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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.
| 701. |
Specific heat of a monoatomic gas at constant volume is `315 J kg^(-1) K^(-1)` and at a contant pressure is `525 J kg^(-1) K^(-1)`. Calculate the molar of the gas. |
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Answer» `C_(P) = M xx 525 " and " C_(V) = M xx 315` where, M is the molecular mass. `C_(P) - C_(V) = R (R = 8.314 J K^(-1) mol^(-1))` `M xx 525 - M xx 315 = 8.314` `M (525 - 315) = 8.314` `M = (8.314)/(210) = 0.0396 kg mol^(-1) = 39.6 g mol^(-1)` |
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| 702. |
For an ideal gas number of moles per litre in terms of its pressure `P` gas contant `R` and temperature `T` is .A. `PT//R`B. `PRT`C. `P//RT`D. `RT//P` |
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Answer» Correct Answer - C According to ideal gas equation `PV = nRT : n//V = P//RT` |
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| 703. |
A and `B` are two idential vessels. `A` contains `15 g` ethane at `1 atm` and `298 K`. The vessel `B` contains `75 g` of a gas `X_(2)` at same temperature and pressure. The vapour density of `X_(2)` is :A. `75`B. `150`C. `37.5`D. `45` |
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Answer» Correct Answer - A `(15)/(30)=(75)/(M_(B))` `M_(B)= 150, (VD)_(B)=(150)/(2)= 75` |
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| 704. |
Two perfect monoatomic gases at absolute temperature `T_(1)` and `T_(2)` are mixed. There is no loss of energy. Find the temperature of the mixture if the number of moles in the gases are` n_(1)` and `n_(2)`. |
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Answer» Correct Answer - `T = (n_(1) T_(1) + n_(2) T_(2))/(n_(1) + n_(2))` Let final temperature is T `(3)/(2) n_(1) RT_(1) + (3)/(2) n_(2) RT_(2) = (3)/(2) (n_(1) + n_(2)) RT` `T = (n_(1) T_(1) + n_(2) T_(2))/(n_(1) + n_(2))` |
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| 705. |
Assuming oxygen molecule to be spherical in shape, calculate the volume of a single molecule of oxygen if its radius is 150 pm. Also calculate the percentage of empty space in one mole of the gas at STP. |
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Answer» As oxygen molecule is spherical, therefore, volume of one molecule `=(4)/(3)pir^(3)=(4)/(3)xx(22)/(7)(150xx10^(-10)cm)^(3)=1.41xx10^(-23) cm^(3)` To calculate the empty space in 1 mole of `O_(2)` molecules, let us first calculate the volume occupied by 1 mole of molecules, i.e., `(6.0 22xx10^(23))xx(1.41xx10^(-23)cm^(23)) ` molecules, This will be `=(6.022xx10^(23))xx(1.41xx10^(-23)cm^(3))=8.49" cm"^(3)`. But volume occupied by 1 mole of molecules at S.T.P.`=22400" cm"^(3)` `:. " "` Empty space=22400-8.49`" cm"^(3)=22391.51" cm"^(3)` `:. ` % age of empty space `=(22391.51)/(22400)xx100=99.9%` |
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| 706. |
One mole of a monoatomic 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 byA. B. C. D. |
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Answer» Correct Answer - A When the atoms (e.g. H atoms) are very close to each other, there are repulsions and the potential energy of the system is high. When they come closer, attraction takes place,potential energy decreases till at a particular distance (called bond length), the potential energy is lowest (being negative, corresponding to minimum in the curve). Here, forces of attraction and repulsion balance each other. On further increasing the interatomic distance, the interatomic attractions decrease, therefore, potential energy begins to increase and at very large distances, there are no forces of attraction or repulasion and potential energy increases to zero. |
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| 707. |
