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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.
| 551. |
What is molar volume of an ideal gas under N.T.P. conditions ? |
| Answer» The molar volume of an ideal gas under `N.T.P` conditions is `22.4 dm^(3)`. | |
| 552. |
what volume of hydrogen gas , at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass=10.8) from the reduction of boron trichloride by hydrogen ?A. `67.2 L`B. `44.8 L`C. `22.4 L`D. `89.6 L` |
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Answer» Correct Answer - A No. moles of boron `= (21.6)/(10.8) = 2` `2BCl_(3) + 3H_(2) rarr 6HCl + 2B` For the formation of `2` moles of B, volume of `H_(2)` required `= 3 "mol" = 3 xx 22.4 = 67.2 L` |
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| 553. |
Assertion: Sulphur dioxide and chlorine are bleaching agents. Reason: Both are reducing agents.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 not reducing agents, chlorine is oxidising agent. |
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| 554. |
For real gases, `(PV)/(nRT)`........ |
| Answer» Correct Answer - `(PV)/(nRT) != 1` | |
| 555. |
Assertion: On cooling, the brown colour of nitrogen dioxide disappears. Reason: On cooling, `NO_(2)` undergoes dimerisation resulting in the pairing of the odd electron in `NO_(2)`.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 `underset((Brown colour))(NO_(2)+NO_(2)) hArrunderset((Colourless))(N_(2)O_(4))` |
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| 556. |
A compound formed by elements `A` and `B` crystallises in a cubic structure where `A` atoms are present at the corners of a cube and the `B` atoms are present at the face centres.The formula of the compound is |
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Answer» An atom at the corner of the cubic contributes `1//8` to the unit cell. Hence, number of atoms of A in the unit cell `=8xx(1)/(8) =1` At atoms at the face of the cube contributes `(1)/(2)` to the unit cell. Hence, number of atoms B in the unit cell `=6xx(1)/(2)=3` Thus, the formula is `AB_(3)`. |
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| 557. |
Containers `A` and `B` have same, gases. Pressure, volume and temperature of `A` are all twice that of `B`, then the ratio of number of molecules of `A` and `B` are |
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Answer» Correct Answer - B According to gas law `PV= nRT, n=(PV)/(RT)` `(n_(A))/(n_(B))=((P_(1)V_(1))/(RT_(1)))/((P_(2)V_(2))/(RT_(2))),(n_(A))/(n_(B))=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2)V_(2))` `(n_(A))/(n_(B))=(2Pxx2V)/(2T)xx(T)/(PV), (n_A)/(n_(B))=(2)/(1)` |
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| 558. |
For a definite amount of gas, pressure and volume are increased to triple of the initial amount, ThereforeA. Temperature increased to nine times of its initial valueB. Temperature increased to thrice of its initial valueC. Temperature remains unalteredD. Temperature reduced to thrice of its initial value |
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Answer» Correct Answer - A `(PV)/(T)= "constant" implies(PV)/(T_(1))=(3P3V)/(T_(2))` `T_(2) =9T_(1)` Temperature increased to nine times of its initial value. |
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| 559. |
Substance which is weakly repelled by a magnetic field isA. `O_2`B. `H_2O`C. `CrO_2`D. `Fe_3O_4` |
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Answer» Correct Answer - B Substance which is weakly repelled by a magnetic field is diamagnetic , e.g., `H_2O`. |
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| 560. |
Statement-1. Ammonia has lower molecular mass than `N_(2)`. Statement-2. At a given temperature te rate of diffusion is inversely proportional to the square root of its density.A. Statement-1 is true, statement-2 is also true, statement-2 is correct explanation of statement-1B. Statement-1 is true, statement-2 is also true, statement-2 is not correct explanation of statement-1C. Statement-1 is true, statement-2 is true.D. Statement-1 is false, statement-2 is true. |
