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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.
| 1. |
When is deviation more in the behaviour of a gas from the ideal gas equation `PV=nRT`?A. At high temperature and low pressureB. At low temperature and high pressureC. At high temperature and high pressureD. At low temperature and low pressure |
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Answer» Correct Answer - B |
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| 2. |
The density of a gas filled in electric lamp is `0.75kg//m^(3)` . After the lamp has been switched on, the pressure in it increases from `4xx10^(4)` Pa to `9xx10^(4)` Pa. What is increases in `u_("rms")` ?A. 100B. 200C. 300D. None of these |
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Answer» Correct Answer - b (b) `U_(1)=sqrt((3P_(1))/(d_(1)))` `:. /_U_("rms")=sqrt((3)/(d))(sqrt(P_(2))-sqrt(P_(1)))` `=sqrt((3)/(0.75))(300-200)` `=sqrt(4)xx100=200` |
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| 3. |
In a glass tube columns of water and mercury appear as shown. This is best attributed to the differences in their:A. densitiesB. molar massesC. surface tensionsD. viscosities |
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Answer» Correct Answer - c |
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| 4. |
The volume of real gases often exceed those calculated by the ideal gas equation. These deviations are best attributed to the:A. attractive forces between the molecuels in real gases.B. dissociation of the molecuels in real gases.C. kinatic energy of the molecules in real gases.D. volumes of the molecules in rel gases. |
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Answer» Correct Answer - d |
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| 5. |
A container holds 3 liter of `N_(2(g))` and `H_(2)O_((l))` at `29^(@)C` The prssure is found to be 1 atm The water in container is instantaneously electrolysed to give `H_(2)` and `O_(2)` following the reaction `H_(2)O_(l) rarrH_(2(g)) +(1)/(2)O_(2_(g))` At the end of electrolysis the pressure was found to be `1.86atm` calculate the amount of water present in the container If the aqueous tension of water at `29^(@)C` is `0.04 atm ` . |
| Answer» Correct Answer - `1.24 g` | |
| 6. |
Match of following (where `U_(rms)`=root mean square speed `U_(av)` =average speed `U_(mp)` = most probable speed) `{:(,"List-I",,"List-II"),("(a)",U_(rms)//U_(av),,(p)1.22),("(b)",U_(av)//U_(mp),,"(q)1.13"),((c),U_(rms)//U_(mp),,"(r)1.08):}` |
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Answer» Correct Answer - (a-r);(b-q);(c-p) |
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| 7. |
The `u_(rms)` of a gas at `300K` is `3R^(1//2)` The molar mass of the gas in `kg mo1^(-1)` is .A. `1.0 xx 10^(-1)`B. `1.0 xx 10^(-2)`C. `1.0 xx 10^(-3)`D. `1.0` |
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Answer» `u_(rms) = sqrt((3RT)/(M))` `:. 3R^(1//2)=sqrt((3RT)/(M))` or `M = (3xxR xx 300)/(R xx 3 xx3) = 100g mo1^(-1)` `=10^(-1) kg mo1^(-1)` . |
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| 8. |
The cirtical constant for water are `374^(@)C` 218 atm and 0.0566 liter `mo1^(-1)` Calculate a,b and `R` . |
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Answer» Given `T_(C) = 374^(@) C =374 + 273 = 647K` `P_(C) = 218 atm` `V_(C) = 0.0566 "litre" mo1^(-1)` `b = (V_(C))/(3) = (0.0566)/(3) = 0.0189 "litre" mo1^(-1)` `a =3P_(C) V_(C)^(2) =3 xx 218 xx (0.0566)^(2)` ` =2.095 litre^(2) "atm mo1"^(-2)` `R = (8)/(3) (P_(C)V_(C))/(T_(C)) = (8 xx 218 xx 0.0566)/(3 xx 647)` `0.05086 "litre atm"K^(-1) mo1^(-1)` . |
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| 9. |
The density of a gas `A` is twice that of a gas `B` at the same temperature. The molecular mass of gas `B` is thrice that of `A`. The ratio of the pressure acting on `A` and `B` will beA. `6:1`B. `7:6`C. `2:5`D. `1:4` |
