InterviewSolution
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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. |
In a crude model of a rotating diatomic molecule of chlorine `(Cl_2)`, the two `(C l)` atoms are `2.0 xx 10^-10 m` apart and rotate about their centre of mass with angular speed `omega = 2.0 xx 10^12 "rad" //s`. What is the rotational kinetic energy of one molecule of `C l_(2)` , Which has a molar mass of `70.0 g // mol` ? .A. `2.32xx10^(-20)J`B. `2.32xx10^(-21)J`C. `2.32xx10^(-19)J`D. `2.32xx10^(-22)J` |
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Answer» Correct Answer - B Let =moment of inertia `l=2(mr^(2))` `m=(70xx10^(-3)/(2xx6.02xx10^(23)) = 5.81xx10^(-26)kg` `r=(2xx10^(-10)/(2)=1xx10^(-10)m` `l=1.16xx10^(-45)k//m^(2)` `therefore k=(1)/(2)//omega^(2)` `=(1)/(2)(1.16xx10^(-45))(2xx10^(12))6(2)` `=2.32xx10^(-21)J` |
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| 2. |
Four molecules of ags have speeds 1,2,3 and 4 `km//s`.The volue of the root mean square speed of the gas molecules isA. `(1)/(2)sqrt(15)km//s`B. `(1)/(2)sqrt(10)km//s`C. `2.56 km//s`D. `sqrt(15/2)km//s` |
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Answer» Correct Answer - D `becauseV_(rms)=`sqrt(((V_(1)^(2)+V_(2)^(2)+…+V_(n)^(2))/(n))` `V_(rms)= sqrt((1^(2)+2^(2)+3^(2)+4^(2)/(4))=sqrt(((1+4+9+16)/(4))` `=sqrt((30)/(4))=sqrt((15)/(2))km//s` `therefore` " "`V_(rms)=sqrt((15)/(2))km//s` |
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| 3. |
The average kinetic energy of gas molecule at `27^(@)C` is `6.21xx10^(-21)` J. Its average kinetic energy at `127^(@)C` will beA. `12.2xx10^(-21)J`B. `8.28xx10^(-21)J`C. `10.35xx10^(-21)J`D. `11.35xx10^(-21)J` |
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Answer» Correct Answer - B as, KE `prop`T `rArr E_(127)/E_(27)`=((127+273))/((27+273))` `E_(127) =6.21xx120^(-21)xx(400)/(300)` `=.28xx10^(-21) J` |
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| 4. |
For gas at a temperature `T` the root-mean-square speed `v_(rms)`, the most probable speed `v_(mp)`, and the average speed `v_(av)` obey the relationshipA. `V_(mp)gtV_(av)gtV_(ms)`B. `V_(ms)gtV_(av)gtV_(mp)`C. `V_(av)gtV_(mp)gtV_(ms)`D. `V_(mp)gtV_(ms)gtV_(av)` |
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Answer» Correct Answer - B `V_(rms)`=`sqrt((3kT)/(m)),V_(av)=sqrt((8kT)/(pim))` `V_(mp)=sqrt((2KT)/(m))` `therefore V_(ms)gtV_(av)gtV_(amp)` |
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| 5. |
12 g of gas occupy a volume of `4xx10^(-3) m^(3)` at a temperature of `76^(@)C`.after the gas is heated at constant pressure, its density becomes `6xx10^(-4) g//cm^(3)`.What is the temperature to which the gas was heated?A. 1000 KB. 1400 KC. 1200 KD. 800 K |
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Answer» Correct Answer - B Volume at `7^(@)C = 4xx10^(-3)m^(3)` Let it be heated to x K. `Volume =(12)/(6xx10^(-4))cm^(3)=2xx10^(4)cm^(3)` Since, `V/T=constant` `therefore` " " `(4xx10^(-3))/(280)=(2xx10^(-2))/(x)` `therefore`" " `x=(2xx10^(-2)xx280)/(4xx10^(-3)=1400 K` |
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| 6. |
A vessel contains a mixture of nitrogen of mass 7 g and carbon dioxide of mass 11 g at temperature 290 K and perssure 1 atm.Find the density of the mixture.A. `1.1 g//L`B. `1.2 g//L`C. `1.515 g//L`D. `1.6 g//L` |
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Answer» Correct Answer - C )`P_(1)V=(7)/(28)xxRxx290` `p_(2)V=(11)/(44)xxRxx290` `therefore (P_(1)+P_(2))V=((7)/(28)+(11)/(44))Rxx290` `before P_(1)+P_(2)=1 atm` `V=(((7)/(28)+(11)/(44))xx8.3xx10^(7)xx290)/(1.013xx10^(6))` `therefore Density `= (mass)/(volume)=(18)/(V)=1.515 g//L` |
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| 7. |
One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the mixture of gas (in atm) isA. 1.5B. 2.5C. 2D. 4 |
