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.
| 51. |
For an ideal gas, interval energy can only be translational K.E. Why ? |
| Answer» In an ideal gas, there is no force of attraction//repulsion amongst the molecules. Therefore, there cannot be internal potential energy. Hence the internal energy can only be translational K.E. | |
| 52. |
Calculate the mean free path of nitogen at `27^(@)C` when pressure is 1.0 atm. Given, diameter of nitogen molecule = `1.5 Å, k = 1.38 xx 10^(-23) JK^(-1)`. If the average speed of nitrogen molecules is `675 ms^(-1)`, find the time taken by the molecule between two successive collsions and the frequency of collisions. |
|
Answer» Here, `T = 27^(@) C = 27 + 273 = 300 K, P = 1 atm = 1.01 xx 10^(5) N//m^(2)` `d = 1.5 Å = 1.5 xx 10^(-10) m, k_(B) = 1.38 xx 10^(-23) JK^(-1) , lambda = ?` from `lambda = (k_(B)t)/(sqrt(2) pi d^(2)P) = (1.38 xx 10^(-23) xx 300)/(1.414 xx 3.14(1.5 xx 10^(-10))^(2) xx 1.01 xx 10^(5)) = 4.1 xx 10^(-7)m` Time interval between two successive collisions `t = (dis t a n ce)/(speed) = (lambda)/(upsilon) = (4.1 xx 10^(-7))/(675) = 0.006xx10^(-7)s` Collision frequency `= 1/t = 1/(0.006 xx 10^(-7)) = 1.67 xx 10^(9) s^(-1)`. |
|
| 53. |
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^(5)`B. 10C. `10^(7)`D. `10^(11)` |
|
Answer» Correct Answer - A From `PV = n RT = nN(R/N)T = n N kT` `(nN)/(V) = P/(kT) ` = number of `"molecules" //m^(3)` `:.` Number of `"molecules"//m^(3) = (P)/(kT) xx 10^(-6)` `=( 4 xx 10^(-10) xx 10^(-6))/(1.38 xx 10^(-23) xx 300) ~= 10^(5)`. |
|
| 54. |
The diameter of a gas molecule is `2.4 xx 10^(-10) m`. Calculate the mean free path at NTP. Given Boltzmann constant `k = 1.38 xx 10^(-23) J molecule^(-1) K^(-1)`. |
|
Answer» Correct Answer - `1.46 xx 10^(-7)m`. Here, `d=2.4 xx 10^(-10)m, lambda = ?` `k =1.38 xx 10^(-23)J "molecule"^(-1) K^(-1)` At, NTP. `T = 273 K, P = 1.01 xx 10^(5) Nm^(-2)` `lambda = (kT)/(sqrt(2) pi d^(2)P)` `=(1.38 xx 10^(-23) xx 273)/(1.414 xx 3.14 (2.4 xx 10^(-10))xx 1.01 xx 10^(5))` `lambda = (1.38 xx 273 xx 10^(-1)xx10^(-7))/(1.414 xx 3.14 xx 2.4 xx 1.01)` `= 1.46 xx 10^(-7)m`. |
|
| 55. |
At what temperature, pressure remaining unchanged, will the rms velocity of hydrogen be doubled its value at NTP ? |
|
Answer» Correct Answer - `819^(@)C` `(C_t)/(C_0) = sqrt((t+273)/(273)) or 2 = sqrt((t+273)/(273))` On solving, we get `t = 819^(@)C` |
|
| 56. |
Calculate the rms speed of smoke particles of mass `5 xx 10^(-17) kg` in their Brownian motion in air at NTP. Given `k_(B) = 1.38 xx 10^(-23) J//K` |
|
Answer» Correct Answer - `1.5 cm//s` `C_(rms) = ? M = 5 xx 10^(-17) kg` `k_(B) = 1.38 xx 10^(-23) J//K T = 273 K` `C_(rms)sqrt((3kT)/(m)) = sqrt((3 xx 1.38 xx 10^(-23) xx 273)/(5 xx 10^(-17)))` `C_(rms) = sqrt((4.14 xx 273)/(5)) xx 10^(-2)m//s = 1.5 cm//s` |
|
| 57. |
An assembly of smoke particle in air at `NTP` is under consideration. If the mass of each particles is `5 xx 10^(-17) kg`. Then the rms speed is (Given: `k = 1.38 xx 19^(-23) J K^(-1)`) |
|
Answer» Here, `m= 5 xx 10^(-17) kg , k=1.38 xx 10^(-23) J//K`. `T=0^(@)C = (0+273)K 273 K, upsilon_(rms) = ?` `upsilon_(rms) = sqrt((3RT)/(M)) = sqrt((3kNT)/(mN)) =sqrt((3kT)/(m))` `= sqrt((3xx 1.38 xx 10^(-23) xx 273)/(5 xx 10^(-17)))` `upsilon_(rms) = 1.5 xx 10^(-2) m//s = 1.5 cm//s`. |