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Answer» Correct Answer - a |
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| 10. |
`n_(1)` and `n_(2)` moles of two ideal gases (mo1 wt `m_(1)` and `m_(2))` respectively at temperature `T_(1)K` and `T_(2)K` are mixed Assuming that no loss of energy the temperature of mixture becomes .A. `n_(1)T_(1)+n_(2)T_(2)`B. `(n_(1)T_(1)+n_(2)T_(2))/(T_(1)+T_(2))`C. `(n_(2)T_(1)+n_(1)T_(2))/(n_(1)+n_(2))`D. `(T_(1)xxT_(2))/(n_(1)xxn_(2))` |
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Answer» Let final temperture be `T` then `KE_(I) +KE_(II) =KE_("mixture")` `(3)/(2)n_(1) RT_(1) + (3)/(2)n_(2) RT_(2) = (3)/(2) (n_(1)+n_(n_(2))RT)` `T =(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))` . |
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| 11. |
Match the items of colums I and II. A. `{:(P,Q,R,S),(1,2,4,3):}`B. `{:(P,Q,R,S),(3,4,2,1):}`C. `{:(P,Q,R,S),(2,1,4,3):}`D. `{:(P,Q,R,S),(1,2,3,4):}` |
| Answer» Correct Answer - b | |
| 12. |
A spherical air bubble is rising from the depth of a lake when pressure is P atm and temperature is T K. The percentage increase in its radius when it comes to the surface of lake will be: (Assume temperature and pressure at the surface to be respectively `2 TK` and `(P)/(4)`)A. 1B. 0.5C. 0.4D. 2 |
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Answer» Correct Answer - a |
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| 13. |
`O_(2)` gas is placed in a 4 litre container containing 3 L of liquid water as shown and total pressure exerted by gases is 720mm Hg. What will be the pressure of `O_(2)(g)` if given container is attached to a empty container of 3 litre at same temperature? Given : [V.P. of `H_(2)O` at `27^(@)C=20mm` of Hg] A. 175 mm of HgB. 350 mm of HgC. 200 mm of HgD. 800 mm of Hg |
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Answer» Correct Answer - a |
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| 14. |
At 0.821 atm and at `177^(@)C`, gas `SO_(3)` occupies 45L Moles of neutrons present in `SO_(3)` are: (Take `N_(A)=6xx10^(23)`)A. 40B. `24xx10^(24)`C. `24xx10^(23)`D. `6xx10^(24)` |
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Answer» Correct Answer - a |
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| 15. |
`6xx10^(22)` gas molecules each of mass `10^(-34)kg` are taken in a vessel of 10 litre. What is the pressure exerted by gas molecules ? The root mean square speed of gas molecules is `100 m//s`.A. 20 PaB. `2xx10^(4) Pa`C. `2xx10^(5) Pa`D. `2xx10^(7) Pa` |
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Answer» Correct Answer - b (b) `P=(1)/(3)(nmc^(2))/(V)` `=(1)/(3)xx(6xx10^(22)xx10^(-24)xx(100)^(2))/(10xx10^(-3))` `=2xx10^(4)"Pa"` |
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| 16. |
`6xx10^(22)` gas molecules each of mass `10^(-24)kg` are taken in a vessel of 10 litre. What is the pressure exerted by gas molecules? The average velocity of the gas molecules is 92.62m/sec.A. `2xx10^(5)Pa`B. 20PaC. `2xx10^(6)Pa`D. `2xx10^(4)Pa` |
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Answer» Correct Answer - d |
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| 17. |
Calculate molecular diameter for a gas if its molar exclued volume is `3.2 pi ml`. (in nenometer) (Take `N_(A) = 6.0 xx 10^(23))` |
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Answer» Correct Answer - 2 `4N(4)/(3)pir^(3) = 3.2 xx pi xx 10^(-3)` `4 xx 6 xx 10^(23) xx (4)/(3) pir^(3) = 3.2 pir^(3) = 3.2 pi xx 10^(-3)` `r^(3) = 10^(-9)m` `r = 1 nm` ltbr. `d = 2r = 2nm` |
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| 18. |
Suppose a gas sample in all have `6xx10^(23)` molecules Each `1//3rd` of the molecules have rms speed `10^(4) cm//sec, 2 xx 10^(4) cm//sec` and `3 xx 10^(4) cm//sec` Calculate the rms speed of gas molecules in sample . |
