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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.
| 351. |
Match gases under specified conditions listed in Coloum-I with their properties//laws in Column-II. `{:(,"Column-I",,"Column-II"),("(a)","Hydrogen gas(P=200 atm T=273K)",,(p)"Compressibility factor "ne1),("(b)","Hydrogen gas (p~0,T=273K)",,"(q)At tractive forces are dominant"),("(c)",CO_(2)(P=1 atm,T=273K),,"(r) PV=nRT"),("(d)","Real gases with very large molar volume",,"(s)P(V-nb)=nRT"):}` |
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Answer» Correct Answer - (a-p,s);(b-r);(c-p,q);(d-r) |
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| 352. |
Assertion: Most probable velocity is the velocity possessed by maximum fraction of molecules at the same temperature. Reason: On collision, more and more molecules acquire higher speed at the same temperature.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 - c |
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| 353. |
Temperature at which `r.m.s` speed of `O_(2)` is equal to that of neon at `300 K` is:A. 280KB. 480KC. 680KD. 180K |
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Answer» Correct Answer - b |
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| 354. |
Two inflated ballons I and II (thin skin) having volume 600 mL and 1500 mL at 300 K are taken as shown in diagram. If maximum volume of inner and outer balloons are 800 mL and 1800 mL respectively then find the balloon which will burst first on gradual heating. A. inner balloonB. outer balloonC. both simultaneouslyD. unpredictable |
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Answer» Correct Answer - b (b) Case `I-` Suppose inner balloon burst first first `(600)/(300)=(800)/(T_(2))impliesT_(2)=400K` Case `II-`Suppose inner balloon burst first `(1500)/(300)=(1800)/(T_(2))," "T_(2)=360K` |
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| 355. |
The density of gas `A` is twice that of B at the same temperature the molecular weight of gas B is twice that of A. The ratio of pressure of gas A and B will be :A. `1:6`B. `1:1`C. `4:1`D. `1:4` |
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Answer» Correct Answer - c (c) `d_(A)=2d_(B)," "3M_(A)=M_(B)," " PM=dRT` `=(P_(A))/(P_(B))xx(M_(A))/(M_(B))xx(d_(A))/(d_(B))xx(RT)/(RT)` `=(P_(A))/(P_(B))xx(1)/(2)=2` `(P_(A))/(P_(B))=(4)/(1)` |
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| 356. |
Sulphur and fluorine form `SF_(6)` and `S_(2)F_(10)`, both of which are gases at `30^(@)C`. When an equimolar mixture of them is allowed to effuse through a pinhole, what is the ratio `SF_(6)//S_(2)F_(10)` in the first sample that escapes? `{:("Molar mass",g"mol"^(-1)),(SF_(6),146),(S_(2)F_(10),254):}`A. `(1.32)/(1)`B. `(1.74)/(1)`C. `(3.03)/(1)`D. `(3.48)/(1)` |
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Answer» Correct Answer - a |
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| 357. |
For which of the following gasses should the correction for the molecular volume be largest : `CO,CO_(2),NH_(3) " or " SF_(6)` ?A. COB. `CO_(2)`C. `NH_(3)`D. `SF_(6)` |
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Answer» Correct Answer - d (d) `V_(i)=V-V(ex)` `V_(ex)=n.b` `b prop "size" ` |
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| 358. |
A gas mixture contains equal number of molecules of `N_(2)` and `SF_(6)` Some of it is passed through gaseous effussion apparatus Calculate how many molcules of `N_(2)` are pressent in the product gas for every 100 molecules of `SF_(6)` ? (At wt of `F=20)` . |
| Answer» Correct Answer - 233 | |
| 359. |
A gaseous mixture was prepared by taking equal moles of `CO` and `N_(2)`. If the total pressure of the mixture was found to be `1` atomosphere, the partical pressure of the nitrogen `(N_(2))` in the mixture isA. 1 atmB. 0.5 atmC. 0.8 atmD. 0.9 atm |