|
| 58. |
Show that the rms velocity of `O_(2)` molecule is `sqrt(2)` times that of `SO_(2)`. Atomic weight of sulphur is 32 and atomic weight of oxygen is 16. |
|
Answer» Correct Answer - `sqrt(2)C_(SO_2)` As rms velocity at a given temperature `prop 1/(sqrt(M))` `:. (C_(OXY))/(C_(SO_2)) = sqrt((M_(SO_2))/(M_(OXY))) = sqrt((32 + 2 xx 16)/(32)) = sqrt(2)` `C_(OXY)= sqrt(2)C_(SO_2)`. |
|
| 59. |
If root mean sqauare velocity of the molecules of hydrogen at NTP is `1.84 kms^(-1)`, calculate the rms velocity of oxygen molecules at NTP. Molecular weights of hydrogen and oxygen are 2 and 32 respectively. |
|
Answer» Correct Answer - `0.46 km s^(-1)` `P=1/3 M/V C^(2)` or `MC^(2) = 3 PV = 3 RT = a "constant" at STP` Hence , `C prop 1//sqrt(M)` Thus `(C_0)/(C_H) = sqrt((M_H)/(M_0)) = sqrt((2)/(32)) = 1/4` or `C_(0)= (C_H)/(4) = 1.84/4 = 0.46 km s^(-1)` |
|
| 60. |
Find the temperature at which oxygen molecules would have the same rms speed as of hydrogen molecules at `300 K`.A. 600 KB. 2400 KC. 1200 KD. 4800 K |
|
Answer» Correct Answer - D `C_("oxy") = sqrt((3RT_(OXY))/(M_(OXY))) = C_(H) = sqrt((3RT_(H))/(M_(H)))` `:. (T_(OXY))/(M_(OXY)) = (T_H)/(M_H)` `:. T_(OXY) = (T_H)/(M_H) xx M_(OXY) = (300 xx 32)/(2) = 4800K`. |
|
| 61. |
Calculate the rms velocity of oxygen molecules at S.T.P. The molecular weight of oxygen is 32. |
|
Answer» Here, `P=1 atm = 1.013 xx 10^(5) N//m^(2)` `M= 32 g = 32 xx 10^(-3) kg` V = molar volume at STP `= 22.4 litre = 22.4 xx 10^(-3) m^(3)` `C_(rms) =?` `C_(rms) = sqrt((3RT)/(M))= sqrt((3PV)/(M))` `= sqrt((3 xx 1.013 xx 10^(5) xx 22.4 xx 10^(-3))/(32 xx 10^(-3)))` `= 461.23 m//s`. |
|
| 62. |
The mean kinetic energy per unit volume of gas (E) is related to average pressure P, exerted by the gas isA. `E = (2)/(3) P`B. `E = (3)/(2) P`C. E = PD. `E = (5)/(4) P` |
|
Answer» Correct Answer - B `E = (3)/(2) P` |
|
| 63. |
What is the princiopal source of heating in many office bulidings in moderate climates? |
|
Answer» we know that organized energy (usable) degenerates into disorganized energy (non-usable). For example , when electricity is used for lighting bulbs in homes and offices, majorportion of electrical energy is converted into heat energy. Even the small part of energy that exists in the form of light, turns into heat energy. it is this heat energy, which is used to warm many office buildings in moderates climates. This explains why the lights in these buildings are kept on most of the time. It si intersting to note that energy degenerated in the form of heat has no further use. The entire heat energy in a buliding cannot be refused to light even a single bulb and that too without some outside efforts. |
|
| 64. |
What is the average velocity of the molecules of an ideal gas ? |
| Answer» Average velocity of the molecules of an ideal gas is zero, because molecules posses all sorts of velocities in all possible direction, whose vector sum would be zero. | |
| 65. |