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Answer» `u_(rms) = sqrt((sumn_(1)u_(1)^(2))/(sumn)) =sqrt((u_(1)^(2) xx n_(1) + u_(2)^(2)xxn_(2)+u_(3)^(2)xxn_(3))/(n_(1)+n_(2)+n_(3)))` `:.u_(rms) =sqrt((2xx10^(23)xx(10^(4))+2xx10^(23)xx(2xx10^(4))^(2)+2xx10^(23)xx(3xx10^(4))^(2))/(6xx10^(23)))` `=2.16 xx 10^(4) cm//`sec . |
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| 19. |
500ml of `NH_(3)` contains `6.00 xx 10^(23)` molecules at S.T.P. How many molecules are present in 100 ml of `CO_(2)` at S.T.P.A. `6 xx 10^(23)`B. `1.5 xx 10^(23)`C. `1.2 xx 10^(23)`D. None of these |
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Answer» Correct Answer - C `(PV_(1))/(n_(2)RT)=(PV_(2))/(n_(2)RT)` or , `(V_(1))/(n_(1))=(V_(2))/(n_(2))` or , `(500)/(1)=(100)/(n_(2))` or , `n_(2)=(1)/(5)` `[6.0 xx 10^(23)` molecules of `NH_(3)=1` mole ] 1 mole of `CO_(2)` contains `6.0 xx 10^(23)` molecules `:. (1)/(5)` mole of `CO_(2)` contains `1.2 xx 10^(23)` molecules |
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| 20. |
At `0^(@)C` and one atm pressure, a gas occupies 100 cc. If the pressure is increased to one and a half-time and temprature is increased by one-third of absolute temperature, then final volume of the gas will be:A. 80 ccB. 88.9 ccC. 66.7 ccD. 100 cc |
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Answer» Correct Answer - B `V_(2)=(P_(1)V_(1)T_(2))/(P_(2)T_(1))impliesP_(1)=P, T_(1)=273K` `P_(2) = (3)/(2)P, T_(2)= T_(1)+(T_(1))/(3)=(4)/(3) xx 273K` `V_(2)=(2P)/(3P) xx(4)/(3) (273)/(273)xx100 c c=(800)/(9) c c = 88.888 c c` |
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| 21. |
A gas deviates from ideal behavious at a high pressure because its moleculesA. Have kinetic energyB. Are bound by covalent bondC. Attract one anotherD. Show the Tyndall effect |
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Answer» Correct Answer - C Some attraction forces exist between the molecules of real gases. When a molecule approaches the wall of the container it experiences an inward pull as a result of attractive by the neighbouring molecules inside the vessel. Therefore the observed pressure is less than the ideal pressure and hence gas deviates from ideal behaviour at high pressure. |
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| 22. |
Under which of the following sets of conditions is a real gas expected to deviate from ideal behaviour? (I) High pressure, small volume (II) High temperature , low pressure (III) Low temperature, high pressureA. only IB. only IIC. only IIID. I and III both |
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Answer» Correct Answer - d (d) A real gas can behave like an ideal gas at low pressure and high temperature. |
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| 23. |
For a certain gas which deviates a little from ideal behaviour, a plot between P/p vs P was found to be non- linear . The intercept on y-axis will be :A. `(RT)/(M)`B. `(M)/(RT)`C. `(MZ)/(RT)`D. `(R)/("TM")` |
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Answer» Correct Answer - a (a) `P(V-nb)=nRT` `PV-P" n "b=nRT " "d=("Mass")/("Volume")` `PV=nRT+P" n "b" " n=("Mass")/(M)` `P("Mass")/(d)=nRT+P" n "b ` `(P)/(d)=(n)/("Mass")RT+(P" n "b)/("Mass")` `(P)/(d)=(RT)/(M)+(Pb)/(M)` |
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| 24. |
Calculate the volume occupied by 16gram `O_(2)` at `300K` and `8.31 Mpa` if `(P_(C)V_(C))/(RT_(C))=3//8` and `(P_(r)V_(r))/(T_(r))=2.21` ( Given `:R=8.314 MPa//K-mol )`A. `120.31mL`B. `124.31mL`C. `248.62mL`D. none of these |
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Answer» Correct Answer - 2 `(PV_(m))/(RT)=(3)?(8)xx2.21," "V_(m)=((3)/(8)xx2.21)xx(RT)/(P),` `V_(m)=(3)/(8)xx2.21xx(8.314xx300)/(8.314)=124.31mL` |
