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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. |
The gases obey the different gas laws only theoretically. Practically all of them show some deviation from these laws. These are called real gases. The deviation are maximum under high pressyre and at low temperature. These are comparatively small when the conditions are reversed. It has been found that the easily liquefiable gases show more deviations from the ideal gas beheviour as compared to the gases which are liqufied with diffculty. Gas deviates from ideal gas beheviour because moleculesA. are coloulessB. attract each otherC. contain covalent bondD. show Brownian Movement. |
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Answer» Correct Answer - B is the correct answer. |
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| 2. |
The gases obey the different gas laws only theoretically. Practically all of them show some deviation from these laws. These are called real gases. The deviation are maximum under high pressyre and at low temperature. These are comparatively small when the conditions are reversed. It has been found that the easily liquefiable gases show more deviations from the ideal gas beheviour as compared to the gases which are liqufied with diffculty. An ideal gas is one which obeys the gas laws underA. a few selected experimental conditionsB. all experimental conditionsC. low pressure aloneD. high pressure alone. |
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Answer» Correct Answer - B is the correct answer. |
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| 3. |
Match the following gas laws with the equation representing them. |
| Answer» Correct Answer - A::B::C::D | |
| 4. |
The density of mercury is `13.6 g mL^(-1)`. Calculate the approximate diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom. |
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Answer» Let the diameter of one mercury atom be `a`. Volume of mercury molecule `=a^(3)` (volume of cube) Volume of `N` mercury molecules `=a^(3)xxN` Density `=(Mass)/(Volume)=(200)/(a^(3)xxN)=13.6` `a^(3)=(200)/(13.6xxN)=(200)/(13.6xx6.023xx10^(23))` or `a=3sqrt((200)/(13.6xx6.023xx10^(3)))` `=2.9004xx10^(-8)cm` |
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| 5. |
The behaviour of matter in different state is governed by various physical law. According to you, what are the factors that determine the state of matter ? |
| Answer» The state of a matter is govered by four factors. Three are temperature, pressure, volume and mass. | |
| 6. |
When is deviation more in the behaviour of a gas from the ideal gas equation `PV=nRT`?A. At high temperature and low pressure.B. At low temperature and high pressure.C. At high temperature and high pressure.D. At low temperature and low pressure. |
| Answer» At low temperature and high pressure, gases deviate more from ideal condition. | |
| 7. |
The behaviour of ideal gas is goverened by various gas laws which are described by mathematical statements as given below: (`i`) `PV=k` (constant) at constant `n` and `T` (`ii`) `V//T=k_(2)` (constant) at constant `n` and `P` (`iii`) `V//n=k_(3)` (constant) at constant `T` and `P` (`iv`) `PV=nRT` (`v`) `P//T=k_(4) (constant) at constant `n` and `V` Answer the following The value of `k_(2)` isA. Independent of nature and amount of gasB. Depends on temperature and pressure conditionsC. Depends on pressure and amount of gasD. Depends only on nature of gas |
| Answer» Depends only on `T` and `P` | |
| 8. |
The behaviour of ideal gas is goverened by various gas laws which are described by mathematical statements as given below: (`i`) `PV=k` (constant) at constant `n` and `T` (`ii`) `V//T=k_(2)` (constant) at constant `n` and `P` (`iii`) `V//n=k_(3)` (constant) at constant `T` and `P` (`iv`) `PV=nRT` (`v`) `P//T=k_(4)` (constant) at constant `n` and `V` Answer the following A cylinder of `10 L` capacity at `300 K` containing the gas is used to fill balloons till finally the cylinder recorded a pressure of `10 m` bar. The number of `He` atoms still present in the cylinder isA. `4.82xx10^(21)`B. `2.41xx10^(23)`C. `2.41xx10^(21)`D. `4.82xx10^(23)` |