If gas molecules undergo inelastic collision with the walls of container when temperature of gas is same as that of walls of container, then (a) the temperature of the gas will decrease the pressure of the gas will increase ( c) neither the temperature nor the pressure will change. (d) the temperature of the gas will increaseA. the temperature of the gas will decreaseB. the pressure of the gas will increaseC. neither the temperature nor the pressure will changeD. the temperature of the gas will increase |
|
Answer» Correct Answer - C When the temperature of the gas is the same as that of the walls of the container, there is no exchange of energy. Hence, the molecules will return with the same average speed, whether the collisions are elastic or inelastic or inelastic and consequently neither the temperature nor the pressure will change. |
|
| 66. |
Two gases A and B each at temperature T. Pressure P and volume V, are mixed. If the mixture be at the same temperature T and its volume also be V, then what should be its pressure ? Explain. |
| Answer» As, `P= 1/3 M/V C^(2) prop T`, therefore if V and T are constants, then `P prop M`. As M becomes 2M, therefore, P becomes 2P. | |
| 67. |
Two gases each at temperature T, volume V and pressure P are mixed such that temperature of mixture is T and volume is V. What will be the pressure of the mixture ? |
|
Answer» From kinetic theory of gases, `P= 1/3 M/V C^(2)`, but `C^(2) prop T` `:. P prop (MT)/(V)`, Both T and V remain unchanged but M is double , therefore pressure P becomes twice. |
|
| 68. |
A vessel is filled with a mixture of two different gases. State with reason (i) will the mean K.E. Per molecule of both the gases be equal ?(ii) Will the root mean square velocities of the molecules be equal (iii) will the pressure be equal ? |
|
Answer» (i) Yes, because the mean K.E. Per molicule ` (=3/2 k_(B)T)` depends only upon the temperature. (ii) No, Because for different gases, rms velocity depends upon the mass of the gas molecules. (iii) No definite idea about pressure because masses of the gases are not given. |
|
| 69. |
Two different gases at the same temperature have equal root mean square velocities. |
| Answer» No, when the two gases have exaclty the same temperature, the average kinetic energy per molecule `(=1/2 mC^(2)=3/2 kT)` for each gas is the same. But as the different gases may have molecules of different masses, the rms speed (C) of molecules of different gases shall be different. | |
| 70. |
Assertion : The root mean square velocity of molecules of a gas having Maxwellian distribution of velocities is higher than their most probable velocity, at any temperature. Reason : A very small no. of molecules of a gas which possess very large velocities increase root mean square velocity, without affecting most probable velocity.A. If, both Assertion and reason are ture and the Reason is the correct explanation of the Assertion.B. If both, Assertion and reason are true but Reason is not a correct expationation of the Assertion.C. If Assertion is true but the Reason is false.D. If Both, Assertion and reason are false. |
|
Answer» Correct Answer - A `C_(rms) = sqrt((3kT)/(m)) C_(mp) = sqrt(2kT)/(m) :. C_(rms) gt C_(mp)` Most proable speed is that which is possessed by large number of molecules. There are molecules whose speed is greater than this speed adn others. Whose speed is less than this value. that is why rms speed of all the molecules becomes greater than the most probabel speed. |