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| 25. |
Calculate the volume occupied by `2.0` mole of `N_(2) at 200K` and `8.21 atm` pressure, if `(P_(C)V_(C))/(RT_(C)) = (3)/(8)` and `(P_(r)V_(r))/(T_(r)) = 2.4`. |
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Answer» Correct Answer - `3.6 L` `(P_(r)V_(r))/(T_(r)) =(P)/(P_(c)) (V_(m))/(V_(c))(T_(c))/(T) = (PV_(m))/(RT) .(RT_(c))/(P_(c)V_(c))` `Z = (PV_(m))/(RT) = 2.4 xx ((3)/(8))` so `V_(m,real) = (0.0821 xx 200)/(8.21) xx0.9 = 1.8 L` so volume of two moles `= 3.6L`. |
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| 26. |
Sample of different gases are given at different conditions in column-I and column-II consisting of translational kinetic energy of these gases at given conditions. `{:(,"Column-I",,"Column-II"),("(P)",2mole SO_(2)(g)at 700K,,"(1)Maximum K.E. per gram"),("(Q)",1 mole SO_(3)(g)at 400K,,"(2)Maximum total K.E."),("(R)",4 mole CH_(4)(g)at 300K,,"(3)Maximum K.E. per gram"),("(S)",2.5 mole He(g)at 450 K,,"(4)Minimum K.E. per molecule"):}`A. `{:("P Q R S"),("4 3 1 2"):}`B. `{:("P Q R S"),("2 3 1 4"):}`C. `{:("P Q R S"),("1 2 3 4"):}`D. `{:("P Q R S"),("2 3 4 1"):}` |
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Answer» Correct Answer - (d) |
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| 27. |
STATEMENT-1 : The average translational kinetic energy per molecule of the gas per degree of freedom is 1/2 KT. STATEMENT-2 : For every molecule there are three rotational degree of freedom.A. If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
| Answer» Correct Answer - C | |
| 28. |
The translational kinetic energy of `10^(20)` molecules of nitrogen at a certain temperature is 0.629J. What is the temperature in `.^(@)C`?A. `43.3^(@)C`B. `23^(@)C`C. `30.^(@)C`D. `15.8^(@)C` |
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Answer» Correct Answer - c |
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| 29. |
Volume occupied by an ideal gas at one atmospheric pressure and `0^@C` is V ml. Its volume at 273 K will beA. V mlB. `V//2` mlC. 2VD. None of these |
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Answer» Correct Answer - A `0^(@)C` is equivalent to 273K i.e., conditions are same so volume will be V ml. |
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| 30. |
In kinetic theory of gases, only translational motion of molecules is considered because:A. there is no intermolecular forces.B. the molecules are considred rigid spheres of negligible volume.C. different molecules may travel at different speedsD. in normal conditions, rotational and vibrational motion is not observed in gas molecules. |
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Answer» Correct Answer - b |
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| 31. |
The densities of hydrogen and oxygen are 0.09 and 1.44 g `L^(-1)`. If the rate of diffusion of hydrogen is 1 then that of oxygen in the same units will beA. 4B. `1//4`C. 16D. `1//16` |
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Answer» Correct Answer - B `r_(O)=r_(H)sqrt((d_(H))/(d_(O)))=1 sqrt((0.09)/(1.44))=sqrt((1)/(16))=(1)/(4)` |
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| 32. |
Statement-1: The diffusion rate of oxygen is smaller than that of nitrogen under same conditions of T and P. Statement-2: Molecular mass of nitrogen is smaller than that of oxygen.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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| 33. |
What is the ratio of diffusion rate of oxygen to hydrogen?A. `1:4`B. `4:1`C. `1:8`D. `8:1` |
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Answer» Correct Answer - A `M_(1)=32g ` for `O_(2),M_(2)=2g `for `H_(2)` `(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1))), (r_(1))/(r_(2))=sqrt((2)/(32))=sqrt((1)/(16))=(1)/(4)` |
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| 34. |
