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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.
| 401. |
If the absolute temperature of a sample of gas is increased by a factor of 1.5, by what ratio does the average molecular speed of the molecules increases?A. 1.2B. 1.5C. 2.2D. `3.0` |
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Answer» Correct Answer - a |
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| 402. |
Calculate relative rates of effusion of `SO_(2)` to `CH_(4)` in the mixture obtained by effusing out a mixture with initial molar ratio`(n_(SO_(2)))/(n_(CH_(4)))=(8)/(1)` for three effusing steps.A. `2:1`B. `1:4`C. `1:2`D. none of these |
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Answer» Correct Answer - c (c) `overset((n_(SO_(2))^(3))/(3))((n_(CH_(4)))/(8//1))=((16)/(64))^(3//2)xx=(1)/(1),` `(r_(SO_(2)))/(r_(CH_(4)))=(1)/(1).sqrt((16)/(64))=(1)/(2)` |
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| 403. |
One litre of a gaseous mixture of two gases mixture of two gases effuses in 311 seconds while 2 litre of oxygen takes 20 minutes. The vapour density of gaseous mixture containing `CH_(4)` and `H_(2)` is:A. 4B. 4.3C. 3.4D. 5 |
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Answer» Correct Answer - b |
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| 404. |
A gaseous mixture comprising of equal moles of `H_(2)//O_(2)//M` (M mass=128) was subjected to series of effusion steps. What will be the number effusion steps required so as to change the composition to: One in which `H_(2):O_(2)` is 4096:1. What will be the composition of this mixture (w.r.t. all the gases)?A. 4,4096:16:1B. `6,4096xx64:64:1`C. 4,2048:8:1D. 5,4096:16:1 |
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Answer» Correct Answer - b |
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| 405. |
A gaseous mixture comprising of equal moles of `H_(2)//O_(2)//M` (M mass=128) was subjected to series of effusion steps. What will be the number effusion steps required so as to change the composition to: One in which lightest: Heaviest gas is `4096:1` What will be the composition of this mixture (w.r.t all the gases)?A. 4,4096:16:1B. `6,4096xx64:64:1`C. 4,2048:8:1D. `5,4096:16:1` |
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Answer» Correct Answer - a |
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| 406. |
One litre gaseous mixture is effused in 4.5 minutes and 30 seconds while 1 litre of oxygen takes 10 minutes for effusion. The gaseous mixture contains in it ethane and hydrogen. Calculate vapour density of gaseous mixture. |
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Answer» Correct Answer - 4 |
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| 407. |
A vertical hollow cylinder of height `1.52m` is fitted with a movable piston of negligible mass and thickness. The lower half of the cylinder contains an ideal gas and the upper half of the mercury comes out of the cylinder Find the temperature assuming the thermal expansion of mercury to be negligible . |
| Answer» Correct Answer - `337.5K` | |
| 408. |
A motorist inflates the tyres of her car to a pressure of 180 kPa on a day when the temperature is `-8^(@)C`. When she arrives at her destination, the tyres pressure increased to 245 kPa. What is the temperature of the tyres if we assume that the tyres expand by 7%?A. 265KB. 360.1KC. 385.9KD. 383.55K |
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Answer» Correct Answer - c |
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| 409. |
At `400K` the root mean square (rms) speed of a gas x (mol. wt. =40) is equal to the most probable speed of gas y at `60K` Find the molecular weight of gas y . |
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Answer» Correct Answer - 4 `u_(rms) =u_(MP)` `sqrt((3RT)/(M_(x)))=sqrt((2RT)/(M_(y)))` `:.sqrt((3Rxx400)/(40)) =sqrt((2Rxx60)/(M_(y)))` `M_(y) =4` . |
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| 410. |