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Answer» Correct Answer - A Statement-2 is the correct explanation for statement-1. |
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| 561. |
the interaction energy of London force is inversely proportional to sixth power of the distance between two interaction particles but their mahnitude depends uponA. charge on interacting particesB. mass of interacting particlesC. polarsablity of interacting particlesD. strength of permanent dipoles in the particles. |
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Answer» Correct Answer - C The magnitude to London forces depends upon the polarisability of the interacting particles. |
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| 562. |
Assertion: A gas can be liquefied at `T=T_(c ) and P lt P_(c)` Reason: A gas can be liquefied when `T lt T_(c ) and P lt P_(c )`.A. If both assertion and reason are true and the reason is the 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 assertion is false but reason is true. |
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Answer» Correct Answer - D At `T=T_(c )`, liquefaction is possible only at `P=P_(c )`. |
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| 563. |
Which of the following statements is wrong for gases?A. Gases do not have a definite shape and volumeB. Volume of the gas is equal to the volume of the container the gasC. Confined gas exerts uniform pressure on the walls of its container in all directions.D. Mass of the gas cannot be determined by weighing a container in which it is enclosed |
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Answer» Correct Answer - D The mass of gas can be determined by weighing the contianer filled with gas and again Weighing this container after removing the gas. The difference between the two weights gives the mass of the gas. |
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| 564. |
Which of the following statements is wrong for gases?A. Gases do not gave a definite shape and volumeB. Volume of the gas is equal to volume of container confining the gasC. Confined gas exerts uniform pressure on the walls of its container in all directionsD. Mass of gas cannot be determined by weighing a container in which it is enclosed |
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Answer» Correct Answer - D Mass of gas can be determined by weighing a container in which it is enclosed as follows: Mass of the gas = mass of the cylinder including gas - mass of empty cyliner So, it is a wrong statement |
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| 565. |
The intermetallic compounds `LiAg` crystallises in cubic lattice in which both lithium and silver have coordination number of eight ,the crystal class isA. simple cubeB. body centred cubeC. face centred cubeD. None of the above |
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Answer» Correct Answer - B In body centered cubic, each atom/ion has a coordination number of 8. |
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| 566. |
Schottky defect in a crystal is observed whenA. an ion leaves its normal site and occupies an interstitial siteB. unequal number of cations and anions are missing from the latticeC. density of the crystal is increasedD. equal number of cations and anions are missing from the lattice |
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Answer» Correct Answer - D Schottky defect in crystals is observed when equal number of cations and anions are missing from the lattice. So, the cytstal remains neutral, e.g. NaCl. |
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| 567. |