At 1 atm and 273 K the density of gas, whose molecular weight is 45, is:A. `44.8 gm // `litreB. 11.4 gm`//` litreC. 2 gm `//` litreD. 3 gm `//` litre |
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Answer» Correct Answer - C The density of gas `=("Molecular wt. of Metal")/("Volume")=(45)/(22.4)=2 gmlitre^(-1)` |
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| 35. |
Two flasks A and B have equal volumes. A is maintained at 300 K and B at 600 K. A contains `H_(2)` gas, B has an equal mass of `CO_(2)` gas. Find the ratio of total K.E. of gases in flask A to that of B.A. `1:2`B. `11:1`C. `33:2`D. `55:7` |
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Answer» Correct Answer - b (b) `KE_(T)=(3)/(2)nRT=(3)/(2)(W)/(M)RT` `KEprop(T)/(M)` `(KE_(1))/(KE_(2))=(T_(1))/(M_(1))(M_(2))/(T_(2))=(300)/(2)xx(44)/(600)=(11)/(1)` |
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| 36. |
When a jar containing gaseous mixture of equal volumes of `CO_(2)` and `H_(2)` is placed in a solution of sodium hydroxide, the solution level willA. RiseB. FallC. Remain constantD. Become zero |
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Answer» Correct Answer - A Solution level will rise , due to absorption of `CO_(2)` by sodium hydroxide. `2NaOH +CO_(2) rarr Na_(2)CO_(3)+H_(2)O` |
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| 37. |
A good vacuum produced in common lab apparatus corresponds to `10^(-6)` torr at `25^(@)C`. Calculate number of molecules per cubic centimeter at this T and P. In scientific notation, `x xx 10^(y)`. Find the value of `y`. |
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Answer» Correct Answer - 10 |
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| 38. |
A jar contains a gas and a few drops of water at `TK` The pressure in the jar is `830 mm` of Hg The temperature of the jar is reduced by `1%` The vapour pressure of water at two temperatures are 30 and 25 mm og Hg Calculate the new pressure in the jar . |
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Answer» At `T K,P_(gas) =P_("dry gas") + P_("moisture")` `:. P_("dry gas") = 830 -30 =800 mm` Now at new temperature `T_(1) =T - (T)/(100) = 0.99T` Since `V_(1) =V_(2), (P)/(T) =const` or `(P_(1))/(T_(1)) = (P_(2))/(T_(2))` `:. P_("dry gas") = (800 xx 0.99T)/(T) = 792 mm` `:. P_(gas) = P_(dry gas) + P_(moisture)` `= 792 + 25 =817 mm pf Hg` . |
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| 39. |
A vacuum pump has a cylinder of volume upsilon and is connected to a vessel of volume `V` to pump out air from the vessel The initial pressure of gas in vessel is `P` Show that after n strockes the pressure in vessel is reduced to `P_(n) = P[(V)/(V+v)]^(n)` . |
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Answer» Let pressure of gas left in vessle after I operation be `p_(1)` Let `n_(1)` moles are removed from vessel after I operation Let n mole were present initially Thus initial state `PV =nRT` `:. P_(1)V =(n-n_(1))RT` `P_(1)V =nRT -n_(1)RT` `P_(1) V =PV -n_(1)RT` The `n_(1)` mole taken out has volume v at pressure `P_(10` Thus By eps (3) and (4) `P_(1) V =PV-P_(1)upsilon` or `P_(1) = P[(V)/(V +upsilon)]..(5)` Similarly for II operation `P_(2) =P_(1) [(V)/(V +upsilon)]=P[(V)/(V+upsilon)]^(2)` Thus for n operation `P_(n) =PP_(2) =P_(1) [(V)/(V +upsilon)]^(n)` . |
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| 40. |
The density of steam at `100^(@)C` and `1 xx10^(5)` Pa is `0.6 kgm^(-3)` The compressibility factor for steam is .A. `0.867`B. `1`C. `0.967`D. `0.767` |
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Answer» `Z =(PV)/(nRT) = (PVm)/(wRT) = (Pxxm)/(rhoxxRT)` `= (1 xx 10^(5)xx18xx10^(-3))/(0.6 xx 8.314 xx373)=0.967` . |
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| 41. |