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Answer» `n=(PV)/(RT)=(10^(-2)xx10)/(0.083xx300)` `N=nxxN_(A)=(10^(-2)xx10xx6.023xx10^(23))/(0.083xx300)=2.41xx10^(21)` |
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| 9. |
Oxygen is present in a `1 L` flask at a pressure of `7.6xx10^(-10)mm Hg`. Calculate the number of oxygen molecules in the flask at `0^(@)C`.A. `2.7xx10^(9) "molecules"`B. `2.7xx10^(10) "molecules"`C. `2.7xx10^(11) "molecules"`D. `2.7 xx10^(12) "molecules"` |
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Answer» Correct Answer - B `PV=nRT` `n=(PV)/(RT)implies (7.6xx10^(-10))/(760)xx(1)/(0.82xx273)` `nimplies 0.0446xx10^(-12)` |
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| 10. |
Consider the adjacent diagram. Initially, flask `A` contained oxygen gas at `27^(@)C` and `950 mm` of `Hg`, and flask `B` contained neon gas at `27^(@)C` and `900 mm`. Finally, two flask were joined by means of a narrow tube of negligible volume equipped with a stopcock and gases were allowed to mixup freely. The final pressure in the combined system was found to be `910 mm` of `Hg`. What is the correct relationship between volumes of the two flasks?A. `V_(B)=2V_(A)`B. `V_(B)=4V_(A)`C. `V_(B)=5V_(A)`D. `V_(B)=5.5V_(A)` |
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Answer» Given, Initial condition `P_(A)=950mm`, `V_(A)=?`, `T_(A)=300 K` `P_(B)=900mm`, `V_(B)=?`, `T_(A)=300 K` `:. n_(total)=(PV)/(RT)=(950xxV_(A))/(Rxx300)+(900xxV_(B))/(Rxx300)` Final condition `P_(A)=910`, `V_(A)=? T_(A)=300` `P_(B)=910`, `V_(B)=? T_(B)=300` `:. n_(total)=(PV)/(RT)=(910xxV_(A))/(Rxx300)+(900xxV_(B))/(Rxx300)` Since `n_(total)=` constant `:. (950xxV_(A))/(Rxx300)+(900xxV_(B))/(Rxx300)=(910xxV_(A))/(Rxx300)+(910xxV_(B))/(Rxx300)` Therefore, `V_(B)=4V_(A)` |
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| 11. |
Oxygen is present in a `1 L` flask at a pressure of `7.6xx10^(-10)mm Hg`. Calculate the number of oxygen molecules in the flask at `0^(@)C`. |
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Answer» Correct Answer - `2.7xx10^(10)gmol^(-1)` |
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| 12. |
The atmospheric pressure on a certain day is `732 mm Hg`. What is the pressure in `kPa`?A. `60 kPa`B. `80.6 kPa`C. `77.8 kPa`D. `97.6 kPa` |
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Answer» Correct Answer - D The relation between atmospheres and pascals is 1 atm = 101325 Pa `= 1.01325 xx 10^(5) Pa` Since `1000 Pa = 1 Pa ("kilopascal")` 1 atm = `1.01325 xx 10^(2) kPa` Thus, the atmospheric pressure in `kPa` is Pressure = `(732 mm Hg) xx ((1 "atm")/(760 mm Hg))xx ((1.01325 xx 10^(2) kPa)/(1 "atm"))` ` = 97.6 kPa` |
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| 13. |
Which of the temperature scales has merged as a result of the study on gases?A. Centigrade scaleB. Fahrenheit scaleC. Kelvin scaleD. Celsius scale |
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Answer» Correct Answer - C The kelvin scale of temperature emerged during the effect of temperature on the volume of gases. |
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| 14. |
A helium atom is two times heavier than a hydrogen molecule. At `298 K`, the average kinetic energy of a helium atom isA. two times that of hydrogen moleculeB. Same as that of the hydrogen moleculeC. four times that of a hydrogen moleculeD. half that of a hydrogen molecule |
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Answer» Correct Answer - B The average kinetic enegy of an atom is given as `(3)/(2)kT`. `:.` It does not depend on mass of the atom. |
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| 15. |