| Answer» Correct Answer - B | |
| 360. |
A gas such as carbon monoxide would be most likely to obey the ideal gas law atA. High temperature and low pressuresB. Low temperature and high pressueC. High temperature and high pressuresD. Low temperature and low pressures |
| Answer» Correct Answer - A | |
| 361. |
Dominance of strong repulsive forces among the molecules of the gas (`Z =` compressibility factor)A. Depends on Z and indicated by `Z=1`B. Depdens on Z are indicated by `Z gt1`C. Depends on Z and indicated by `Z lt 1`D. Is independent of Z |
| Answer» Correct Answer - C | |
| 362. |
A rigid vessel of volume ` 0.50m^(3)` containing `H_(2) ` at `20.5^(@)C` and a pressure of `611 xx 10^(3)` Pa is connected to a second rigid vessel of volume 0.75`m^(3)` containing Ar at `31.2^(@)C` at a pressure of `433xx10^(3)` Pa. A value separating the two vessels is opened and both are cooled to a temperature of `14.5^(@)C`. What is the final pressure in the vessels?A. `2xx10^(5)`B. `3.22xx10^(5)" Pa"`C. 4840 paD. `4.84xx10^(5)" Pa"` |
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Answer» Correct Answer - d (d) `n_(T)=n_(2)+n_(2) ` `(Pxx1.25)/(Rxx287.5)=(611xx10^(3)xx0.5)/(293.5 R)+(433xx10^(3)xx0.75)/(204.5R)` `P=4.84xx10^(5)Pa` |
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| 363. |
A gas of volume 100cc is kept in a vessel at pressure `10^(4)` Pa maintained at temperature `24^(@)C`. If now the pressure is increased to `10^(5)` Pa, keeping the temperature constant, then the volume of the becomes:A. 10ccB. 100ccC. 1ccD. 1000cc |
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Answer» Correct Answer - a |
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| 364. |
A gaseous molecule has an empirical formula `CH_(3)` Its density is `2.06g//litre` at the same temperature and pressure at which oxygen has density of `2.21g//litre` Calculate the molcular formula of gas . |
| Answer» Correct Answer - `C_(2)H_(6)` | |
| 365. |
In an experiment, it was found that for a gas at constnat temperature, PV=C. The value of C depends on:A. atmospheric pressureB. quantity of gasC. molecular weight of gasD. volume of chamber |
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Answer» Correct Answer - b |
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| 366. |
If the mean free-path of gaseous molecule is `60cm` at a pressure of `1xx10^(4)mm` mercury, what will be its mean free-path when the pressure is increased to `1 xx10^(-2)` mm mercury ? .A. `6.0 xx 10^(-1) cm`B. `6.0 cm`C. `6.0 xx 10^(-2) cm`D. `6.0 xx 10^(-3) cm` |
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Answer» Mean free path `prop (1)/("Pressure")` `(l_(1))/(l_(2)) = (P_(2))/(P_(1)) implies l_(2) = (l_(1)P_(1))/(P_(2))` ` = (60 cm xx 1 xx 10^(4)mm)/(1xx10^(-2)mm)` ` = 6 xx 10^(-1) cm` . |
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| 367. |
Pure hydrogen sulphide is stored in a tank of 100 litre capacity at `20^(@)`C and 2 atm pressure. The mass of the gas will beA. 34gB. 340 gC. 282.4 gD. 28.24g |
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Answer» Correct Answer - C `n=(PV)/(RT)=(m)/(M)` `m=(MPV)/(RT)=(34xx2xx100)/(0.082xx293)=282.4gm` |
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| 368. |
Calculate the volume occupied by `96 g CH_(4)` at 16 atm and `27^(@)C(R -0.08` litre atm `K^(-1) mol^(-1))` . |
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Answer» Correct Answer - 9 Given `P =16atm, m =16` `T = 27 +273 = 300K, V =? ` `:. PV = (w)/(m)RT` `:. V = (wRT)/(Pm) = (96 xx 0.08 xx 300)/(16 xx 16) =9 litre` . |
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| 369. |
The volume of helium is `44.8 L` atA. `100^(@)C` and 1 atmB. `0^(@)C` 1 atmC. `0^(@)C` and 0.5 atmD. `100^(@)C` and 0.5 atm |
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Answer» Correct Answer - c |
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| 370. |