An LPG cylinder containing containing 15 kg butane at `27^(@)C` and 10 atm pressure is leaking. After one day, its pressure decreased to 8 atm. The quantity of the gas leaked isA. 1 kgB. 2 kgC. 3 kgD. 4 kg |
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Answer» Correct Answer - C PV=nRT. Here, T and V are constant. Hence, `(P_(1))/(P_(2))=(n_(1))/(n_(2))=(w_(1))/(w_(2))`. `:. (10)/(8)=(15)/(w_(2))" or " w_(2)=12" kg"`. Hence, gas leaked=15-12=3 kg |
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| 568. |
Which one of the following is a molecular crystal?A. Rock saltB. QuartzC. Dry iceD. Diamond |
| Answer» Correct Answer - c | |
| 569. |
In a tetragonal crystalA. a=b=c,`alpha=beta=90^@ ne gamma`B. `alpha=beta=gamma=90^@, a=bnec`C. `alpha=beta=gamma=90^@, a ne b ne c`D. `alpha=beta=90^@, gamma=120^@, a = b ne c` |
| Answer» Correct Answer - b | |
| 570. |
At which of the following four conditions will the density of nitrogen be the largest?A. `STP`B. `273 K` and `2 atm`C. `546 K` and `1 atm`D. `546 K` and `2 atm` |
| Answer» The density of a gas is given by `rho=PM//RT`. Obviously, the choice that has a greater `P//T` will have greater density. | |
| 571. |
The units of the van der Waals constant `a` areA. `mol L^(-1)`B. `atm L^(-2) mol^(2)`C. `atm L^(2) mol^(-2)`D. `L mol^(-1)` |
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Answer» Correct Answer - C The correction term for pressure has the units of atmosphere. Thus, `(an^(2))/(V^(2)) rArr atm` or `a rArr (atm V^(2))/(n^(2)) rArr (atm L^(2))/(mol^(2))` The correction term for volume has the units of liter. Thus, `nb rArr Liter` `b rArr ("Liter")/(n) rArr ("Liter")/("mole")` |
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| 572. |
By X-ray diffraction methods, the unit length of NaCl is observed to be 0.5627 nm. The density of NaCl is found to be `2.164 g cm^(-3)`. What type of defect exists in the crystal ?Calculate the percentage of `Na^+` and `Cl^-` ions missing. |
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Answer» Calculated density , `rho=(ZxxM)/(a^3xxN_0)=(4xx58.5 "g mol"^(-1))/((0.5627xx10^(-7) cm)^3 xx (6.022xx10^23 "mol"^(-1))=2.1809 " g cm"^(-3)` Observed density =`2.164 "g cm"^(-3)` As observed density is less than theoretically calculated value, this means that some `Na^+` and `Cl^-` ions missing from their lattice site, i.e., there is Schottky defect. Actual formula units of NaCl per unit cell can be calculated as follows : `Z=(a^3xxrhoxxN_0)/M=((0.5627xx10^(-7)cm)^3 xx(2.164 cm^(-3))xx(6.022xx10^23 mol^(-1)))/(58.5 "g mol"^(-1))`=3.968 `therefore` Formula units missing per unit cell=4-3.968=0.032 `therefore` % missing =`0.032/4xx100`=0.8% |
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| 573. |
What will be the pressure of the gas mixture when `0.5 L` of `H_(2)` at `0.8` bar `2.0 L` of oxygen at `0.7` bar are introduced in a `1L` vessel at `27^(@)C` ? |
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Answer» Correct Answer - 1.8 bar Let the partial pressure of `H_(2)` in the vessel be `p_(H_2).` Now, `p_(1)=0.8"bar" " "p_(2)=p_(H_2)` ? `V_(1)=0.5 " "V_(2)=1L` It is known that, `p_(1)V_(1)=p_(2)V_(2)` `rArr p_(2)=(p_(1)V_(1))/(V_(2)) `rArr p_(H_2)=(0.8xx0.5)/(1) =0.4 bar Now, let the partial pressure of `O_(2)` in the vessel be `p_(O_2).` Now, `p_(1)=0.7"bar" " "p_(2)=p_(o_2)=` ? `V_(1)=2.0L " "V_(2)=1L` `p_(1)V_(1)=p_(2)V_(2)` `rArrp_(2)=(p_(1)V_(1))/(V_(2))` `rArrp_(o_2)=(0.7xx20)/(1)` = 0.4 bar Total prassure of the gas mixture in the vessel can be obtained as: `p_("total")=p_(H_2)+p_(O_2)` = 0.4+1.4 =1.8 bar Hence, the total preesure of the gaseous mixture in the vesses is 1.8 bar |
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| 574. |
1L of a gas is at a pressure of `10^(-6)` mm of Hg at `25^(@)C`. How many molecules are present in the vessel.A. `3.2xx10^(6)`B. `3.2xx10^(13)`C. `3.2xx10^(10)`D. `3.2xx10^(4)` |
