, `v_(B)^(2)=v^(2)=(3kT)/(2m)" "...(iii)` Dividing eqn. (ii) by eqn. (iii), we get `(w^(2))/(v^(2))=(KT)/((m/(3kT))/(2m))=2/3` |
|
| 39. |
A cylinder contained `10kg`of gas at pressure `10^(7) N / m^(2)`. The quantity of gas taken out of cylinder if final pressure is `2.5 xx 10^(6) N//m^(2)` is (Assume temperature of gas is constant)A. `9.5 ` kgB. `7.5` kgC. `14.2` kgD. zero |
|
Answer» Correct Answer - B `(P_(1))/(P_(2))=(rho_(1))/(rho_(2))=(m_(1))/(m_(2))` `m_(2)=(P_(2))/(P_(1))xxm_(1)=(2.5xx10^(6))/(10^(7))xx10=2.5 kg` Mass of gas taken out `=10-2.5=7.5 kg` |
|
| 40. |
A gas at 300 K has pressure `4 xx 10^(-10) N//m^(2)`. IF `k = 1.38 xx 10^(-23) J//K`, the number of `"molecule"// cm^(3)` is of the order ofA. `10^(3)`B. `10^(5)`C. `10^(6)`D. `10^(9)` |
|
Answer» Correct Answer - B using, `PV=nRT =nN((R)/(N))T=nNt_(B)T` `(nN)/(V)=(P)/(k_(B)T)=` number of molecules per `m^(3)` `therefore` Number of mulecules `//cm^(3)=(P)/(k_(B)T)xx10^(-6)` `=(4xx10^(-10)xx10^(-6))/(1.38xx10^(-23)xx300)=10^(5)` |
|
| 41. |
Fig shows of `PV//T` versus P for `1.00 xx 10^(-3) kg` of oxygen gas at two different temperatures. (a) What does the dotted plot signify ? (b) Which is true : `T_(1) lt T_(2) or T_(2) lt T_(1) ?` ( c) What is the value of `PV//T` where the curves meet on the Y-axis ? (d) If we obtained similar plot for `1.00 xx 10^(-3) kg` of hydrogen, would we get the same value of `PV//T` at the point where the curves meet on the y-axis ? If not, what mass of hydrogen yield the same value of `PV//T` (for low pressure high temperature region of the plot) ? (Molecular mass of `H = 2.02 u`, of `O = 32.0 u, R = 8.31 J "mol"^(-1) K^(-1)` |
|
Answer» (a) The dotted plot in the graph signifies the ideal behaviour of the gas, `PV/T` i.e the ratio is equal .`muR(mu` is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas. (b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature `T_1 `is closer to the dotted plot than the curve of the gas at temperature `T_2`. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, `T_1 gt T_2` is true for the given plot. (c) The value of the ratio PV/T, where the two curves meet, is `muR`. This is because the ideal gas equation is given as: `PV =mu RT` `(PV)/(T)=mu R` Where P is the pressure T is the temprature V is the volume `mu `is the number of moles R is the universal consatn Molecular mass of oxgen =32.0 g Mass of oxygen `=1xx10^(-3) kg =1g` ` R = 8.314 J "mole" ^-1 K^-1` ` therefor (PV)/(T)=(1)/(32)xx8.314` ` =0.26 J K^-1` Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is (d) If we obtain similar plots for `1.00 × 10^(–3)` kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u). We have: `(PV)/(T)=0.26 J K^-1` ` R=8.314 J "mole" ^-1K^-1` Molecular mass (M) of `H_2` =2.02 u `(PV)/(T)=mu R `at constant temperature Where `mu =(m)/(M)` m=Mass of` H_2` `therefore m=(PV)/(T)xx(M)/R` `=(0.26xx2.02)/(8.31)` `= 6.3 × 10^–2 g = 6.3 × 10^–5 ` kg of `H_2 `will yield the same value of PV/T. |
|
| 42. |
A cylinder of fixed capacity `44.8` litres constains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by `15^(@)C` ? Given `R=8.31 j mol e^(-1) K^(-1)` . (For monoatomic gas, `C_(v)=3 R//2`)A. 265 JB. `310.10 J`C. `373.95 J`D. `387.97 J` |
|
Answer» Correct Answer - C Since one mole of any ideal gas at STP occupies a volume of 22.4 litre. Therefore, cylinder of fixed capacity 44.8 litre must contain 2 moles of helium at STP. For helium, `C_(V)=3/2R` (monatomic) `therefore` Heat needed to raise the temperature, Q= number of moles `xx` molar specific heat `xx` raise in temperature `=2xx3/2Rxx15=45R=45xx8.31J=373.95J` |