Statement-1: Gases like `N_(2),O_(2)` behave as ideal gases at high temperature and low pressure. Statement-2: Molecular probable velocity is the velocity possessed by maximum fraction of molecules at the same temperature. Statement-2:Molecular interactions diminish at high temperature and low pressure.A. Statement-I is True, Statement-II is True : Statement-II is a correct explanation for Statement-IB. Statement-I is True, Statement-II is True : Statement-II is `NOT` a correct explanation for Statement-IC. Statement-I is True, Statement-II is False.D. Statement-I is False, Statement-II is True. |
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Answer» Correct Answer - a |
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| 411. |
The rates of diffusion of `SO_(3),CO_(2),PCl_(3)` and `SO_(2)` are in the following order:A. `PCl_(3)gtSO_(3)gtCO_(2)`B. `CO_(2)gtSO_(2)gtPCl_(3)gtSO_(3)`C. `SO_(2)gtSO_(3)gtPCl_(3)gtCO_(2)`D. `CO_(2) gt SO_(2) gt SO_(3) gt PCl_(3)` |
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Answer» Correct Answer - d |
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| 412. |
The ratio of rates of diffusion of `SO_(2)`, `O_(2)` and `CH_(4)` isA. `1:sqrt(2):2`B. `1:2:4`C. `2:sqrt(2):1`D. `1:2:sqrt(2)` |
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Answer» Correct Answer - a |
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| 413. |
The rates of diffusion of two gases `A` and `B` are in the the ratio `1:4` A mixture contains these gases `A` and `B` in the ration `2:3` The ration of mole fraction of the gases `A` and `B` in the mixture is (assume that `P_(A) =P_(B))` .A. `1:24`B. `1:18`C. `1:12`D. `1:6` |
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Answer» `(r_(A))/(r_(B)) =sqrt((M_(B))/M_(A)) =(1)/(4)` `(M_(B))/(M_(A)) = (1)/(16)` Now `(n_A)/(n_(B)) = (n_(A))/((n_(A)+n_(B))) xx (n_(A+n_(B)))/(n_(B))` ` =(w_(A))/(m_(A)) xx (M_(B))/(w_(B)) = (2)/(3) xx (1)/(16) =1/(24)` . |
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| 414. |
Which of the following statement (s) is//are correct ? .A. Molecules of different ideal gases have the same kinetic energy at a given temperature .B. The total kinetic energy for two moles of an ideal gas is equl to `3RT`C. The ratio of specific heat at contant pressure and the specific heat at constant volume for noble gases is 1.33 .D. The gas with a larger value of the ration of critical temperature to crtical pressure `(T_(c)//P_(c))` will have larger value of excluded volume . |
| Answer» The ratio of specific heat at constnat pressure and the specific heat of constant volume for noble gases is 1.66 . | |
| 415. |
A 2.00 litre evacuated container has a mass of 1050.0g. When the container is filled with an unknown gas at 800mm Hg pressure and `25.0^(@)C` the mass is 1052.4g. What is the molar mass of the gas (in `g"mol"^(-1)`)?A. 28B. 31C. 54D. 56 |
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Answer» Correct Answer - a |
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| 416. |
A cylindrical container with moveable piston initially hold 3.0 mole of a gas at 8.0 atm pressure and a volume of 5.0 litre If piston is moved to create a volume 10.0 litre while simultaneously with drawing 1.5 mole of gas Calculate the final pressure . |
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Answer» Correct Answer - 2 `PV =nRT` or `(PV)/(n) =RT` `P_(1) =8atm, V_(1) =5.0` litre `n_(1) =3.0` mole `P_(2) =?` `V_(2) =10` litre `n_(2) =1.5` mole Now `(P_(1)V_(1))/(n_(1)) =(P_(2)V_(2))/(n_(2))` `(8xx5)/(3)=(P_(2)xx10)/(1.5)` `P_(2) =2atm` . |
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| 417. |
To an evacuated vessel movable piston under external pressure of 1 atm, 0.1 mol of He and 1.0 mol of an unknown compound ( vapour pressure 0.68 atm, at `0^(@)C`) are introduced. Considering the ideal gas behaviour , the total volume ( in litre) of the gases at `0^(@)C` is close to |
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Answer» Correct Answer - 7 `P_(He)=1-0.68=0.32 atm` `V=?` `n=0.1, V= (nRT)/(P) = (0.1 xx 0.0821 xx 273)/(0.32)= 7` |