At identical temperature and pressure the rate of diffusion of hydrogen gas is `3sqrt3` times that of a hydrocarbon having molecular formula `C_(n)H_(2n-n)` What is the value of n ? .A. 1B. 4C. 3D. 8 |
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Answer» Correct Answer - b (b) `(r_(H_(2)))/(r_("H.C"))=sqrt((M)/(2))=3sqrt(3)` `(M)/(2)=9xx3` `M=54` `n=4` |
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| 42. |
Ar and He are both gases at room temperture. How do the average moleculr velocities (V) of their atoms compare at this temperature?A. `V_(He)=10V_(Ar)`B. `V_(Ar)=10V_(He)`C. `V_(He)=3V_(Ar)`D. `V_(Ar)=3V_(He)` |
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Answer» Correct Answer - c |
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| 43. |
An ideal gas at 650 Torr occupies a bulb of unknown volune. A certain amount of gas is withdrawn and found to occupy `1.52 cm^(3)` at one atm. The pressure of the gas remaining in the bulb is 600 Torr. Calculate volume of the bulb (in mL) taking temperature constant. Give answer excluding decimal places. |
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Answer» Correct Answer - 23 |
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| 44. |
When an ideal diatomic gas is heated at constant pressure the fraction of the heat energy supplied which increases the internal energy of the gas is .A. `(2)/(5)`B. `(3)/(5)`C. `(3)/(7)`D. `(5)/(7)` |
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Answer» `C_(p) =C_(v) + w, w =R` and `C_(v) = (3)/(2) R +R = (5)/(2) R` for diatomic gas Thus `(5)/(2) R` factor of `C_(p)((7)/(2)R)` is used in increasing internal energy or heat supplied to increase internal energy of gas at constnat `P` is `(5//2R)/(7//2R)= (5)/(7) .` . |
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| 45. |
A gas in an open container is treated from` 27^(@)C` to `127^(@)C` . The fraction of original amount of gas remaining in the container will beA. `(3)/(4)`B. `(1)/(4)`C. `(1)/(2)`D. `(1)/(8)` |
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Answer» Correct Answer - 1 On heating a gas in a open container (P and V constant ) moles of gas go out Thus `n_(1)T_(1)=n_(2)T_(2)` or`" "n_(1)xx300=n_(2)xx400` `n_(2)=(3n_(1))/(4)` `:. " "` Fraction of moles & gas left `=(n_(2))/(n_(1))=(3n_(2))/(4n_(1))=(3)/(4)` |
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| 46. |
The volume of 1 litre is equal toA. `1000 cm^(3)`B. `100cm^(3)`C. `10 dm^(3)`D. `10^(6) cm^(3)` |
| Answer» Correct Answer - D | |
| 47. |
Pressure remaining the same, the volume of a given mass of an ideal gas increases for every degree centigrade rise in temperature by define fraction of its volume atA. `0^(@)`CB. Its critical temperatureC. Absolute zeroD. Its Boyle temperature |
| Answer» Correct Answer - A | |
| 48. |
Assertion `:` 4.58 mm and `0.0098^(@)C` is known to be triple point of water. Reason `:` At this pressure and temperature all the three states i.e., water , ice and vapour exist simulataneously.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 |
| Answer» Correct Answer - A | |
| 49. |
Under critical states of a gas for one mole of a gas, compressibility factor is :A. `3//8`B. `8//3`C. `1`D. `1//4` |
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Answer» Correct Answer - A For `1mol` of gas `Z = (P_(C)V_(C))/(RT_(C))` (Under critical condition) But, `P_(C) = (a)/(27b^(2)) , V_(C) = 3b, T_(C) = (8a)/(27Rb)` `:. Z = ((a)/(27b^(2))) xx (2b)/(R ) xx (27Rb)/(8a) = (3)/(8)` Hence, (A)`. |
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| 50. |
Select incorrect statement for real gases :A. In low pressure region repulsive forces dominatesB. Volume of gas particles is not negligible in low pressure regionC. Gases behaves as an ideal gas at low pressure and low temperatureD. In high pressure region attractive forces dominates |
| Answer» Correct Answer - a,b,c,d | |