A mixture of gases in a cyliner at `760 mm` pressure contains `65 %` nitrogen, `15%` oxygen, and `20%` carbon dioxide by volume. What is the partial pressure of each gas in `mm`? |
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Answer» `P_(N_(2))=(760xx65)/(100)=494 mm` `P_(O_(2))=(760xx15)/(100)=114mm` `P_(CO_(2))=(760xx20)/(100)=152mm` |
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| 16. |
A mixture of hydrogen and oxygen at one bar pressue contains `20%` by weight of hydrogen . Partial pressure of hydrogen will beA. `0.2 ba r`B. `0.4 ba r`C. `0.6 ba r`D. `0.8 ba r` |
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Answer» Correct Answer - D Weight of `H_(2)= 20 g` in `100 g` mixture, Weight of `O_(2)= 80g` `:.` Moles of `O_(2)= (80)/(32)=(5)/(2)` `:.` Total moles `=10+(5)/(2)=(25)/(2)` `:. P_(H_(2))= P_(T)xx` mole fraction of `H_(2)` `=1xx(10)/(25//2)= 0.8 "bar"` |
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| 17. |
A mixture of hydrogen and oxygen at `1` bar pressure contains `20%` of hydrogen by weight. Calculate the partial pressure of hydrogen. |
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Answer» Correct Answer - 0.8 bar Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g. Then, the number of moles of dihydrogen, `n_(H_2)=(20)/(2)=10` moles and the number of moles of dioxygen, `n_(O_2)=(80)/(32)=2.5` moles Given, Total pressure os the mixture, `p_("total")` = 1 bar Then partial pressure of dihydrogen, `P_(H_2)=(n_(H_2))/(n_(H_2)+n_(O_2))xxP_("total")` `=(10)/(10+2.5)xx1` = 0.8 bar Hence, the partial pressure of dihydrogen is 0.8 bar. |
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| 18. |
What is the coordination number of sodium in sodium oxide (`Na_2O`)?A. 6B. 4C. 8D. 2 |
| Answer» Correct Answer - c | |
| 19. |
The oxide `Tl_n Ca_2 Ba_2 Cu_3 O_10` is found to be superconductor at 125 K. The value of n is |
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Answer» Correct Answer - 2 As given molecule is neutral , sum of oxidation states =0 . Hence, n(+3)+2(+2)+2(+2)+3(+2)+10(-2)=0 or 3n +4 +4 +6-20=0 or 3 n =6 or n=2 |
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| 20. |
What will be the happen to the volume of a bubble of air found under water in a lake where the temperature is `15^(@)C` and the pressure is 1.5 atm, if the bubble then rises to the surface where the temperature is `25^(@)C` and pressure is 1.0 atm ?A. Its volume will become greater by a factor of 2.5B. Its volume will become greater by a factor of 1.6C. Its volume will become greater by a factor of 1.1D. Its volume will become smaller by a factor of 0.70 |
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Answer» Correct Answer - B `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `(1.5"atm"xxV)/(288" K")=(1"atm"xxV_(2))/(298" K"):. V_(2)=1.6" V"` |
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| 21. |
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 the radius when it comes to the surface of a lake will be (Assume temperature and pressure at the surface to be, respectively, `2T K` and `P//4`.)A. `100%`B. `50%`C. `40%`D. `200%` |
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Answer» Correct Answer - A `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2)) , V_(1) =` initial volume, `V_(2) =` final volume `(P V_(1))/(T) = (P)/(4) xx (V_(2))/(2T)` `V_(1) = (V_(2))/(8)` `V_(2) = 8V_(1)` `V_(2) = 8 xx (4)/(3) pi ^(3) = (4)/(3) pi (8r)^(3)` New radius `= 2r` ( `:.` 100% radius will increase) |
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| 22. |
Under the influence of electric field , which of the following statements is true about the movement of electrons and holes in a p-type semiconductor ?A. Electron will move towards are positively charged plate through electron holesB. Holes will appear to be moving towards the negatively charged plateC. Both electrons and holes appear to move towards the positively charged plateD. Movement of electrons is not related to the movement of holes |