|
| 71. |
The velocities of three molecules are 3v, 4v and 5v. Calculate their root mean square velocity. |
|
Answer» `C = sqrt(((3upsilon)^(2)+(4 upsilon)^(2)+(5 upsilon)^(2))/(3)) = sqrt((50)/(3)) upsilon` `=4.08 upsilon`. |
|
| 72. |
The temperature of an open room of volume `30 m^(3)` increases from `17^(@)C to 27^(@)C` due to sunshine. The atmospheric pressure in the room remains `1 xx 10^(5) Pa`. If `n_(i) and n_(f)` are the number of molecules in the room before and after heating then `n_(f)` and `n_(i)` will beA. `2.5 xx 10^(25)`B. `-2.5 xx 10^(25)`C. `-1.61 xx 10^(23)`D. `1.38 xx 10^(23)` |
|
Answer» Correct Answer - B Here, `V_(1) = 30m^(3), T_(1) = 17+273 = 290 K` `P_(1) = 1 xx 10^(5)Pa` `V_(2) = 30 m^(3) , T_(2) = 27+273 = 300K`, `P_(2) = 1 xx 10^(5) Pa`. Let `N_(1),N_(2)` be the no, of moles of a gas at temperature `T_(1) and T_(2)` respectively. then `N_(1) =(P_(1)V_(1))/(RT_(1)) = ((1 xx 10^(5)) xx 30)/(83 xx 290) = 1.24 xx 10^(3)` or `N_(2) = (P_(2)V_(2))/(RT_(2)) = ((1 xx 10^(5)) xx 30)/(83 xx 300) = 1.20 xx 10^(3)` ltbr. Change in the number of moles `N_(2)-N_(1) =(1.20-1.24) xx 10^(3) = -0.04 xx 10^(3)` Change in the number of molecules `n_(f)-n_(i) = (N_(2)-N_(1)) xx (6.023 xx 10^(23))` `= - (0.04 xx 10^(3)) xx (6.023 xx 10^(23))` =-2.5 xx 10^(25)`. |
|
| 73. |
A gas in a vessel is at the pressure `P_(0)`. If the masses of all the molecules be made half and their speeds be made double, then find the resultant pressure. |
|
Answer» `P_(0) = 1/3 (mn C^(2))/(V)` and `P_(1) = 1/3 ((m//2)n)/(V) (2C)^(2) = 2P_(0)`. |
|
| 74. |
A mixture of helium and hydrogen gases is filled in a vessel at `30^(@) C`. Compare the rms velocities of molecules of the two gases. Atomic weights of hydrogen and helium are 1 and 4 respectively. |
| Answer» `(C_1)/(C_2) =1` as rms velocity depends only on temperature of the gas. | |
| 75. |
The speeds of 4 molecules of a gas in arbitary units are 2, 3, 4 and 5. what is the ratio of rms speed and mean speed of the gas molecules. |
|
Answer» Correct Answer - 1 `upsilon_(rms) = sqrt((2^(2)+3^(2)+4^(2)+5^(2))/(4))` `= sqrt((54)/(4)) = sqrt(13.5) = 3.67` `upsilon_(mean) = (2+3+4+5)/(4) = 3.5` `(upsilon_(rms))/(upsilon_(mean)) = (3.67)/(3.5) = 1.05 ~~1`. |
|
| 76. |
The ratio of rms velocities of two gases at the same temperature is `1: sqrt(2)`. Compare the vapour densities of the gases. |
|
Answer» `(C_1)/(C_2) = sqrt((M_2)/(M_1)) = sqrt((rho_2)/(rho_1)) = 1/(sqrt(2))` `:. (rho_2)/(rho_1) = 1/2 or rho_(1) : rho_(2) = 2:1`. |
|
| 77. |
The molecules of a given mass of 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? |
|
Answer» Here, `C_(1) = 100 m//s, T_(1) = 27^(@)C = (27+273)K = 300K` `P_(1) = 1.00 atm. C_(2) = ?, T_(2) = 127^(@)C = (127+273)K = 400K , P_(2)= 2.0 atm`. from `(P_(1)V_(1))/(T_1) = (P_(2)V_(2))/(T_2) , (V_1)/(V_2) = (P_2)/(P_1)* (T_1)/(T_2) = 2 xx 300/400 = 3/2` Again `P_(1)=1/3 M/(V_1) C_(1)^(2)` and `P_(2) = 1/3 M/(V_2) C_(2)^(2) :. (C_(2)^(2))/(C_(1)^(2))*(V_1)/(V_2) = (P_2)/(P_1)` `C_(2)^(2) = C_(1)^(2) xx (P_2)/(P_1) xx(V_2)/(V_1) = (100)^(2) xx 2 xx 2/3` `C_(2) = (100 xx 2)/(sqrt(3)) m//s`. |