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 Equation of stage `p(v-b) = RT` indicates that inter molecular forces of attraction or repulsion are absent. As a result. Interatomic potential `V_(r)` remains constant on increasing pressure in the beginning. As the molecular come very close on increasing pressure. their electronic and nuclear repulsion leading to interatomic potential increases abrupty. |
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| 708. |
At 1 atmospheric pressure and `0 .^(@)C` , certain mass of a gas measures `0.4 L` . Keeping the pressure constant , if the temperature is increased is increased to `273 .^(@)C` , what will be its volume ?A. 0.8 LB. 22.4 LC. 54.6 LD. 0.4 L |
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Answer» Correct Answer - A `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) rArr (1xx0.4)/(273)=(1xx V_(2))/(546)` `V_(2)=(1xx0.4xx546)/(273xx1)=0.8 L` |
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| 709. |
Two flasks of equal volume connected by a narrow tube of negliglible volume are filled with `N_(2)` gas. When both are immersed in boiling water, the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in an ice-water mixture keeping the other in boiling water. |
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Answer» Temp. of the gas when both the flasks are immersed in boiling water`=100^(@)C=373" K"`. Pressure=0.5 atm Average temp. of the gas when one flask is immersed in ice and the other in boiling water `=(0+100)/(2)=50^(@)C=323" K"` As volume remains constant, `(P_(1))/(T_(1))=(P_(2))/(T_(2))` `(0.5)/(373)=(P_(2))/(323) " or " P_(2)=(323)/(373)xx0.5=0.433" atm"`. |
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| 710. |
Given that the co-volume of `O_(2)` gas is `0.318 L mol^(-1)`. Calculate the radius of `O_(2)` molecule. |
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Answer» 26 K or `52.85^(@)C` `V=5dm^(3)` `P=10.0` bar `b=0.0564Lmol^(-1)` `a=6.7` bar `L^(2)mol^(-2)` is `n=(w)/(M)=(128)/(64)` `triangleT=((P+(an^(2))/(V))(V-nb))/(Rn)` |
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| 711. |
A `10.0 L` container is filled with a gas to a pressure of `2.00 atm` at `0^(@)C`. At what temperature will the pressure inside the container be `2.50 atm` ? |
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Answer» As the volume of the container remains constant, applying pressure-temperature law,viz. `(P_(1))/(T_(1))=(P_(2))/(T_(2))` We get `(2 atm)/(273 K)=(2.50 atm)/(T_(2))` or `T_(2)=341 K=341-273^(@)C=68^(@)C` |
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| 712. |
Which of the following is true about the number of molecules in `A` and `B`?A. Flask `A` contains eight times more molecules than flasks `B`.B. Flask `B` contains eight times more molecules than flask `A`.C. Both flasks contain an equal number if molecules.D. Flasks `A` contains four times more molecules than flasks `B`. |
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Answer» `n_(A)=(m)/(2)`, `n_(B)=(m)/(16)`, `(n_(A))/(n_(B))= 8 since` `n_(A) prop N`, `:. (N_(A))/(N_(B))=8` (`n=` number of moles)(`N=` number of molecules) |
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| 713. |
Value of gas constant `R` is .A. `0.082 J "litre" atm`B. `0.987 cal "mol"^(-1) K^(-1)`C. `8.314 J "mol"^(-1) K^(-1)`D. `83 erg mol^(-1) K^(-1)` |
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Answer» Correct Answer - C is the correct answer. |
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| 714. |
At 298 k, which of the following gases has the lowest average molecular speed?A. `CO_(2)" at"" 0.20" " atm"`B. `He " at " 0.40 " atm"`C. `CH_(4) " at" " 0.80"" atm"`D. `NO" at"" 1.00"" atm"` |
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Answer» Correct Answer - A `u_(av)=sqrt((8pV)/(piM)` `u_(av)propsqrt(p/(M))=sqrt(0.20/44)=0.067" for " CO_(2)` `=sqrt(0.40)/(4)=0.32` for He `=sqrt((0.80)/(16))=0.22" for " CH_(4)` `=sqrt((1)/(30))=0.18" for "NO` |
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| 715. |
Select the correct statements.A. Gases tend to behave non-ideally at low temperatures and high pressuresB. Gases tend to behave ideally at high temperatures and low pressuresC. The extend to which Z deviates from 1 is a measure of the non-ideality of a gasD. All of the above statements are correct |