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Answer» Correct Answer - C Ideal gas equation is given by `pV=nRT` " "..(i) For oxygen ,`P =1atm,V=1L,n=n_(0)` Therefore,Eq (i) becomes `therefore` " "` 1xx1=nO_(2)RT rArr n_(O_(2))=1/RT` and `0.5xx2=n_(N_(2))RT rArr n_(N_(2))=1/RT` For mixture of gas, `P_(mix)V_(mix)=n_(mix)RT` Here, `n_(mix)=n_(O_(2))+n_(N_(2))` `therefore` " "`(P_(mix)V_(mix))/(RT)=1/RT+1/RT` `rArr P_(mix)V_(mix)=2` |
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| 8. |
Two perifect gases at absolute temperatures `T_(1) and T_(2)` are mixed. The absolute temperature of the mixture is T. there is no loss of energy.If `m_(1) and m_(2)` are masses of molecules and `n_(1) and n_(2)` are number of molecules, thenA. `T=(T_(1)+T_(2))/(2)`B. `T=(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))`C. `T=(n_(1)T_(1)+n_(2)T_(2))/(T_(1)+T_(2))`D. None of the above |
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Answer» Correct Answer - B Average KE per molecule of a gas `=(3)/(2)KT` Before mixing two gases, the averge KE of all molecules of 2 gases `=(3)/(2)kn_(1)T_(1)+(3)/(2)kn_(2)T_(2)` after mixing the average KE of both gases `=(3)/(2)k(n_(1)+n_(2))T` where,T is temperature of mixture `therefore` No loss of enegy. `therefore` " " `(3)/(2)k(n_(1)+n_(2))T=(3)/(2)kn_(1)T_(1)+(3)/(2)kn_(2)T_(2)` `T=(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))` |
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| 9. |
How many cylinders of hydrogen at atmospheric pressure are required to fill a ballon whose volume is `500 m6(3)` if hydrogen is stored in cylinders of volume 0.05 `m^(3)` at an absolute presure of `15xx10^(5)` Pa?A. 700B. 675C. 605D. 710 |
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Answer» Correct Answer - B `15xx10^(5)xxV_(2)=500xx1.013xx10^(5)` `V_(2)=33.77 m^(3)` `therefore` Number of cylinders `=33.77/0.05=675` |
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| 10. |
Figure shows graphs of pressure vs density for an ideal gas at two temperature `T_(1) and T_(2)`. Which of the following is correct? A. `T_(1)gtT_(2)`B. `T_(1)=T_(2)`C. `T_(1)ltT_(2)`D. any three is possible |
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Answer» Correct Answer - A According to ideal lgals equaltion, `pV=nRT` `pV=(m)/(M)RT` or `p=(rhoRT)/(M)` or `p=(RT)/(M)rho` Hence, more temperature means more slope. |
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| 11. |
In an ideal gas without preferred direction of motion of molecules,A. `V_(x)=V_(y)=V_(z)`B. `V_(x)^(2)=V_(y)^(2)=V_(z)^(2)`C. `barV_(x)^(2)=barV_(y)^(2)=barV_(z)^(2)`D. None of the above |
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Answer» Correct Answer - C An isotropic gas is one which have same properties throughout and their molecules are all moving random directions, so only average values of square of their velocity components are equal. |
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| 12. |
Given is the graph between `(PV)/T` and P for 1 gm of oxygen gas at two different temperatures `T_(1) and T_(2)` Fig. Given, density of oxygen `= 1.427 kg m^(-3)`. The value of `(PV)//(T)` at the point A and the relation between `T_(1) and T_(2)` are respectively : A. `0.256 JK^(-1) and T_(1)ltT_(2)`B. `8.314 J mol^(-1)K^(-1) and T_(1)ltT_(2)`C. `0.256 JK^(-1) and T_(1)gtT_(2)`D. `4.28 JK^(-1) and T_(1)ltT_(2)` |
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Answer» Correct Answer - C `pV =nRt =(m)/(M)RT` where, m=mass of gas and `(m)/(M) =n =number of moles` `rArr (pV)/(T)=nR=constant for all values of P (ideally it is a straight line) `rArr (pV)/(T)=(1g)/(32g)xx8.31J mol^(-1)K^(-1)=0.256 JK^(-1)` Also `T_(1)gtT_(2)` |
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| 13. |
The figure shown, the p-V diagram of two different masses `m_(1) and m_(2)` drawn at constant temperature T, then ltbr. A. `m_(1)gtm_(2)`B. `m_(2)gtm_(1)`C. `m_(1)=m_(2)`D. insufficient data |
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Answer» Correct Answer - B We know that, `pV =nRT=(m)/(M)RT` `m=(pV)(M)/(RT)` m`prop`pv `p_(2)V_(2)gtP_(1)V_(1)` From the curve, `m_(2)gtm_(1)` |