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Answer» Correct Answer - a,b (c ) is wrong because electrons move towards anode while holes move towards cathode. (d) is also not true for the same reason |
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| 23. |
Which of the following statements are true about metals ?A. Valence band overlaps with conduction bandB. The gap between valence band and conduction band is negligibleC. The gap between valence band and conduction band cannot de determinedD. Valence band may remain partially filled . |
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Answer» Correct Answer - a,b,d Only (c ) is true |
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| 24. |
Boiling point is the temperature at which the vapour pressure becomes equal to.......... |
| Answer» Atmospheric pressure | |
| 25. |
Assertion: `C_(P)-C_(V)=R` for an ideal gas. Reason: `((delE)/(delV))_(T)=0` for an ideal gas.A. If both (`A`) and (`R`) are correct and (`R`) is the correct explanation of (`A`).B. If both (`A`) and (`R`) are correct, but (`R`) is not the correct explanation of (`A`).C. If (`A`) is correct, but (`R`) is incorrect.D. If (`A`) is incorrect, but (`R`) is correct. |
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Answer» Correct Answer - A::B::C::D Both are correct, but reason is not the correct explanation of assertion. `C_(P)-C_(V)=R` or `((delH)/(delT))_(P)-((delU)/(delT))_(V)=R` |
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| 26. |
Why is FeO(s) not formed in stoichiometric composition ? |
| Answer» This is because some `Fe^(2+)` ions in FeO are replaced by `Fe^(3+)` ions. As `3 Fe^(2+)` ions are replaced by `2 Fe^(3+)` ions to maintain electrical neutrality , there is less amount of metal as compared to stoichiometric proportion. | |
| 27. |
CsCl has cubic structure. Its density is `3.99 g cm^(-3)`. What is the distance between `Cs^+` and `Cl^-` ions ? (At. Mass of Cs =133) |
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Answer» CsCl has BCC structure. It has one formula unit in the unit cell. So Z=1 `rho=(ZxxM)/(a^3xxN_0)` or `a^3=(ZxxM)/(rhoxxN_0)=(1xx(133+35.5) "g mol"^(-1))/(3.99 g cm^(-3)xx6.02xx10^23 mol^(-1))=70.15xx10^(-24) cm^3` `a=(70.15)^(1//3)xx10^(-8) cm^(3) =(70.15)^(1//3)xx10^2` pm (1pm=`10^(-10)` cm) For BCC structure, Interionic distance (d)`=(sqrt3a)2=1.732/2xx412.4=357` pm ltbr gt [To solve `(70.15)^(1//3)`, put `x=(70.15)^(1//3)`. Then log x=`1/3 log 70.15 =1/3xx1.8460=0.6153` x=Antilog 0.6153=4.124] |
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| 28. |
Under critical conditions, the compressibility factor for a gas is .A. `3//8`B. `8//3`C. `1`D. `1//4` |
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Answer» Correct Answer - A `P(c) V_(c) = (3)/(8) RT_(c)` `Z = (P_(c) V_(c))/(RT_(c)) = (3)/(8)` |
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| 29. |
Assertion: `SO_(2)` gas is easily liquefied while `H_(2)` is not. Reason: `SO_(2)` has low critical temperature while`H_(2)` has high critical temperature.A. If both (`A`) and (`R`) are correct and (`R`) is the correct explanation of (`A`).B. If both (`A`) and (`R`) are correct, but (`R`) is not the correct explanation of (`A`).C. If (`A`) is correct, but (`R`) is incorrect.D. If (`A`) is incorrect, but (`R`) is correct. |
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Answer» Correct Answer - A::B::C::D `H_(2)` has weak intermolecular attraction. Hence, `H_(2)` is not easily liquified. |
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| 30. |
`NH_(3)` gas is liquefied more easily than `N_(2)`. HenceA. van der Waals constant `a` and `b` of `NH_(3)gt` that of `N_(2)`B. van der Waals constant `a` and `b` of `NH_(3)lt` that of `N_(2)`C. `a(NH_(3))gta(N_(2))` but `b(NH_(3))ltb(N_(2))`D. `a(NH_(3))lta(N_(2))` but `b(NH_(3))gtb(N_(2))` |