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Answer» Correct Answer - D All the given statements are correct. |
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| 716. |
Assertion : The gases show ideal behaviour when the volume occupied is large so that the volume of the molecules can be neglected in comparison to it . Reason : The behaviour of the gas becomes more ideal when pressure is very low .A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false . |
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Answer» Correct Answer - B For 1 mole of real gas, `(P+(a)/(V^(2)))(V-b)=RT` when V is very large, the terms b and `a//V^(2)` are negligible and the van der Waals equation becomes ideal gas equation, PV = RT hence, real gases behave like ideal gas. |
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| 717. |
Under what conditions to gases show maximum deviations from ideal gas behaviour ?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 At low temperature and high pressure, the volume of the particles is not negligible as compared to the volume of the gas. Also, the intermolecular forces start acting upon the molecules. Hence, they deviate from ideal behaviour. |
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| 718. |
The surface tension of water at `20^(@)C` is `73` dynes `cm^(-1)`, The minimum value of work done needed to increases surface are of water from `2 cm^(2)` to `5cm^(2)` isA. `192` dynes `cm`B. `219` dynes `cm`C. `921` dynes `cm`D. `912` dynes `cm` |
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Answer» Correct Answer - B `w= gammadA= 73xx3= 219` dynes cm |
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| 719. |
Which of the following gases follows non-ideal behaviour?A. `N_(2)` gas having density `1.25g" L"^(-1)` at STPB. 2.8 g CO gas in 0.1 L flask exerting a pressure of 24.63 atm at 300 kC. 1.6 g `CH_(4)` in 0.5 L flask at 273 K exerting a pressure of 4 atmD. 0.1 g `H_(2)` gas at STP occupies volume of 1.12 L |
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Answer» Correct Answer - C Non-ideal behaviour is followed if `pVnenRT` `p=(n)/(V)RT=(w)/(mV)RT=(dRT)/(m)` LHS,p=1 atm `RHS,(dRT)/(m)=(1.25g" L"^(_1)xx0.0821" L "atm " mol"^(-1)K^(_1)xx273K)/(28g" mol"^(-1)` Thus, `p=(dRT)/(m)=1` Thus, ideal behaviour `pV=nRT` `24.63xx0.1=(2.8)/(2.8)xx0.0821xx300 ` `2.463=2.463` Thus,ideal behaviour `pV=nRT` `4xx0.5=(1.6)/(16)xx0.0821xx273` 2=2.24 Thus,`pVnenRT` Thus, non-ideal behaviour `0.1 g H_(2)=0.05 " mol "H_(2)` `=0.05xx22.4L` at STP=1.12 L at STP Thus,ideal behaviour. |
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| 720. |
For ideal gases, `Z = 1` at all temperature and pressure. |
| Answer» Correct Answer - True | |
| 721. |
A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have equalA. IntermolecularB. kinetic energyC. Total energyD. Potential energy |
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Answer» Correct Answer - B `KE` of gas and liquid molecules at any temperature is given by `(3)/(2)RT`. At equilibrium both have same `T` and thus same `KE`. |
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| 722. |
A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have equalA. total energyB. potential energyC. intermolecular forcesD. kinetic energy |
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Answer» Correct Answer - D At equilibrium , the rate of evaporation is equal to the rate of condensation which is possible if on the average , the molecules in the two phases have equal kinetic energies. |
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| 723. |
How is the pressure of a gas related to its density at a particular temperature ? |
| Answer» Correct Answer - `d=(MP)/(RT)` | |
| 724. |
A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have equalA. Intermolecular forcesB. Potential energyC. Kinetic energyD. Total energy |