A valve between a 5 litre tank in which the gas pressure is 9 atm and a 10 litre tank containing gas at 6 atm is opened and pressure equilibration ensures at a constant temperature. What is the final pressure (in atm) in the two tanks ? |
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Answer» Correct Answer - `0007` |
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| 371. |
Vibrational energy isA. Partially potential and partially kineticB. Only potenitalC. Only kineticD. None of the above |
| Answer» Correct Answer - A | |
| 372. |
The intercept of the line drawm for log `P` (P in atm) and log `(1)/(V)` (V in litre) for 1 mole of an ideal gas at `27^(@)C` is equal to .A. `log 2.463`B. `log 24.63`C. `log 22.4`D. `log 2.24` |
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Answer» `PV =K` or `log P =log K +log .(1)/(V)` Also `PV = RT = 0.0821 xx 300 = 24.63` `:. K =24.63` . |
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| 373. |
`100 g` of an ideal gas (m01 wt 40) is pressent in a cylinder at `27^(@)C` and 2 atm pressure During transportation cylinder The valve attached to cylinder cannot keep the pressure greater than 2 atm and therefore `10g` of gas leaked out through cylinder Calculate (i) the volme of cylinder before and after dent (ii) the pressure inside the cylinder . |
| Answer» Correct Answer - `30.79 litre, 27.71 litre 2.22 atm` | |
| 374. |
The average kinetic energy of an ideal gas per molecule in SI units at `25^(@)C` will beA. `6.17xx 10^(-21)Kj`B. `6.17 xx 10^(-21) J`C. `6.17 xx 10^(-20) J`D. `6.17 xx 10^(-20) J` |
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Answer» Correct Answer - B Average kinetic energy per molecule. `bar(K.E.) =(3)/(2) KT = (3)/(2) xx (8.314)/(6.02 xx 10^(23))xx298` `= 6.17 xx 10^(-21) J` |
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| 375. |
The average kinetic energy of one molecule of an ideal gas at `27^(@)C` and 1 atm pressure atA. `900 cal K^(-1)` molecule `^(-1)`B. `6.21 xx 10^(-21) JK^(-1)` molecule `^(-1)`C. `336.7 JK^(-1)` molecule `^(-1)`D. `3741.3 jk ^(-1) "molecule"^(-1)` |
| Answer» Correct Answer - B | |
| 376. |
One mole of a monoatomic real gas satisfies the equation `p(V-b)=RT` where `b` is a constant. The relationship of interatomic potential `V(r)` and interatomic distance `r` for gas is given byA. B. C. D. |
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Answer» Correct Answer - C `P(V-b)=RT` `implies PV-Pb=RT implies (PV)/(RT)=(Pv)/(RT)+1` `implies Z= 1+(PV)/(RT)` Hence `Z gt 1` at all pressures This means, repulsive tendencies will be dominant when interatomic distance are small This means, interatomic potential is never negative but becomes positive at small interatomic distances. Hence answer is (C ). `P(V-b)=RT` `implies PV-Pb= RTimplies (PV)/(RT) =(Pb)/(RT)+1` `implies Z= 1+(Pb)/(RT)` |
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| 377. |
One mole of a monoatomic real gas satisfies the equation `p(V -b) = RT` where `b` is a constant. The relationship of interatomic potential `V(r )` and interatomic distance `r` for the gas is given by |
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Answer» Correct Answer - C `P (V-b) = RT` `rArr PV = Pb = RT` `rArr (PV)/(RT) = (Pb)/(RT) +1` `rArr Z = 1+(Pb)/(RT)` Hence `Z gt 1` at all pressure. This means, repulsive tendencies will be dominant when interatomic distance are small. This means, interatomic potential is never negative but becomes positive at small interatomic distances. Hence answer is (C ) |
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| 378. |
The ratio among most probable velocity, mean velocity and root mean velocity is given byA. `1:2:3`B. `1 sqrt(2) : sqrt(3)`C. `sqrt(2) : sqrt(3) : sqrt( 8 //pi)`D. `sqrt(2) : sqrt( 8//pi) : sqrt(3)` |