|
| 43. |
A flask contains argon and chlorine in the ratio 2:1 by mass. The temperature of the mixture is `27^(@)C`. Obtain the ratio of `(i)` average kinetic energy per molecule, and `(ii)` root mean square speed of the molecules of two gases. Atomic mass of argon = 39.9 u, Molecular mass of chlorine = 70.9 u. |
|
Answer» The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to `(3//2) k_(B)T`. It depends only on temperature, and is independent of the nature of the gas. (i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1:1. (ii) Now lt `1//2 mv_(rms)2 `= average kinetic energy per molecule = (3/2) ) kBT where m is the mass of a molecule of the gas. Therefore, `=(V_(rms)^(2))_(Ar)/(V_(rms)^(2))_(Cl)=(m)_(cl)/(m)_(Ar)=70.9/39.9=1.77` where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides, `(V_(rms))_(Ar)/(V_(rms))_(Cl)=1.33` You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains unaltered. |
|
| 44. |
Explain why (a) there is no atmosphere on moon (b) there is fall in temperature with altitude |
|
Answer» The moon has small gravitiational force (pull) and hence , the escape velocity is small . The value of escape velocity for the moon is 4.6 `km//s` The air molecules have large range of speeds . Even though the rms speed of the air molecules is smaller than the escape velocity on the moon a signrificant number of molecules have speed greater than escape veloity and they escape. Now rest of the molecules arrange the speed distribution for the equilibrium temperature . Again a significant number of molecules escape as their speed exceed escape speed. Hence, over a long time the moon has lost most of its atmosphere. `At 300 K, V_("rms")= sqrt((3KT)/(m)) = sqrt((3xx1.38 xx10^(-23) xx 300)/(7.3 xx 10^(-26)) )=1.7 km//s` `V_(es) " for moon " =4.6 km//s " "[V_(es) ="escape velocity"]` (b). As the molecules move higher , their potential energy increase and hence kinetic energy decreases and hence, temperature reduces. At greater height more volume is availble and gas expands and hence, some cooling takes place. Note. We should not relate temperature directly with potential energy .it is directly related with kinetic energy of hte molecules. |
|
| 45. |
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the r.m.s. speed of a helium gas atom at `-20^(@) C` ? (Atomic mass of Ar = 39.9 u, of He = 4.0 u).A. `2.52xx10^(3)K`B. `2.52xx10^(2)K`C. `4.03xx10^(3)K`D. `4.03xx10^(2)K` |
|
Answer» Correct Answer - A Let 1 and 2 represent for Argon atom and Helium atom. rms speed of Argon, `v_(rms_(1))=sqrt((3RT_(1))/(M_(1)))` rms speed of Helium `v_(rms_(2))=sqrt((3RT_(2))/(M_(2)))` According to question, `v_(rms_(1))=v_(rms_(2))` `thereforesqrt((3RT_(1))/(M_(1)))=sqrt((3RT_(2))/(M_(2))),(T_(1))/(M_(1))=(T_(2))/(M_(2))` `or T_(1)(T_(2))/(M_(2))xxM_(1)=(253)/(4)xx39.9=2.52xx10^(3)K` |
|
| 46. |
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the r.m.s. speed of a helium gas atom at `-20^(@) C` ? (Atomic mass of Ar = 39.9 u, of He = 4.0 u). |
|