| Answer» A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have equal knietic energy. Total energy and potential energy are different because in two phases, the intermolecular attractions are different. | |
| 725. |
(a) Name the type of intermolecular forces existing in the molecules of `BCl_(3),NCl_(3)` and `NHCl_(2)`. (b) Which of these is most likely to exist in the condensed state and which one is least likely? |
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Answer» (a) `BCl_(3)`, being symmetrical molecule has London forces only. `NCl_(3)`,being pyraimdal has a net dipole moment, i.e, it is polar and hence dipole-dipole forces are present in addition to London forces. `NHCl_(2)` has hydrogen bonding becauseof the presence of N-H bond in addition to London forces. (b) Due to presence of hydrogen bonding which is a strong bond compared to dipole-dipole and London forces, `NHCl_(2)` is most likely to be in comdensed phase. `BCl_(3)` has weakest intermolecular forces and hence is least likely to be in condensed phase, i.e., it is likely to be gaseous. |
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| 726. |
`127 mL` of a certain gas diffuses in the same time as `100 mL` of chlorine under the same conditions. Calculate the molecular weight of the gas. |
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Answer» `(V_(1))/(V_(2))=sqrt((d_(2))/(d_(1)))=sqrt((M_(2))/(M_(1)))` `(127)/(100)=sqrt((71)/(M_(1)))` `:. M_(1)=43.58` |
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| 727. |
The rate of diffusion of a gas isA. direction proportional to its densityB. directly proportional to its molecular weightC. directly proportinal to the square root of its molecular weightD. inversely proportional to the square root of its molecular weight |
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Answer» Correct Answer - D |
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| 728. |
The average veloctiy of an ideal gas molecule at `27^(@)C` is `0.3 m s^(-1)`. The average velocity at `927^(@)C` will beA. `0.6m//s`B. `0.3m//s`C. `0.9m//s`D. `3.0m//s` |
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Answer» Correct Answer - A |
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| 729. |
The average veloctiy of an ideal gas molecule at `27^(@)C` is `0.3 m s^(-1)`. The average velocity at `927^(@)C` will beA. `0.6 m s^(-1)`B. `0.3 m s^(-1)`C. `0.9 m s^(-1)`D. `3.0 m s^(-1)` |
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Answer» `(v_(1))/(v_(2))=sqrt((T_(1))/(T_(2)))` `T_(1)=273^(@)C=300K` `T_(2)=927^(@)C=1200K` `V_(1)=0.3 m s^(-1)` `V_(2)=sqrt((T_(2))/(T_(1))xxV_(1)=sqrt(1200)/(300)xx0.3=0.6m s^(-1)` |
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| 730. |
A liquid can exist onlyA. between triple point and critical temperatureB. at any temperature above the melting pointC. between melting point and critical temperatureD. between boiling and melting temperature. |
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Answer» Correct Answer - D A liquid below its melting point, is present in solid state and above its melting point, is present in vapour (gaseous) state, so a liquid can exist between melting point and boiling point. |
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| 731. |
The temperature of the gas is raised from `27^(@)C` to `927^(@)C`, the root mean square velocity isA. `sqrt((927)/(27))` times of the earlier valueB. same as beforeC. halvedD. doubled |
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Answer» Correct Answer - D Root mean square velocity at `T_(1)` temperature, `U_(1)=sqrt((3RT_(2))/(M))=sqrt((3R(27+273))/(M))" ...(i)"` Root mean square velocity at `T_(2)` temperature, `U_(2)=sqrt((3RT_(2))/(M))=sqrt((3R(927+273))/(M))" ...(ii)"` Eq. (i) divided by Eq. (ii) `(U_(1))/(U_(2))=sqrt((27+273)/(927+273))=sqrt((300)/(1200))=(1)/(2)` `U_(2)=2U_(1)` |
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| 732. |