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Answer» Correct Answer - D Most probable velocity `:` mean velocity `: u_(rms)` `=sqrt((2RT)/(M)): sqrt((8RT)/(pi M)): sqrt((3RT)/(M))= sqrt(2) : sqrt((8)/(pi)):sqrt(3)` |
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| 379. |
At what temperature is the average velocity of `O_(2)` molecule equal to the root mean square velocity at `27^(@)C`?A. `80.57^(@)C`B. `80^(@)C`C. `83^(@)C`D. `86.5^(@)C` |
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Answer» Correct Answer - A `u_(av)=v_(rms)` `sqrt((8RT)/(pim))=sqrt((3RT)/(m))` `(8RT)/(pim)=(3RT)/(m)implies (8RT)/(pim)=(3Rxx300)/(m)` `T= 353.57 K` `t= 80.57^(@)C` |
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| 380. |
Root mean square velocity of a gas molecule is proprotional toA. `m^(1//2)`B. `m^(0)`C. `m^(-1//2)`D. `m` |
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Answer» Correct Answer - C `u_(rms)=sqrt((3KT)/("Molecular weight"))i.e., u_(rms)prop (1)/(sqrt(m))prop (m) ^((1)/(2))` |
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| 381. |
A quantity of gas is collected in a graduated tube over the mercury. The volume of gas at `18^(@)C` is 50 mL and the level of mercuty in the tube is 100mm above the outside mercuty level. The barometer reads 750 torr. Hence, volume of gas 1 atm and `0^(@)C` is approximately:A. 22 mLB. 40mLC. 20mLD. 44mL |
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Answer» Correct Answer - b |
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| 382. |
Assertion: A mixture of `He and O_(2)` is used for respiration for deep sea divers. Reason: `He` is soluble in blood.A. If both assertion and reason are true and the reason is correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - C The assertion, that a mixture of helium and oxygen is used for deep sea divers, is correct. The He is not soluble in blood. Therefore, this mixture is used. |
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| 383. |
Two vassels having equal volumes contains `H_(2)` and He at 1 and 2 atm respectively same temperature Select the correct statements .A. `u_(rms)H_(2)=u_(rms)`B. `r_(H_(2))=(r_(He))/(sqrt2)`C. `u_(rms)H_(2)=sqrt2u_(rmsHe)`D. `r_(H_(2))=sqrt8r _(He)` |
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Answer» `u = (3RT)/(M)` `:. (u_(H_(2)))/(u_(He)) = sqrt((M_(He))/(M_(H_(2))))=sqrt((4)/(2)) =sqrt2` Also `:. (r_(H_(2)))/(r_(He)) = sqrt((M_(He))/(M_(H_(2))))xx(P_(H_(2)))/(P_(he))` `=sqrt((4)/(2)) xx (1)/(2) = (1)/sqrt2` . |
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| 384. |
Which of the following statement is incorrect ? .A. Joule-Thomson coefficient is zero at inversion temperature of a real gas .B. Ideal gas do not show Joule-Thomson effectC. Inversion temperature of `H_(2)` and he is very-very lowD. Joule Thomson coefficient ` mu =((deltaP)/(deltaT))_(H)` . |
| Answer» Joule-Thomson coefficeint `mu=((deltaT)/(deltaP))_(H),mu=0` at inversion temperature Also ideal gas does not show Joule -Thomson effect `H_(2)` and He have low value of `T_(i)` and thus show heating effect if subjected for Joule-Thomson effect . | |
| 385. |
Which of the following is a character of a gas at Boyle temperature?A. the effects of the repusive and attractive intermolecular forces just offset each otherB. the repulsive intermolecular forces ar stronger than the attractive intermolecular forcesC. the repulsive intermolecular forces ar weaker than the attractive intermolecular forcesD. `b-(a)/(RT)gt0` |
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Answer» Correct Answer - a (a) `T_(B)` is a temperature at which gas behave like an ideal gas mean attractive and repulsing forces just offset each other. |
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| 386. |
Indicate the correct statement for equal volumes of `N_(2)(g)` and `CO_(2)(g)` at `25^(@)C` and 1 atm.A. The average translational K.E. per molecule is the same for `N_(2)` and `CO_(2)`B. The rms speed is same for both `N_(2)` and `CO_(2)`C. The density of `N_(2)` is less than that of `CO_(2)`D. The total translational K.E. of both `N_(2)` and `CO_(2)` is the same |