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| 14. |
Choose the correct relation between the rms speed `(V_(rms))` of the gas molecules and the velocity of sound in that gas `(V_(s))` in identical situations of pressure and temperature.A. `V_(ms)=V_(s)`B. `V_(ms)=sqrt((3/gamma))^(v_(s))`C. `C.V_(rms)=sqrt((gamma/3))^(v_(s))`D. `gammav_(rms)=3V_(s)` |
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Answer» Correct Answer - B `V_(rms)=sqrt(((3RT)/(m)))`and `V_(s)=sqrt(((gammaRT)/(m))` `therefore V_(rms)=sqrt(((3)/(gamma))V_(s)` |
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| 15. |
In case of molecules of an ideal gas, which of the following , average velocities cannot be zero?A. `ltbarVlgt`B. `ltbarV^(3)gt`C. `ltbarV^(4)gt`D. `ltbarV^(5)gt` |
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Answer» Correct Answer - C The average velocity of even power quantity connot be zero |
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| 16. |
From the following V-T diagram, what is true about pressure?" "[2009] A. `P_(1)ltP_(2)`B. `P_(1)gtP_(2)`C. `P_(1)=P_(2)`D. Cannot predict |
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Answer» Correct Answer - A As V-T graph is a straight line with positive siope therefore its equation, may be written as `V=aT+b` From standard gas equation, `p=(RT)/(V)=(RT)/(AT)+b=(R)/(a+b)//T` As, `T_(2)gtT_(1)` `therefore`" "` P_(2)gtP_(1)orPO_(1)ltP_(2)` |
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| 17. |
A vessel containing 1 mole of `O_(2)` gas (molar mass 32) at tempeature T. The pressure of the gas is P.An identical vessel containing onle mole of He gas (molar mass 4) at temperature 2 T has a pressure of [2013]A. `P//8`B. PC. PD. ` 8P` |
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Answer» Correct Answer - C Applying gas equation,`pV=nRT` We can write,`p_(1)V=n_(1)RT_(1)` and `P_(2)V=n_(2)RT_(2)` `rArr p_(2)/p_(1)=n_(2)/nm_(1)xxT_(2)/T_(1)` `=1/1xx(2T)/(T)=2` `rArr P_(2)=2P` |
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| 18. |
Two ballons are filled, one with pure He gas and other by air, repectively. If the pressure and temperature of these ballons are same then the number of molecules per unit volume is:A. more in He filled ballonB. same in both balloonsC. more in air fillled ballonD. in the ratio 1:4 |
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Answer» Correct Answer - B Assuming the bllons have the same volume , as `pV=nRT. If p,V andT are the same , n the number of moles present will be the same , whether it is He or air Hence,number of molecules per unit volume will be same in both the bolloons |
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| 19. |
The pressure of a gas kept in an isothermal container is `200Kpa`. If half the gas is removed from it, the pressure will beA. 100 kPaB. 200 kPaC. 400 kPaD. 800 kPa |
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Answer» Correct Answer - A for isothermal process, temperature remains constant According to ideal gas equation, `pV=nRT` For a given container ,V remains constant. `therefore` " " p`prop`n When half of gas is removed, number of moles becomes half.Hence, p[ressure becomes half. |
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| 20. |
If the intermolecules forces vanish away, the volume occupied by the molecules contained in `4.5 kg` water at stantard temperature and pressure will be given byA. `5.6m^(3)`B. `4.5m^(3)`C. `11.2m^(3)`D. `5.6 L` |
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Answer» Correct Answer - A If internmolecular forces vanish aqay, the liquid will convert into vapours. Molecullar mass of water vapours `=18g =18xx10^(-3)kg` So,`18xx10^(-3)kg` of water vapour will occupy `22400 m^(3)` then, 4.5 kg of water vapour will occupy `(22.4xx10^(-3))/(18xx106(-3))xx4.5=5.6 m^(3)` |
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