Which of the following defines the critical temperature of a substances gtA. The temperature above which a substance can exist only as gasB. The maximum temperature at which a gas can be liquefiedC. The temperature above which a liquid cannot existD. All of these |
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Answer» Correct Answer - D Critical temperature `(T_(c))` is the temperature above which the gas can never be liquefied howsoever high the pressure may be. |
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| 733. |
Which of the following were called permanent gases ? (i) `H_(2)` , (ii) `O_(2)` (iii) `N_(2)` , (iv) `CH_(4)`A. `(i),(ii),(iii)`B. `(i),(ii),(iii),(iv)`C. `(ii),(iii),(iv)`D. `(i),(ii)` |
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Answer» Correct Answer - B They were called permanent gases as they could not be liquefied at room temperature by applying any amount of pressure. The reason for th efailure to liqefy the so-called permanent gases is now clear: They had not been cooled sufficiently before applying pressure. |
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| 734. |
Which of the following gases can be liquefied easily ?A. `NH_(3)`B. `H_(2)`C. `N_(2)`D. `O_(2)` |
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Answer» Correct Answer - A `NH_(3)` can be liquefied easily as its molecules can interact through intermolecular hydrogen bonding which is possible because an `H` atom is directly bonded to a highly electronegative `N` atom with a lone pair. |
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| 735. |
The number of atoms is 100 g of a fcc crystal with density ="10.0 g/cm"^3` and cell edge equal to 200 pm is equal toA. `5xx10^24`B. `5xx10^25`C. `6xx10^23`D. `2xx10^25` |
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Answer» Correct Answer - A `rho=(ZxxM)/(a^3xx10^(-30)xxN_0)` or `M=(10xx(200)^3xx10^(-30)xx6xx10^23)/4=12` Thus, 12 g contain=`N_0=6xx10^23` atoms `therefore` 100 g will contain `=(6xx10^23)/12xx100=5xx10^24` |
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| 736. |
It is possible to liquefy a gasA. at a temperature above critical temperature and at a pressure above critical pressureB. at a temperature at critical temperature and at a pressure lower than critical pressureC. at critical temperature and a pressure equal to critical pressureD. at a temperature above critical temperature and pressure below critical pressure. |
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Answer» Correct Answer - C Liquefaction of gases takes place at a temperature either equal to critical temperature or at a temperature lower than critical temperature. Also at the critical temperature, the pressure would be equal to critical pressure. |
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| 737. |
The graph between PV vs P at constant temperature is linear parallel to the pressure axis. |
| Answer» Correct Answer - True | |
| 738. |
The correct order of temperature for a real gas is `{:("Boyle temp.","Critical temp.","Inversion temp.",),((I),(II),(III),):}`A. `III gt I gt II`B. `I gt II gt III`C. `II gt I gt III`D. `I gt III gt II` |
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Answer» Correct Answer - A `T_(i)=(2a)/(Rb), T_(c )=(8a)/(27Rb), T_(B)=(a)/(Rb)` |
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| 739. |
Out of `CO_(2)` and `NH_(3)` gases, which is expectd to show more deviation from the ideal gas behaviour ? |
| Answer» `NH_(3)` is expected to show ore deviation because it cane easily liquefied being of polar nature. This shows that the attractive force in the molecules of `NH_(3)` are more than the forces present in `CO_(2)` molecules. | |
| 740. |
Real gases show deviation from ideal behaviour at low temperature and high pressure. |
| Answer» Correct Answer - True | |
| 741. |
A real gas most closely approaches the behaviour of an ideal gas at:A. `15 atm` and `200 K`B. `1 atm` and `273 K`C. `0.5 atm` and `500 K`D. `15 atm` and `500 K` |
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Answer» Correct Answer - C Lowest pressure and highest temperature |