| Answer» Gases which can be liquefied easily have high value of `a` and low value of `b`. So the answer is (`c`). | |
| 31. |
The compressibility factor (Z) for a gas is less than one.What does it signify ? |
| Answer» This indicates tht the gas can be compressed more as compared to an ideal gas. | |
| 32. |
`XmL` of `H_(2)` gas effuses through a hole in a container is 5 second. The time taken for the effusion of the same volume of the gas specified below under identical conditions is .A. `10` seconds: `He`B. `20` seconds: `O_(2)`C. `25` seconds: `CO`D. `35` seconds: `CO_(2)` |
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Answer» Correct Answer - B `(r_(H_(2)))/(r_(H_(e)))=sqrt((4)/(2))=sqrt(2)` `:.` (a) in incorrect `(r_(H_(2)))/(r_(O_(2)))=sqrt((32)/(2))=4` (b) is correct `(r_(H_(2)))/(r_(C_(O)))=sqrt((28)/(2))=sqrt(14)` (c) is correct `(r_(H_(2)))/(r_(Co_(2)))=sqrt((2)/(44))=sqrt((1)/(12))` (d) is correct. |
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| 33. |
`NH_(3)` is liquefied more easily than `N_(2)`. HenceA. a and b o f `NH_(3) gt` that of `N_(2)`B. `a(NH_(3)) gt a(N_(2))` but `b(NH_(3))lt b(N_(2))`C. `a(NH_(3)) lt a(N_(2))` but `b(NH_(3)) gt b(N_(2))`D. none |
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Answer» Correct Answer - B More is the value of a mass, greater is the liquefaction and bigger is the size of molecule bigger is the `b` value |
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| 34. |
A vessel is filled with a mixture of oxygen and nitrogen. At what ratio of partial pressures will the mass of gases be identical?A. `P(O_(2))= 0.785P(N_(2))`B. `P(O_(2))= 8.75 P(N_(2))`C. `P(O_(2))= 11.4P(N_(2))`D. `P(O_(2))= 0.875P(N_(2))` |
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Answer» Correct Answer - D `PV= nRT` `PV=(w)/(M)RT` `P_(O_(2))V=(w)/(32)RT " " P_(N_(2))V=(w)/(28)RT` `(P_(O_(2)))/(P_(N_(2)))=(28)/(32) " "P_(O_(2))= 0.875 P_(N_(2))` |
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| 35. |
A vessel is filled with a mixture of oxygen and nitrogen. At what ratio of partial pressures will the mass of gases be identical?A. `P(O_(2))=0.785P(N_(2))`B. `P(O_(2))=8.75P(N_(2))`C. `P(O_(2))=11.4P(N_(2))`D. `P(O_(2))=0.875P(N_(2))` |
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Answer» `PV=nRT` `PV=(w)/(m)RT` `P_(O_(2))V=(w)/(32)RT` ….(`i`) `P_(O_(2))V=(w)/(28)RT` ….(`ii`) `(P_(O_(2)))/(P_(N_(2)))=(28)/(32)` `P_(O_(2))=0.875 P_(N_(2))` |
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| 36. |
Two bulbs `A` and `B` of equal capacity are filled with `He` and `SO_(2)`, respectively, at the same temperature. (a) If the pressures in the two bulbs are same, what will be the ratio of the velocities of the molecules of the two gases? (b)At what temperature will the velocity of `SO_(2)` molecules become half of the velocity of `He` molecules at `27^(@)C`? (c) How will the velocities be affected if the volume of `B` becomes four times that of `A`? (d) How will the velocities be affected if half of the molecules of `SO_(2)` are removed from `B`? |
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Answer» Let the velocities of `He` and `SO_(2)` be `u_(1)` and `u_(2)`. (`a`) `u_(1)=sqrt((3RT)/(M_(He))), u_(2)=sqrt((3RT)/(M_(SO_(2))))` `(u_(1))/(u_(2))=sqrt(M_(SO_(2))/(M_(He)))=sqrt((64)/(4))=4` (`b`) Let the velocity of `He` be `u_(1)` Velcotiy of `SO_(2)=(u_(1))/(2)` `u_(1)=sqrt((3RT_(1))/(M_(He))), (u_(1))/(2)=sqrt((3RT_(2))/(M_(SO_(2)))` `sqrt((3RT_(1))/(M_(He)))=2xxsqrt((3RT_(2))/(M_(SO_(2)))` `sqrt((T_(1))/M_(He))=2xxsqrt((T_(2))/(M_(SO_(2))))` `(T_(1))/M_(He)=4xxs(T_(2))/(M_(SO_(2)))` `T_(2)=(300xx64)/(4xx4)=1200 K` Temperature `=1200-273=927^(@)C` (`c`) Velocity given by the realtion `u=sqrt((3PV)/(M))`. When volume increases by four times, pressure becomes `1//4` and `PV` remains constant. So there will be no change in the velocities. (`d`) Since volume do not depend on the number of molecules, there will be no change in velocities. |