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| 418. |
0.5 L of evacuated container is filled by gas upto 1 atm exactly, by connecting it to 20 litre cylinder initially at 1.2 atm. How many evacuated containers can be filled ? |
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Answer» Correct Answer - 8 |
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| 419. |
A 5.00L evacuated cylinder is charged with 25.5g of `NH_(3)` and 36.4 g of HCl. Calculate the final pressure at `85.0^(@)C` after the two compounds have reacted completely: `NH_(3)(g)+HCl(g)rarrNH_(4)CI(s)`A. 2.94 atmB. 5.88atmC. 8.82atmD. 14.7atm |
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Answer» Correct Answer - a |
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| 420. |
What is the relationship between the average velocity `(v)`, root mean square velocity `(u)` and most probable velocityA. `alpha : v : u : : 1: 1.128: 1.224`B. `alpha : v:u : : 1.128: 1: 1.224`C. `alpha :v:u: : 1.128 : 1.224 :1`D. `alpha : v : u : : 1.124: 1.228:1` |
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Answer» Correct Answer - A `V_(av): V _(rms) : V_("most probable") = V: U : alpha` `sqrt((8RT)/(piM)): sqrt((3R)/(M)): sqrt((2RT)/(M))` `alpha : V:U = sqrt(2) : sqrt((8)/(pi)): sqrt(3) = 1: 1.128 : 1.224` |
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| 421. |
A 2.00L balloon at `20.0^(@)C` and 7.45mm Hg floats an altitude where the temperature is `10.0^(@)C` and the air pressure is 700mm Hg. What is the new volume of the balloon?A. 0.94LB. 1.06LC. 2.06LD. 2.20L |
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Answer» Correct Answer - c |
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| 422. |
A bubble of gas released at the bottom of a lake increases to four times its original volume when it reaches the surface. Assuming that atmospheric pressure is equivalent to the pressure exerted by a column of water 10 m high, what is the depth of the lake?A. 80mB. 90mC. 40mD. 70m |
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Answer» Correct Answer - D Since volume of bubble at bottom of lake becomes 8 times. Its means the pressure at the bottom of the lake is 8 times the pressure at the surface Therefore, pressure due to depth of lake. `=8 xx 10-10= 70m`. |
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| 423. |
A high altitude balloon contains 6.81 g of helium in `1.16xx10^(4)" L at"-23^(@)C.` Assuming ideal gas behaviour, how many grams of helium would have to be added to increase the pressure to `4.0xx10^(-3)` atm?A. 1.27 gB. 1.58 gC. 2.68 gD. 2.23 g |
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Answer» Correct Answer - d (d) `P_(1)=(6.81)/(4)xx(0.0821xx250)/(1.1xx104)` `=3.012xx10^(-3)` `P_(f)=4xx10^(-3)` `/_Puarr=1xx10^(-3)` `n_("He-added")=(10^(-3)xx1.16xx10^(4))/(0.0821xx250)` `wt=n_("He")xx4=2.24"gm".` |
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| 424. |
A He atom at 300 K is released from the surface of the earth to travel upwards. Assuming that it undergoes no collision with other molecules, how high will it be before coming to the rest?A. 9.53 mB. 9.5 mC. 953 mD. `9.53xx10^(4)` m |
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Answer» Correct Answer - d (d) `(3)/(2)KT=mgh` `h=(3)/(2)xx(1.38xx10^(-23)xx300)/(9.81xx4xx1.66xx10^(-27))` `=9.53xx10^(4)m` |
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| 425. |
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the moleculesA. Exert no attractive forces on each otherB. Do work equal to loss of `KE`C. Collide without loss of energyD. Are above the inversion temperature . |
| Answer» In absence of attractive forces, energy is not needed to separate molecules from each other on expansion . | |
| 426. |
11 moles of `N_(2)` and 12 moles of `H_(2)` mixture reacted in 2.0 litre vessel at 800 K. After equilibrium was reached, 6 mole of `H_(2)` was present. 3.58 litre of liquid water is injected in equibrium mixture and resultant gaseous mixture suddenly cooled to 300K. What is the final pressure of gaseous mixture? Neglect vapour pressure of liquid solution. Assume (i) all `NH_(3)` dissolved in water (ii) no change in volume of liquid (iii) At 300 K no reaction takes place between `N_(2)` and `H_(2)` A. 18.47 atmB. 60 atmC. 22.5 atmD. 45 atm |