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Answer» (b) `(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2))Rightarrow (10^(-6x)xx1000)/(298)=(760xxV_(2))/(273)` Number of molecules `(6.02xx10^(23))/(22400)xx1.2xx10^(-6)` `3.2xx10^(13)` |
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| 575. |
An ideal gas cannot be liquedfied becauseA. its critical temperature is always above `0^(@)C`B. its molecules are relavtively smaller in sizeC. it solidifies before becoming a liquidD. forces operating between its molecules are negligible |
| Answer» (d) In the ideal gas, the intermolecular forces of attraction are negative and hence, it cannot be liquefied. | |
| 576. |
Actual graph for the given parameters in (Q.25) will be A. `I, III`B. `I, II`C. `II`D. `I` |
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Answer» Correct Answer - C As `P` increases, `PV` also increases, hence graph `II` |
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| 577. |
At low pressure, the van der Waals equation is reduced toA. `Z=(pV_(m))/(RT)=1-(ap)/(RT)`B. `Z=(pV_(m))/(RT)=1+(b)/(RT)P`C. `pV_(m)=RT`D. `Z=(pV_(m))/(RT)=1-(a)/(RT)` |
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Answer» Correct Answer - A When pressure is low `[P+(a)/(V^(2))](V-b)=RT` or `PV=RT+Pb-(a)/(V^(2)) or (PV)/(RT)=1-(a)/(VRT)` `Z= -(a)/(VRT)( :. (PV)/(RT)=Z)` |
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| 578. |
The van der Waals equation for `CH_(4)` at low pressure isA. `PV=RT-Pb`B. `PV=RT-(a)/(V)`C. `PV=RT+(a)/(V)`D. `PV=RT+Pb` |
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Answer» The van der Waals gas equation for `1 mol` of gas is `(P+(a)/(V^(2)))(V-b)=RT` At low `P`, volume is high. So `(V-b)=V` `:. (P+(a)/(V^(2)))V=RT` or `PV=RT-(a)/(V)` Hence, the correct choice is (`b`) |
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| 579. |
For a real gas , the compressibility factor Z has different values at different temperatures and pressures . Which of the following is not correct under the given conditions ?A. Z lt 1 at very low pressure .B. Z gt 1 at high pressure .C. Z = 1 under all conditions .D. Z = 1 at intermediate pressure |
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Answer» Correct Answer - C Z = 1 under all conditions for an ideal gas. |
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| 580. |
The density of solid argon is `1.65 "g mL"^(-1)` at `-233^@C` . If the argon atom is assumed to be sphere of radius `1.54xx10^(-8)` cm, what percentage of solid argon is apparently empty space ? (At. Mass of Ar =40) |
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Answer» Density of `1.65 "g mL"^(-1)` means that 1.65 g of solid argon has volume =1 mL Volume of one atom of Ar =`4/3pir^3` No. of atom in 1.65 g or 1 mL of solid Ar=`(1.65/40"mol")(6.023xx10^23 "mol"^(-1))=(1.65xx6.023xx10^23)/40` `therefore` Total volume of all atoms Ar in 1 mL of Ar =`4/3pir^3 xx (1.65xx6.023xx10^23)/40` `=4/3xx22/7xx(1.54xx10^(-8))^3 xx (1.65xx6.023xx10^23)/40` `=0.380 cm^3` or 0.380 mL Volume of solid Ar = 1 mL `therefore` Empty space =(1-0.380) mL =0.620 mL `therefore` % empty space =`0.620/1xx100` =62% |
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| 581. |
Density of a gas is found to be `5.46//dm^(3)` at `27^(@)C` at 2 bar pressure What will be its density at `STP` ? .A. ` 3.0 g dm^(-3)`B. `5.0 g dm^(-3)`C. `6.0 g dm^(-3)`D. `10.82 g dm^(-3)` |
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Answer» Correct Answer - A `d=(PM)/(RT)` or `d prop (P)/(T)` `(d_(1))/(d_(2))=(P_(1))/(P_(2))xx(T_(2))/(T_(1))` or `d_(2)=(d_(1)xxP_(2)xxT_(1))/(P_(1)xxT_(2))` `d_(2)=(5.46xx1xx300)/(2xx273)=3.0 g dm^(-3)` |
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| 582. |
Select the correct statements. (`I`) Greater is humidity, lesser will be rate of evaporation of water. (`II`) Greater is humidity, lesser will be density of air. (`III`) If room temperature `=` dew point, realtive humidity `= 100%`. (`IV`) Dew point is the temperature at which the gas a given atmospheric condition becomes staturted with `H_(2)O` (`v`)A. `I`, `II`B. `II`,`IV`C. AllD. None |