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Answer» Correct Answer - a,c,d (a,c,d) `KE=(3)/(2)RT(1"mole")U_(rms)=sqrt((3RT)/(M_(w)))` `d=(PM_(w))/(RT)` |
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| 387. |
A mixture of helium of neon gases is collected over water at `28.0^(@)C` and 745 mmHg. If the partial pressure of helium is 368 mmHg, what is the partial pressure of neon?A. 348.7 mmHgB. 377 mmHgC. 384.7 mmHgD. none of these |
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Answer» Correct Answer - a (a) `P_(T)=P_(He)+P_(Ne)+V.PH_(2)O` `745=368+P_(Ne)+28.3` `P_(Ne)=348.7 mm " of " Hg` |
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| 388. |
Equal moles of `CO,B_(2)H_(6),H_(2)` and `CH_(4)` are placed in a container If a hole was made in container after 5 minute partial pressure of gases in container would be (at.wt of `C,O,B` and `H` are `12,16,1` and 1 respectively) .A. `P_(CO ) gt P_(B_(2)H_(6)) gt P_(H_(2)) gt P_(CH_(4))`B. `P_(CO) =P_(B_(2))H_(6) gt P_(CH_(4)) gt P_(H_(2))`C. `P_(CO) gtP_(B_(2)H_(6)) gt P_(H_(2)) gt P_(CH_(4))`D. `P_(B_(2))H_(6) gt P_(H_(2)) gt P_(CH_(4)) gt P_(CO)` |
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Answer» `CO` and `B_(2)H_(6)` have same molecular mass. The diffusion of lighter gas will be more Thus moles coming out from container will show `H_(2) gtCH_(4) gtCO =B_(2)H_(6)` . |
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| 389. |
80 mL of `O_(2)` takes 2 minutes to pass through the hole. What volume of `SO_(2)` will pass through the hole in 3 minute?A. `(120)/(sqrt(2))` mLB. `120xxsqrt(2)` mLC. `(12)/(sqrt(2))` mLD. None of these |
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Answer» Correct Answer - a (a) `(r_(1))/(r_(2))=(V_(1)//t_(1))/(V_(2)//t_(2))=(V_(1)xxt_(2))/(V_(2)xxt_(1))=sqrt((M_(2))/(M_(1))),` `(80xx3)/(V_(2)xx2)=sqrt((64)/(32))=sqrt(2), " " V_(2)=(120)/(sqrt(2))` |
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| 390. |
Which statement is not a principle (postulate) of kinetic molecular theory?A. The molecules of a gas are in rapid random motion.B. The molecules of an ideal gas exhibit. No attractive forces.C. The collisions of gaseous molecules with one another and the walls of their container are elastic.D. Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. |
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Answer» Correct Answer - d |
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| 391. |
Which is not one of the postulates of the kinetic molecular theory?A. At a constant temp. all the particle have the same speed.B. There are no force of attraction between molecules.C. Gas particles move in a straight line between collisions.D. The molecules are in a state of constant random motion. |
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Answer» Correct Answer - a |
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| 392. |
Dry ice is solid carbon dioxide. A 0.050 g sample of dry ice is placed in an evacuated 4.6 L vessel at `30^(@)C`. Calculate the pressure inside the vessel after all the dry has been converted to `CO_(2)` gas.A. 6.14 atmB. 0.614 atmC. 0.0614 atmD. `6.14xx10^(-3)` atm |
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Answer» Correct Answer - d (d) `P=(nRT)/(V)` `P=(0.05xx0.0821xx303)/(44xx4.6)=6.14xx10^(-3)atm` |
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| 393. |
A 4.40 g piece of solid `CO_(2)` (dry ice) is allowed to sublime in a balloon. The final volume of the balloon is 1.00 L at 300 K. What is the pressure (atm) of the gas?A. 0.122B. 2.46C. 122D. 24.6 |
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Answer» Correct Answer - b (b) `P=(nRT)/(V)=2.46"atm"` |