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Answer» Correct Answer - c (c) `N_(2)+3H_(2)hArr2NH_(3)` `{:("Initial moles",11,12,0),("at equilibrium",9,6,4):}` moles of `N_(2)` and `H_(2)` present at equilibrium =15 after addition of water `NH_(3)(g)+H_(2)O(l)rarrNH_(4)OH(l)` Volume of vessel available for gaseous mixture fo `N_(2)` and `H_(2)=20-3.58` `implies 16.42 L` Pressure exerted by gaseous mixture at `300 K= (15xx0.821xx300)/(16.42)implies22.5 "atm"` |
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| 427. |
Consider the composite system, which is held at 300 k , shown in the following figure. Assuming ideal gas behavior, calculate the total pressure if the barriers sparating the compartments are removed. Assume that the volume of the barriers is negligible. (Given : R =0.082 atm L/mol. K) A. 1 atmB. 2 atmC. 2.3 atmD. 3.2 atm |
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Answer» Correct Answer - b (b) The number of moles of He, Ne and Xe is given by `n_(He)=(PV)/(RT)=(3)/(RT)," "n_(Ne)=(PV)/(RT)=(10)/(RT),` `n_(Xe)=(PV)/(RT)=(1)/(RT),` `n=n_(He)+n_(Ne)+n_(Xe)=((3+10+1))/(RT)` The total pressure is given by `P=((n_(He)+n_(Ne)+n_(Xe))RT)/(V)` `P_("total")=((3+10+1))/(RT)xx(RT)/(7)=2atm` |
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| 428. |
The Joule -Thomson coefficient for a gas is zero at .A. inversion temperatureB. Critical temperatureC. Absolute temperatureD. Below`0^(@)C` |
| Answer» At inversion temperature gases show neither cooling nor heating on subjecting to Joule Thomson effect . | |
| 429. |
The hydrogen was warms up during the Joule -Thomson expansion The temperature at which the Joule Thomson coefficeint becomes negative is called Joule-Thomson inversion temperature . |
| Answer» The temperature at which the Joule-Thomson coefficeint becomes zero is called Joule-Thomson inversion temperature . | |
| 430. |
The kinetic energy of N molecules of `O_(2)` is x joule at `-123^(@)C`. Another sample of `O_(2)` at `27^(@)`C has a kinetic energy of 2x. The latter sample contains molecules of `O_(2)`.A. NB. `(N)/(2)`C. 2ND. 3N |
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Answer» Correct Answer - a |
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| 431. |
Cycle tubes, each of capacity 4 L are to be filled by `N_(2)` gas at `5` atm and `300 K`. The gas is present in a many tubes can be completely inflated by connecting the cylinder with tubes. |
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Answer» Correct Answer - `0004` |
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| 432. |
Calculate the root mean square speed total and average translational kintic energy in joule of the molecules in 9 methane at `27^(@)C` . |
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Answer» `T =27 + 273 = 300 K` `R =8.314 xx 10^(7)` erg `u_("rms")` for `CH_(4)` `= sqrt(((3RT)/(M))) = sqrt(((3 xx 8.314 xx 10^(7) xx 300)/(16)))` ` =6.84 xx 10^(4) cm sec^(-1)` Now `KE//mol` of `CH_(4) = (1)/(2) Mu^(2)` ` = (1)/(2) xx 16 xx (6.84 xx 10^(4))^(2) = 374.28 xx 10^(8) erg mo1^(-1)` `:. KE` for `(1)/(2)` mole `CH_(4) = (374.28 xx 10^(8))/(2) erg` `= (374.28 xx 10^(8))/(2 xx 10^(7))` joule = 1871. 42 joule Average kinetic energy `= (K.E//mo1)/(Av.no) = (374.28 xx 10^(8))/(6.023 xx 10^(23))` `= 62. 14 xx 10^(-15) erg = 62.14 xx 10^(-22)` joule . |
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| 433. |
Assuming the same pressure in each case, calculate the mass of hydrogen (in g) required to inflate a balloon to a certain volme V at `127^(@)C` if 8g helium is required to inflate the balloon to half the volume, `0.50 V` at `27^(@)C`. |
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Answer» Correct Answer - `0006` |
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| 434. |