|
| 78. |
If a molecule of krypton is 2.25 times heavier than a molecule of hydrogen, what would be the ratio of their root mean square velocities in a mixture of equal masses of the two gases ? |
| Answer» `(C_(Kr))/(C_H) = sqrt((M_H)/(M_(Kr))) = sqrt((1)/(2.25)) 10/15 = 2:3`. | |
| 79. |
Two molecules of gas have speeds of `9 xx 10^(6) ms^(-1)` and `1 xx 10^(6) ms^(-1)` respectively. What is the root mean square speed of these molecules? |
|
Answer» Here, `upsilon_(1) = 9 xx 10^(6)m//s^(-1) , upsilon_(2) = 1 xx 10^(6) ms^(-1)` `upsilon_(rms) = sqrt((upsilon_(1)^(2)+upsilon_(2)^(2))/(2)) = sqrt(((9xx10^(6))^(2)+(1xx10^(6))^(2))/(2)) = sqrt(41) xx 10^(6)m//s`. |
|
| 80. |
The molecules of a given mass of a gas have rms velocity of `200 m//s at 27^(@)C and 1.0 xx 10^(5) N//m_(2)` pressure. When the temperature and pressure of the gas are respectively `127^(@)C and 0.05 xx 10^(5) Nm^(-2)`, the rms velocity of its molecules in `ms^(-1)` isA. `(400)/(sqrt(3))`B. `(100sqrt(2))/3`C. `100/3`D. `100 sqrt(2)` |
|
Answer» Correct Answer - A Here, `upsilon_(1) = 200 m//s, T_(1) = 27^(@)C = 27+273` `300 K` `P_(1) = 1.0 xx 10^(5) N//m^(2), T_(2) = 127^(@)C = 127+273` `=400K` `P_(2) = 0.05 xx 10^(5) N//m^(2), upsilon_(2)=?` As `upsilon_(rms) = sqrt((3RT)/(m))` , therefore, `upsilon_(rms) prop sqrt(T)` There is no effect of change in pressure `:. (upsilon_2)/(upsilon_1) = sqrt((T_2)/(T_1)) = sqrt(400/300) = 2/(sqrt(3))` `upsilon_(2) = (2)/(sqrt(3)) upsilon_(1) = (2 xx 200)/(sqrt(3)) = (400)/(sqrt(3))m//s`. |
|
| 81. |
The root mean square velocity of hydrogen molecule at `27^(@)C is (upsilon_(H)` and that of oxygen at `402^(@)C is (upsilon_(0)`, then |
|
Answer» Correct Answer - C `(upsilon_H)/(upsilon_O) = sqrt((T_H)/(T_O)) = sqrt((27+273)/(402+273)) = sqrt(300)/(675)) = sqrt(4/9) = 2/3` or `2upsilon_(O) = 3upsilon_(H)`. |
|
| 82. |
The molar specific heat of a gas as given from the kinetic theory is `(5)/(2) R`. If it is not specified whether it is `C_(p) or C_(upsilon)` one could conclude that the molecules of the gasA. are definitely monoatomicB. are definitely rigid diatomicC. are definitely non-rigid diatomicD. can be monoatomic or rigid diatomic |
|
Answer» Correct Answer - D A monoatomic gas has 3 degree of freedom and diatomic gas has 5 degrees of freedom. For monotomic gas, `C_(upsilon) = 3/2 R and C_(p)=C_(upsilon) + R` `or C_(P) = 3/2 R + R = 5/2R` For a ditomaic gas, `C-(upsilon) =5/2 R, C_(upsilon)+R` =5/2R+ R=7/2 R` Hence we can conclude that the molecules of the gas can b monoatomc and rigid diatomic. |
|
| 83. |
The molar specific heats of an ideal gas at constant pressure and volume are denotes by `C_(P)` and `C_(upsilon)` respectively. If `gamma = (C_(P))/(C_(upsilon))` and `R` is the universal gas constant, then `C_(upsilon)` is equal toA. `gamma R`B. `(1 + gamma)/(1 - gamma)`C. `(R)/(gamma - 1)`D. `(gamma - 1)/R` |
|
Answer» Correct Answer - C `gamma = (C_p)/(C_upsilon)` or `C_(p) = gamma C_(upsilon)` As, `C_(P) = C_(upsilon) = R` , so `gamma C_(upsilon) - C_(upsilon) = R` or `C_(upsilon) = (R)/(gamma - 1)`. |
|
| 84. |