Answer» Temperature of the helium atom, `T_He=-20^(@) C=253 K` Atomic mass of argon,` M_(Ar) = 39.9 u` Atomic mass of helium, `M_(He) = 4.0 u` Let, `(v_(rms))`.Ar be the rms speed of argon. Let `(v_(rms))`He be the rms speed of helium. The rms speed of argon is given by: `(v_(rms))_(Ar)=sqrt((3RT_Ar)/(M_(Ar))` Where, R is the universal gas constant `T_(Ar)` is temperature of argon gas The rms speed of helium is given by: `v_(rms)_He=sqrt((3RT_(He))/(M_(He))` It is given that: `=sqrt((3RT_(Ar))/(M_(Ar)))=sqrt((3RT_(He))/(M_(He)))` ` (T_(Ar))/(M_Ar)=(T_(He))/(M_He)` `T_(Ar)=(T_(He))/(M_(He))xxM_(Ar)` `=253/4xx39.9` ` = 2523.675 = 2.52 xx 10^3 K ` Therefore, the temperature of the argon atom is `2.52 xx 10^3 K`. |
|
| 47. |
A gas is filled in a container at pressure `P_(0)`. If the mass of molecules is halved and their rms speed is doubled, then the resultant pressure would beA. `2P_(0)`B. `4P_(0)`C. `(P_(0))/(4)`D. `(P_(0))/(2)` |
|
Answer» Correct Answer - A Using, `P=1/3rhov_(rms)^(2)=1/3(mN)/(V)v_(rms)^(2)` `thereforePpropmv_(rms)^(2)` As m is halved and `v_(rms)` is doubled then P becomes twice or `2P_(0).` |
|
| 48. |
ABCDEFGH is a hollow cube made of an insulator (figure) face ABCD has positive charge on it. Inside the cube, we have ionised hydrogen. A. will be validB. will not be valid, since the ions would experience forece other than due to collisions with the walls.C. will not be valid since collisions with would not be elasticD. will not be valid because isotropy is lost |
| Answer» Due to presence of external positive charge on the face ABCD. The usual expression for pressure on the basis of kinetic theory will not be valid as ions would also experience electrostatic forces other than the forces due to collisions with the walls of the container. Due to pressence of positive charge the isotropy is also lost. | |
| 49. |
Diatomic molecules like hydrogen have energies due to both translations as well as rotational motion. From the equation in kinetic theory `pV =2/3 E,E " is "`A. The total energy per unit volumeB. only the translational part of energy because rotational energy is very small compared to the translational energy.C. only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentumD. the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero |
|
Answer» According to kinetic theory we assume the walls only exert perpendicular forces on molecules. They do not exert any parallel force, hence there will not be any type of rotation present. The wall producess only change in translational motion. Hence , in the equation `pV =2/3 E" "["where" underset(V= "volume")("p= pressure")]` E is representing only translational part of energy. |
|
| 50. |
We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law in the final state ? (Hydrogen molecules can be consider as spheres of radius `1Å`). |
|
Answer» Assuming hydrogen molecules as spheres of radius `1 Å` So `r= 1 Å=` radius The volume of hydrogen molecules `=4/3 pi r^(3)` `= 4/3 (3.14)(10^(-10))^(3)` `~~ 4 xx 10^(-30) m^(3)` `"Number of moles of " H_(2) =("Mass")/("Molecular mass")` `=(0.5)/(2)= 0.25` Molecules of `H_(2)` present =Number of moles of `H_(2)` present `xx 6.023 xx 10^(23)` `=0.25 xx 6. 023 xx 10^(23)` `:.` Volume of molecules present = Molecules number `xx` volume of each molecule `=0.25 xx 6.023 xx 10^(23) xx 4 xx 10^(-30)` `=6.023 xx 10^(23) xx 10^(-30)` `~~ 6xx 10^(-7) m^(3)` Now. it ideal gas Law is considered to be constant. `p_(i)V_(i) =p_(f)V_(f)` `V_(f) = (p_(i))/(p_(f)) V_(i) = ((1)/(100)) (3 xx 10^(-2))^(3)` `=(27 xx 10^(-6))/(10^(2))` `=2.7 xx 10^(-7) m^(3) " ".........(ii)` Hence on compression the volume of the gas is of the order of the molecular volume [form Eq.(i) and Eq.(ii).]. The intermolecular forces will play role and the gas will deviate from ideal gas behaviour. |
|