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| 394. |
A closed vessel contains equal number of nitrogen and oxygen molecules at pressure of `P mm`. If nitrogen is removed from the system, then the pressure will be:A. PB. 2PC. `P//2`D. `P^(2)` |
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Answer» Correct Answer - C Since no. of moleculares is halved so pressure should also be halved. |
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| 395. |
At room temperature following traction goes to completion `2AB(g)+B_(2)(g)rarr2AB_(2)(s)` `AB_(2)` is solid with negligble vapour pressure below `0^(@)C`. At 300 K, the AB in the smaller flask exerts a pressure of 3 atm and in the larger flask `B_(2)` exerts a pressure of 1 atm at 400 K when they are separated out by a close valve, The gases are mixed by opening the stop cock and after the end of the reaction the flask are cooled to 250 K The final pressure is :A. 0.156 atmB. 0.3125 atmC. 0.625 atmD. 3.2 atm |
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Answer» Correct Answer - c (c) `n_(AB)=(3xx8.21)/(0.0821xx300)=1,` `n_(B_(2))=(1xx8.21)/(0.0821xx400)=0.25` `2AB(g)+B_(2)(g)rarr2AB_(2)(s)` `{:("Initial moles",1,0.25),("After reaction",0.5,0):}` `P_(AB)xx(8.21+8.21)=0.5xx0.0821xx250` `P_(AB)=0.625atm` |
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| 396. |
To an evacauted `504.2 mL` steel container is added `25 g CaCO_(3)` and the temperature is raised to `1500K` causing a complete decomposition of the salt. If the density of `CaO` formed is `3.3 g//cc`, find the accurate pressure developed in the container using the Van der waals equation of state. The van der waals constants for `CO_(2)(g)` are `a = 4 (L^(2)-atm)/(mol^(@)), b = 0.04 (L)/(mol). (ca-40, C -12 ,O -16)`. Report your answer as nearest whole number. |
| Answer» Correct Answer - 62 atm | |
| 397. |
Four par ticles have speed `2, 3, 4` and 5 cm/s respectively Their `RMS` speed is .A. `3.5 cm//s`B. `(27//2) cm//s`C. `sqrt54 cm//s`D. `11.2 m^(3)` |
| Answer» `u_(rms)=sqrt([[2^(2)+3^(2)+4^(2)+5^(2))/(4)])=sqrt[[(54)/(4)]) cms^(-1)` | |
| 398. |
The volume occupied by 8.8 g of `CO_(2)` at `31.1^(@)C` and 1 bar pressure ( in L ) is |
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Answer» Correct Answer - 5 No of mole of `CO_(2) = (8.8)/(44) = 0.2` mole, `T=304.1K , P =1` bar `V= (nRT)/(P) = (0.2 xx 0.083 xx 304.1)/(1) = 5L` |
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| 399. |
Match the entries listed in Column I with appropriate entries listed in Column II. |
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Answer» Correct Answer - `A rarr r; B rarr s; C rarr p; D rarr q` `y = (1)/(V^(2))` or `sqrt(y) =(1)/(V)` `P =x` and `P = ("constant")/(V)` (A) `x=(k)sqrt(y)implies y=k^(1)x^(2)` (B) `V=kT,y=V` and `(1)/(T) =x:. y = (k)/(x)` (C ) `P= kT, PT = KT^(2)` or `y= kx` (D) `v= (c )/(p) implies y = c sqrt(x), y^(2) =cx` |
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| 400. |
A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas ) . The partition separating the two compartments is fixed and is a perfect heat insulator (figure). If the old partitiion is replaced by a new partition which can slide and conduct heat but dows NOT allow the gas to leak across (Figure2) , the volume ( in `m^(3))` of the compartment A after the system attains equilibrium is `"________________"` |
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Answer» Correct Answer - 2.22 `P_(1) = 5` `P_(2)=1` `v_(1)=1` `v_(2)=3` ` T_(1) = 400` `T_(2)= 300` `n_(1)= (5)/(400R) ` `n_(2)=(3)/(300R)` Let volume be `(v+x) v= (3-x) 15-5x= 4+4x` `(P_(A))/(T_(A))=(P_(B))/(T_(B))` `implies (n_(b_(1))xxR)/(v_(b_(1)))=(n_(b_(2))xxR)/(v_(b_(2)))` `implies (5)/(400(4+x))=(3)/(300R(3-x))` `implies 5(3-x) = 4+4x` `implies x= (11)/(9)` `v= 1+x=1+(11)/(9) = ((20)/(9))=2.22` |
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