According to kinetic theory of gases, absolute temperature of a gas is directly proportional to the average kinetic energy of translation per molecules to the gas. Zeroth law of thermodynamics established that temperature is the obly physical quantity that determines whether a given system is in thermal equilibrium with another system. heat flows from a system at higher temperatures to another system at lower temperature till their temperatures becomes equal. temperature corresponds to level in case of liquids, pressure in case of gases potential in case of electricity. Read the above passage and answer the following questions : (i) The absolute temperature of a gas is increased three times. what will be the increase in rms velocity of gas molecules ? What values of life do you learn this concept of temperature ? |
|
Answer» (i) As `T prop 1/2 mC^(2) :. T prop C^(2) or C prop sqrt(T)` When T becomes three times , C becomes `sqrt(3)` times. `:. Increases in rms velocity = `sqrt(3) C - C = (1.732-1) C = 0.732 C = 73.2 %` (ii) Temperature is the degree of hotness of the system. just as a liquid flows higher to lower level, gas flows higher pressure to lower pressure, charge flows from higher to lower potential , in the same way, heat energy flows from higher temperature to lower temperature. thus directions of flow of heat energy is determined by higher level of heating of the system and not by higher heat content of the system. This is so true in every sphere of day life. for example, an educational institution is known not by the number of teaching faculty, but by the level of teaching faculty and the level of infrastructure provided. |
|
| 85. |
The ratio of the specific heats `(C_(P))/(C_(upsilon)) = gamma` in terms of degrees of freedom is given by |
| Answer» Correct Answer - `gamma = (1+2/n)`. | |
| 86. |
Two perfect gases at absolute temperature `T_(1) and T_(2)` are mixed. There is no loss of energy. The masses of the molecules are `m_(1) and m_(2)`. The number of molecules in the gases are `n_(1) and n_(2)`. The temperature of the mixture is |
|
Answer» Before mixing, kinetic energy of all the molecules of two gases `= 3/2 kn_(1)T_(1)+3/2 kn_(2) T_(2)` If T is temperature of the mixuture , the average Ke of all molecules of two gases `= 3/2 k(n_(1)+n_(2))T = 32 k(n_(1)T_(1)+n_(2)T_(2))` `T=(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))`. |
|
| 87. |
Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP. |
|
Answer» As is known , volume occupied by 1 gram mole of gas at NTP = 22400 c c. `:.` Number of molecules in `1 c c` of hydrogen = `(6.023 xx 10^(23))/(22400) = 2.688 xx 10^(19)` As hydrogen is a diatomic gas, each molecule has 5 degree of freedom, therefore, Total number of degree of freedom = `5xx2.688 xx 10^(19) = 1.344 xx10^(20)`. |
|
| 88. |
Calculate the number of molecules is 2 c.c of the perfect gas at `27^(@)C` at a pressure of 10 cm. of mercury. Mean kinetic energy of each molecule at `27^(@)C = 7 xx 10^(-4) erg` and acceleration due to gravity `= 980 cm s^(-2)`. |
|