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| 742. |
300 L of ammonia gas at `20^(@)C` and 20 atm pressure is allowed to expand in a space of 600 L capacity and to a pressure of 1 atm. Calculate the drop in temperature. |
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Answer» `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))=(20" atm"xx300L)/(293K)=(1" atm"xx600L)/(T_(2))` or `T_(2)=29.3K` `therefore` drop in temperature `=293-29.3=263.7K` |
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| 743. |
Which of the following pair of gases will have same rate of diffusion under similar conditions?A. `H_(2)` and `He`B. `CO_(2)` and `N_(2)O`C. `CO` and `C_(2)H_(4)`D. `NO` and `CO` |
| Answer» Because molecular mass is almost same | |
| 744. |
`200 cm^(2)` of a ga at 800 mm pressure is allowed to expand till the pressure is 0.9 atm keeping ath temperature constant. Calculate the volume of the gas. |
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Answer» Using `P_(1)V_(1)=P_(2)V_(2)` `V_(2)=(P_(1)V_(1))/(P_(2))=((800//760))/(0.9)=233.91cm^(3)` |
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| 745. |
A `0.5 dm^(3)` flask contains gas `A` and `1 dm^(3)` flask contains gas `B` at the same temperature. If density of `A= 3g//dm^(3)` and that of `B= 1.5g //dm^(3)` and the molar mass of `A= 1//2` of `B`, the ratio of pressure excerted by gases is:A. `(P_(A))/(P_(B))=2`B. `(P_(A))/(P_(B))=1`C. `(P_(A))/(P_(B))=4`D. `(P_(A))/(P_(B))=3` |
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Answer» Correct Answer - C `V_(A)= 0.5dm^(3)` `V_(B)= 1dm^(3)` `d_(A)= 3g//dm^(3)` `d_(B)= 1.5g//dm^(3)` `M_(A)=(1)/(2)M_(B)` `P_(A)=(D_(A)RT)/(M_(A)), p_(B)(d_(B)RT)/(M_(B))` `(P_(A))/(P_(B))=(d_(A))/(d_(B))xx(M_(B))/(M_(A))implies (3)/(1.5)xx2= 4` |
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| 746. |
A certain gas takes three times as long to effuse out as helium. Its molar mass will beA. `27 u`B. `36 u`C. `64 u`D. `9 u` |
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Answer» Correct Answer - B `(r_(1))/(r_(2))=(V_(1)//t_(1))/(V_(2)//t_(2))=(t_(2))/(t_(1))=sqrt((M_(2))/(M_(1)))` (for equal volumes, `V_(1)=V_(2)`) `implies (M_(2))/(M_(1))=(t_(2)^(2))/(t_(1)^(2))` `implies M_(2)= 4xx(3)^(2)= 36 u` |
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| 747. |
A certain gas takes three times as long to effuse out as helium. Its molecular mass will beA. 64 uB. 9 uC. 27 uD. 36 u |
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Answer» Correct Answer - D `(r_(He))/(r_(X))=sqrt((M_(X))/(M_(He)))" or " (V//t_(He))/(V//t_(X))=sqrt((M_(X))/(M_(He)))` or `(t_(X))/(t_(He))=sqrt((M_(X))/(M_(He)))` `:. (3)/(1)=sqrt((M_(X))/(4))" or "(M_(X))/(4)=9 " or "M_(X)=36` |
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| 748. |
What is the difficulty faced by the mountaineers with respect to the air present around them ? How is this difficulty solved ? |
| Answer» At altitude, the atmospheric pressure is low. Hence, air is less dense. As a result, less oxygen is avaiable for breathing. The person feels uneasiness, headache etc. This is called altitude sickness. This difficulty is solved by carrying oxygen cylinders with them. | |
| 749. |
Which curve in figure represents the curve of ideal gas ? A. B onlyB. C and D onlyC. E and F onlyD. A and B only |
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Answer» Correct Answer - A For ideal gas PV = constant at all pressures |
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| 750. |
Explain why cooking is faster in a pressure cooker.A. loss of heat due to radiation is minimumB. food particles are effectively smashedC. water boils at higher temperature inside the pressure cookerD. food is cooked at constant volume |
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Answer» Correct Answer - C Cooking is fast in a pressure cooker, because water boils at higher temperature inside the pressure cooker. |
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