The averge translational kinetic energy of `8g` methane at `27^(@)C` is .A. `3.1xx10^(-21) J`B. `6.2xx10^(-21) J`C. `4.2xx10^(-21) J`D. `8.4xx10^(-21) J` |
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Answer» Average `K.E = (3)/(2) xx (R )/(N) xx T` ` =(3)/(2) xx (8.314)/(6.023 xx 10^(23)) xx 300` `= 6.21 xx 10^(-21) J` . |
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| 435. |
0.1 mole of argon has pressure P and temperature 7K in the vessel. On keeping the vessel at `50^(@)C` higher temperature, 0.8g of argon was given out to maintain same pressure. The original temperature was: [Ar=40]A. 273KB. 200KC. 100KD. 300K |
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Answer» Correct Answer - b |
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| 436. |
According to kinetic theory of gases in an ideal gas between two successive collisions a gas molecule travles .A. In a straight line pathB. With an accelerated velocityC. In a circular pathD. In a wavy path |
| Answer» It is one of the assumption of kinetic theory of gases . | |
| 437. |
According to kinetic theory of gases,A. Collisions are always elasicB. Heavier molecules transfer more momentum to the wall of the containerC. Ony a small number of molecules have very high velocityD. Between collisions, the molecules move in straight lines with constant velocities |
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Answer» Correct Answer - A::B::C::D (a) Fact (b) `P = MV=Msqrt((3RT)/(M))= sqrt(3MRT)` (c ) Max well distribution (d) Fact |
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| 438. |
According to kinetic theory of gases, for a datomic molecule.A. The pressure exerted by the gas is proportional to mean velocity of the moleculeB. The pressure exerted by the gas is proportional to the root mean velocity of the moleculeC. The root mean square velocity of the molecule is inversely proportional to the temperatureD. The mean translational kinetic energy of the molecule is proportioanl to the absolute temperature |
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Answer» Correct Answer - D Pressure exerted by the gas, `P = (1)/(3) (mn u^(2))/(V)`.....(1) Here, `u=` root mean squarevelocity `m=` mass of a molecule , `n=` No. of molecules of the gas Hence (a) and (b) are clearly wrong. Again `u^(2)=(3RT)/(M)` [explained from (1) ] Here, `M=` Molecular wt. of the gas , Hence `(c )` is wrong Further Average `K.E. =(3)/(2) KT,` Hence, (d) is true. |
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| 439. |
According to kinetic theory of gases, for a diatomic moleculeA. The pressure exerted by the gas is proportional to the mean velocity of the moleculesB. The pressure exerted by the gas is proportional to the root mean square velocity of the moleculesC. The root mean square velocity is inversely proportional to the temperatureD. The mean translational kinetic energy of the molecules is proportional to the absolute temperature. |
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Answer» Correct Answer - D All molecules of an ideal gas show random motion. They collide with each other and walls of container during which they lose or gain energy so they may not have same kinetic energy always. |
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| 440. |
According to kinetic theory of gases, for a diatomic molecule wich is (are) not correct?A. The pressure exerted by the gas is directly proportional to the mean speed of the moleculeB. The pressure exerted by the gas is directly proportional to the root mean square speed of the molecule.C. The root mean square speed of the molecule is inversely proportional to the temperature.D. The mean transitional kinetic energy of the molecule is proportional to the absolute temperature. |
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Answer» Correct Answer - abc |
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| 441. |