Answer» Correct Answer - `5.71 xx 10^(8)` K.E of the given gas at `27^(@)C` `=1/2 MC^(2) = 3/2 PV` `=3/2 xx (10 xx 13.6 xx 980) xx 2 ergs` `:.` No. Of molecules in the given volume `=(K.E. "of given gas")/(K.E."per molecule of a gas")` `(3/2 xx 10 xx 13.6 xx 980 xx2)/(7 xx 10^(-4)` `=5.71 xx 10^(8)` |
|
| 89. |
In a mixture of gases, the average number of degrees of freedom per molecule is 6. the rms speed of the molecules of the gas is C. the velocity of sound in the gas isA. CB. `(2C)/(3)`C. `(3C)/(4)`D. `(2C)/(sqrt(3))` |
|
Answer» Correct Answer - B `upsilon_(rms) = sqrt((3RT)/(M)) and upsilon_(sound) = sqrt((gamma RT)/(M))` `:. (upsilon_(sound))/(upsilon_(rms)) = sqrt((gamma)/(3))` As, `gamma = 1+2/n= 1+ 2/6 = 4/3` `:. upsilon_(sound) = upsilon_(rms)sqrt((gamma)/(3)) = Csqrt((4//3)/(3)) = (2C)/(3)`. |
|
| 90. |
An open glass tube is immersed in mercury in such a way that a lenth of 8cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46cm. What will be length of the air column above mercury in the tube now? (Atmosphere pressure =76cm of Hg) |
|
Answer» Here, initially, `0.05 m = 5 cm ` of tube extends above the mercury level. `:. V_(1) = 5 cm , P_(1) =P =76 cm of Hg c o l`. = atmospheric pressure When tube is raised further by 0.43 m let h be the length of air column above the mercury level. ltbr. `:. P_(2) = 76 - (48-h) = (28 +h)` and `V_(2) = h` As `P_(2) = V_(2) = P_(1)V_(1)` `:. h^(2)+28 h - 380 =0` `h =(-28 +- sqrt(28^(2)+4 xx1 xx 380))/(2)` `(-28 +-sqrt(784+1520))/(2)` `(-28+- 48)/(2) = 10 cm` `h = 0.1 m`. |
|
| 91. |
On which factors does the average KE of gas molecules depend : nature of gas, its temperature , its volume? |
| Answer» Average KE of gas molecules is directly proportional only to absolute temperature of the gas. | |
| 92. |
How is the average KE of a gas molecule related to the temperature of the gas ? |
| Answer» Averaged KE of a gas molecule `1/2 mC^(2) = 3/2 kT`. | |
| 93. |
Calculate (i) rms velocity and (ii) mean kinetic energy of one gram molecule of hydrogen at STP. Given density of hydrogen at STP is `0.09 kg m^(-3)`. |
|
Answer» Here `rho = 0.09 kg m^(-3)` At STP, pressure `P= 0.01 xx 10^(5)Pa`. According to kinetic theory of gases. `P = 1/3 rho C^(2)` or `C= sqrt((3P)/(rho)) = sqrt((3xx 1.01 xx 10^(5))/(0.09)) = 1837.5 ms^(-1)` Volume occupied by one mole of hydrogen at `STP = 22.4 "litres" = 22.4 xx 10^(-3)m^(3)` `:.` Mass of hydrogen `M =` volume `xx` density = ` 22.4 xx 10^(-3) xx 0.09 = 2.016 xx 10^(-3)kg` Average `K.E//mole = 1.2 MC^(2)` `=1/2 xx (2.016 xx 10^(-3)) xx (1837.5)^(2) = 3403.4 J`. |
|
| 94. |
An empty barometric tube 1 m long is lowered vertically (mouth downwards) into a tank of water. What will be the depth above the water level in the tube, when the water has risen 20 cm inside the tube ? Take atmospheric pressure as 10.4 m column of water. |
|
Answer» Correct Answer - `2.6 m of water column` Let 1 sq. Cm be teh area of cross section of the barometric tube. Here, `P= 10.4m` of water `V = 100 xx 1 = 100 c.c.,` `P_(1) = ? , V_(1) = (100-20) = 80 c.c.` As, `P_(1) = PV//V_(1) = 10.4 xx 100//80` `=13 cm` of water column. `:.` Depth of water level `=P_(1)-P` `= 13 = 10.4 = 2.6 m` of water column |
|
| 95. |