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| 37. |
Calculate the root mean square velocity of nitrogen at `27^(@)C` and `70cm` pressure. The density of `Hg` is `13.6 g cm^(-3)`. |
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Answer» Volume `V_(2)` at `27^(@)C` and `70 cm` pressure is given by `(P_(2)V_(2))/(T_(2))=(P_(1)V_(1))/(T_(1))` `(70 cmxxV_(2))/(300 K)=(70 cmxx0.0224 cm^(3))/(273 K)` `:. V_(2)=0.026725 cm^(3)` Pressure `P=((70)/(76))xx1.01325xx10^(5) Nm^(-2)` `=9.332xx10^(4)Nm^(-2)` `:. u_(rms)=((3PV)/(M))^(1//2)` `((3xx9.332xx10^(4) Nm^(-2)xx0.026725 m^(3) mol^(-1))/(28xx10^(-3) kg mol^(-1)))^(1//2)` `=517 ms^(-1)` |
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| 38. |
What percent of a sample of nitrogen must be allowed to escape if its temperature, pressure, and volume are to be changed from `220^(@)C`, `3 atm`, and `1.65 L` to `110^(@)C`, `0.7 atm`, and `1 L`, respectively? |
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Answer» Applying the formula, `m` (mass of the gas) `= (PV xx M)/(RT)`, under boht the conditions, Mass of gas before escaping `= (3.0 xx 1.65 xx 28)/(0.0821 xx 493) = 3.42 g` Mass of gas after escaping `= (0.7 xx 1.0 xx 28)/(0.0821 xx 383) = 0.62 g` Percentage of nitrogen allowed to escape `= ((3.42 - 0.62))/(3.42) xx 100 = 81.87` |
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| 39. |
What percent of a sample of nitrogen must be allowed to escape if its temperature, pressure, and volume are to be changed from `220^(@)C`, `3 atm`, and `1.65 L` to `110^(@)C`, `0.7 atm`, and `1 L`, respectively?A. `41.4%`B. `8.18%`C. `4.14%`D. `81.8%` |
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Answer» Let `P_(1)=3 atm`, `T_(1)=220+273=493 K`,`V_(1)=1.65 L` `P_(2)=0.7 atm`, `T_(2)=110+273=383 K`, `V_(2)=1 L` Using the gas `PV=nRT` equation , we get `n_(1)=(P_(1)V_(1))/(RT_(1))` and `n_(2)=(P_(2)V_(2))/(RT_(2))` Thus, fraction remaining is `(n_(2))/(n_(1))=(P_(2)V_(2))/(RT_(2))xx(RT_(1))/(P_(1)V_(1))=(0.7xx1)/(Rxx383)xx(Rxx493)/(3xx1.65)=0.182` Fraction escaped `=1-0.182=0.818` Percentage escaped `=0.818xx100=81.8%` So the correct choice is (`d`). |
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| 40. |
(A) A closed cylinder containing high pressure gas tends to rise against gravity when the gas is allowed to escape through an orifice at the bottom (R) The velocity of escaping gas develops an upward thrust proportional to the area of cross-section of the orificeA. If both A and R are correct and R is the correct explanation of AB. If both A and R are correct but R is not the correct explanation of AC. If A is correct but R is incorrectD. If A is incorrect but R is correct |
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Answer» Correct Answer - C Assertion is correct but reason is wrong as the velocity of escaping gas is inversely proportional ot the area of cross-section of orifice |
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| 41. |
A gas cylinder contains `370 g` oxygen at `30.0 atm` pressure and `25^(@)C`. What mass of oxygen will escape if the cylinder is first heated to `75^(@)C` and then the valve is held open until gas pressure becomes `1.0 atm`, the temperature being maintained at `75^(@)C`? |
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Answer» `n=(370)/(32)=11.6 mol` `V=(nRT)/(P)=(11.6xx0.082xx298)/(30.0 atm)=9.432 L` The final number of moles `n=(PV)/(RT)=(1.0atmxx9.43)/(0.082xx348)=0.330 mL` Final weight of `O_(2)=0.330 molxx32=10.6 g` Mass of `O_(2)` escaped `=370 g(initial)-10.6 g` `=359 g` |
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| 42. |
A refrigeration tank holding `5.00 L` feron gas `(C_(2)Cl_(2)F_(4))` at `25^(@)C` and `3.00 atm` pressure developed a leak. When the leak was discovered and repaired, the tank has lost `76.0 g` of the gas. What was the pressure of the gas remaining in the tank at `25^(@)C`? |