According to kinetic theory of gases:A. collisions are aways elasticB. heavier molecuels transfer more momentum to the wall of the container.C. only a small number of molecules have very high velocityD. between collisons, the molecules move to straight lines with constant velocities |
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Answer» Correct Answer - acd |
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| 442. |
Four one litre flasks are separately filled with the gases, `O_(2),F_(2),CH_(4)` and `CO_(2)` under the same conditions. The ratio of number of molecules its these gases:A. `2:2:4:3`B. `1:1:1:1`C. `1:2:3:4`D. `2:3:3:4` |
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Answer» Correct Answer - b |
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| 443. |
If a mixture of `CO` and `N_(2)` in equal amount have total 1 atm pressure, then partial pressure of `N_(2)` in the mixture isA. 1 atmB. `0.50` atmC. 2 atmD. 3 atm |
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Answer» Correct Answer - B Partial pressure of a component `=` Mole fraction `xx` Total pressure or , `p_(N_(2))=("No. of moles of "N_(2))/("Total no. of moles")xx 1` or , `p_(N_(2))=(1)/(20xx1=0.5)` |
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| 444. |
There are `6.02 xx 10^(22)` molecules each of `N_(2),O_(2)` and `H_(2)` which are mixed together at `760 mm` and `273 K`. The mass of the mixture in grams is `:`A. 6.2B. 4.12C. 3.09D. 7 |
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Answer» Correct Answer - A `6.02 xx10^(22)` molecules of each `N_(2),O_(2)` and `H_(2)=(6.02xx10^(22))/(6.02 xx 10^(23))` moles of each Weight of mixture `=` weight of 0.1 mole `N_(2) + `weight of 0.1 mole `H_(2)+` weight of 0.1 mole of `O_(2)` `(28 xx 0.1) +(2xx0.1)+(32 xx0.1)=6.2 ` gm |
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| 445. |
There are `6.02 xx 10^(22)` molecules each of `N_(2),O_(2)` and `H_(2)` which are mixed together at `760 mm` and `273 K`. The mass of the mixture in grams is `:`A. `6.2`B. `4.12`C. `3.09`D. 7 |
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Answer» Correct Answer - 1 Moles of each gas `=(6.02xx10^(22))/(N_(A))=0.1` So mass of mixture `=0.1xxM_(N_(2))+0.1xxM_(O_(2))+0.1xxM_(H_(2))` `=0.1xx28+0.1xx32+0.1xx2=6.2 ` grams. |
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| 446. |
The density of a mixturee of `O_(2) and N_(2)` gases at 1 atm and 273 K is 0.0013 gm/mL. If partial pressure of `O_(2)` in the mixture is A, then calculate value of 25 A. |
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Answer» Correct Answer - 7 |
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| 447. |
Some gas is present in J-shaped tube and in this case level of mecury in both limbs of tube differ by `39` cm of Hg. If mercury is replaced by another liquid of density `(3)/(4)` of density of mercury then find the height difference in limbs. Assume temperature to be constant and smaller limb is closed, (density of mercury 13.6 g/mL) |
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Answer» Correct Answer - 52 |
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| 448. |
Helium is often found with methance, `CH_(4)`. How do the diffusion rates of helium and methane compare at the same temperature? Helium diffuses:A. sixteen times as fast as methane.B. four times as fast as methane.C. twice as fast as methane.D. at the same rate as methane. |
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Answer» Correct Answer - c |
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| 449. |
The ratio of the rate of diffusion of helium and methane under indentical conditions of pressure and temperature will beA. 4B. 2C. 1D. `0.5` |
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Answer» Correct Answer - B `(r_(He))/r_(CH_(4))=sqrt((M_(CH_(4)))/(M_(He)))=sqrt((16)/(4))=2` |
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| 450. |
The ratio of the rate of diffusion of helium and methane under indentical conditions of pressure and temperature will beA. `4`B. `2`C. `1`D. `0.5` |
| Answer» `(gammaHe)/(gammaCH_(4)) = sqrt(M_(CH_(4))/(M_(He))) = sqrt((16)/(4)) =2` . | |