1 mole of `H_(2)` gas is contained in box of volume `V= 1.00 m^(3) at T = 300 K`. The gas is heated to a temperature of T = 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)A. same as the pressure initially.B. 2 times the pressure initially.C. 10 times the pressure initiallyD. 20 times the pressure initially |
|
Answer» Correct Answer - D Applying standard gas equation, `(P_(2)V_(2))/(T_2) = (P_(1)V_(1))/(T_1)` `(P_2)/(P_1) = (V_1)/(V_2)* (T_2)/(T_1)` As, `(T_2)/(T_1) = 3000/300 =10`, and every molecule of `H_(2)` splits into hydrogen atoms, dubling the number, therefore volume available to given number of entities becomes half i.e. `V_(2) = 1/2 V_(1)`, therefore, `(P_2)/(P_1) = 2 xx 10 = 20`. |
|
| 96. |
Calculate for hydrogen at `27^(@)` (i) KE of one gram mole of the gas (ii) KE of one gram of the gas (iii) root mean square velocity of the molecule. Given, molecule wt. Of hydrogen = 2. |
|
Answer» Here, `T=27^(@C = (27 + 273) K` `=300 K,M=2`gram, (i) KE of one gram mole `=3/2 RT = 3/2 xx 8.31 xx 300 = 3.74 xx 10^(3)J` (ii) KE of one gram `= (3.74 xx 10^(3))/(2)` `1.87 xx 10^(3)J` (ii) `upsilon-(rms) = sqrt((3RT)/(M)) = sqrt((3 xx 8.31 xx 300)/(2))` `= 61.15 m//s` |
|
| 97. |
A `3000 cm^(3)` tank contains oxygen at `20^(@)C` and the gauge pressure is `2.5 xx 10^(6) Pa`. Find the mass of the oxygen in the tank. Take 1 atm `= 10^(5) Pa`. |
|
Answer» Correct Answer - `103.8 gram` Pressure of oxygen gas, `P_(1) = 2.5xx10^(6) Pa + 1 atm` `=25 xx 10^(5) +10^(5) = 26 xx 10^(5) Pa` `V_(1) = 3000 c.c., T_(1) = 20 + 273 = 293 K`, `P=10^(5)Pa, V =? T=273 K`. `V = (P_(1)V_(1)T)/(T_(1)P) = ((26 xx 10^(5))xx(3000)xx 273)/(293 xx 10^(5)) c.c.` Mass of this volume of oxygen = `(32 V)/(22400)` `=(32 xx 26 xx 3000 xx 273)/(22400 xx 293) = 103.8 gram` |
|
| 98. |
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2 atm. And temperature `17^(@)`. Take the radius of a nitrogen molecule to be roughly `1.0 Å`. Compare the collision time with the time molecule moves freely between two successive collisions. (Molecular mass of nitrogen = 28.0 u). |
|
Answer» Here, `lambda = ? , f = ?, p=2atm = 2 xx 1.013 xx 10^(5) Nm^(-2), T = 17^(@)C = (17+273)K = 290 K` `sigma = 2 xx 1 = 2 Å = 2 xx 10^(-10)m, k=1.38 xx 10^(-23) J "molecule"^(-1) K^(-1), M=28 xx 10^(-3) kg` `lambda = (kT)/(sqrt(2)pi sigma^(2)p) = (1.38 xx 10^(-23)xx290)/(1.414 xx 3.14(2xx10^(-10))^(2)xx 2.026 xx 10^(5)) = (1.38xx29xx10^(-7))/(1.414 xx 3.14 xx 4 xx 2.026) = 1.11 xx 10^(-7)m` `upsilon_(rms) = sqrt((3RT)/(M)) = sqrt((3xx8.31 xx 290)/(28xx10^(-3))) = 508.24 m//s` collision frequency = no. of collisions per second =`(upsilon_(rms))/(lambda) = (508.24)/(1.11 xx 10^(-7)) = 4.58 xx 10^(9)`. |
|
| 99. |
Using the ideal gas equation, determine the value of gas constant R. Given that one gram mole of a gas at S.T.P occupies a volume of 22.4 litres |
|
Answer» Correct Answer - `8.31 J mole^(-1) K^(-1)` Here, `R = ? V = 22.4` litre `= 22.4 xx 10^(3) m^(3)` at S.T.P, `T =273 K`, and `P=1` atmosphere `= 1.013 xx 10^(5)N//m^(2)` For one mole of gas, `PV =RT` `R = (PV)/(T) = (1.013 xx 10^(5) xx 22.4 xx 10^(-3))/(273)` `= 8.31 J mole^(-1) K^(-1)`. |
|
| 100. |
What does universal gas constant R signify? |
| Answer» The universal gas constant R signifies work done on or by one mole of the gas per Kelvin. | |