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Answer» Original gas pressure `n=(PV)/(RT)=(3.00atmxx5.00mL)/(0.082xx348)` `=0.613 mol=0.613xx171 g` `=105 g` originally present Quantity remaining `105-76=29 g=(29)/(171)=0.17 mol` `P=(nRT)/(V)=(0.17xx0.082xx298)/(5.00 L)=0.83 atm` |
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| 43. |
Which of the following property of water can be used to explain the spherical shape of rain droplets ?A. viscosityB. surface tensionC. critical phenomenaD. pressure |
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Answer» Correct Answer - B Surface tension tires to make the surface area minimum for a volume. This is so for spherical shape. |
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| 44. |
A person living in Shimla observed that cooking food without using presssure cooker takes more time. The reason for this observation is that at high altitude :A. pressure increasesB. temperature decreasesC. pressure decreasesD. temperature increases |
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Answer» Correct Answer - C At high altitude, pressure is low. Hence, boiling takes place at lower temperature and, therefore, cooking takes more time. In a pressure cooker, pressure cooker, pressure is high and hence noiling point increases. |
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| 45. |
Assertion: Helium shows only positive deviations from ideal behaviour. Reason: Helium is an inert gas.A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct and reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If assertion and reason are both incorreect. |
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Answer» Correct Answer - C Correct reason : Helium atoms being small in size, the intermolecular forces of attraction can be naglected. |
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| 46. |
Critical temperature for `CO_(2)` and `CH_(4)` are `31.1^(@)C` and `-81.9^(@)C` respectively. Which of these has stronger intermolecular forces and why ? |
| Answer» Higher the critical temperature, more easily the gas can be liquefied, i.e., greater are the intermolecular forces of attraction. Hence, `CO_(2)` has stronger intermolecular forces than `CH_(4)`. | |
| 47. |
Critical density of a gas having molecular mass `39 g mol^(-1)` is `0.1 g cm^(-3)`. Its critical volume in `L mol^(-1)` isA. `0.390`B. `3.90`C. `0.039`D. `39.0` |
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Answer» Correct Answer - A Critical density `= 0.1 g cm^(-3) = 100 gL^(-1)` Critical volume `= ("Molecular mass")/("Critical density")` `= ((39 mol^(-1)))/(100 gL^(-1)) = 0.39 L mol^(-1)` |
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| 48. |
Calculate the total pressure in a 10 L cylinder which contains 0.4 g of helium, cylinder. Assume ideal behaviour of gases.`(R=0.082 L atm K6(-1)mol^(-1))` |
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Answer» Total no. of moles of all gases(n)`=(0.4)/(4)+(1.6)/(32)+(1.4)/(28)=0.2` Volume of the gasesour mixture (V)=10 L Temperature (T)=27+273=300 K, R=0.082 L atm `K^(-1)mol^(-1)` As gases have behaviour, PV=nRT or `P=(nRT)/(V)=(0.2xx0.082xx300)/(10)=0.492 atm` `"Partial pressure of helium "=(0.4//4)/(0.2)xx0.492=0.246 atm` |
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| 49. |
The critical temperature of water is higher than that of `CO_(2)` because the `H_(2)O` molecular hasA. fewer electrons than `O_(2)`B. two covalent bondsC. V-shapeD. diople moment. |
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Answer» Correct Answer - D Attractive forces in `H_(2)O` molecules are more than in `CO_(2)` molecules because of greater polarity (dipole moment). |
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| 50. |
The intermolecular interaction that is dependent on the inverse cube of distance between the molecules isA. London forceB. Hydrogen bondC. lon-ion interactionD. lon-dipole interaction. |
| Answer» Correct Answer - B | |