| Answer» `Z=(P_(C)V_(C))/(RT_(C))=(3)/(8)`, each molecule moves with all together different speed. | |
| 583. |
Which gas shows real behaviour?A. `8 g O_(2)` at `STP` occupies `5.6 L`.B. `1 g H_(2)` in `0.5 L` flask exerts a pressure of `24.63 atm` at `300 K`.C. `1 mol NH_(3)` at `300 K` and `1 atm` occupies volume `22.4 L`.D. `5.6 L` of `CO_(2)` at `STP` is equal to `11 g`. |
| Answer» For real behaviour, `(PV)/(RT)!=1`. | |
| 584. |
A gas cylinder having a volume of 25.0 L contains a mixtue of butane `CH_(3)(CH_(2))_(2) CH_(3)` and isobutane `(CH_(3))_(3) CH` in the ratio of 3 : 1 by moles. If the pressure inside the cylinder is `6.78 xx 10^(6)` pa and the temperature is 298 K, calculate the number of molecular of each gas assuming ideal gas behaviour. (1 atm = 101325 Pa) |
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Answer» Total moles `(n)=(6.78x10^(6)xx25)/(101325xx0.082xx298)=68.37mo` `N_("butane")=(1)/(4)xx68.37xx6.02xx10^(23)=3.07xx10^(25)` `N_("Isobutane")=(1)/(4)xx68.37xx6.02xx10^(23)=1.02xx10^(25)` |
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| 585. |
For which crystal anion-anion contact is valid ?A. NaFB. NaIC. CsBrD. KCl |
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Answer» Correct Answer - B Iodide ion is largest in size and `Na^+` ion is small in size. Hence, `I^(-) - I^-` contact is possible in `Na^+ I^-` (`Na^+` ion present in the void) |
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| 586. |
Each rubidium halide crystallising in the NaCl type lattice has a unit cell length 0.30 Å greater than for corresponding potassium salt (`r_(K^+)`=1.33 Å) of the same halogens . Hence, ionic radius of `Rb^+` isA. 1.03 ÅB. 1.18 ÅC. 1.48 ÅD. 1.63 Å |
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Answer» Correct Answer - C As RbX crystallies in NaCl type lattice, unit cell length of RbX =`2(r_(Rb^+)+r_(X^-))` Unit cell length of KX =`2(r_(K^+) + r_(X^-))` `therefore 2(r_(Rb^+)+r_(X^-))-2(r_(K^+)+r_(X^-))`=0.30 Å or `r_(Rb^+) - r_(K^+)`=0.15 Å or `r_(Rb^+) = r_(K^+)`+ 0.15 Å =1.33 +1.15=1.48 Å |
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| 587. |
The volume of helium is `44.8 L` atA. `100 ^(@)C` and `1 atm`B. `0 ^(@)C` and `1 atm`C. `0 ^(@)C` and `0.5 atm`D. `100 ^(@)C` and `0.5 atm` |
| Answer» Correct Answer - Use `PV=RT` | |
| 588. |
If 1 g of each of the following gases are takes at STP, which of the gases will occupy (a) greatest volume and (b) smallest volume ? `Co, H_(2)O, CH_(4), NO` |
| Answer» The molar mass of the gases are , `CO(28 g mol^(-1)) , H_(2)O (18 g mol^(-1)), CH_(4) (16 g mol^(-1))` and `NO (30 g mol^(-1))`. Since molar volume of all the gas `22.4 L` under S.T.P. conditions,1 gram of `CH_(4)` will occupy maximum volume while one gram of `NO` will occupy minimum volume. | |
| 589. |
The value of `PV` for `5.6 L` of an ideal gas is ……… `RT` at `NTP`. |
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Answer» Correct Answer - 0.25 |
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| 590. |
The value of `PV` for `5.6 L` of an ideal gas is ……… `RT` at `NTP`.A. `0.25`B. `0.30`C. `1.0`D. `0.45` |
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Answer» `V_(ideal)=22.4 L` `:. PV_(ideal)=(5.6)/(22.4)RT=0.25RT` |
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| 591. |
Calculate the pressure of 154 g carbon dioxide in a vessel of 2.0 L capacity at `30^(@)C`, a = 648 L bar atm `K^(-1) mol^(-1), b = 0.0427 L mol^(-1)` |
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Answer» a. `NH_(3)` will have larger value of a because of H bonding b. `N_(2)` should have larger value of b because of larger molecular size. |
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| 592. |
`1` of `NO_(2)` and `7//8 L` of `O_(2)` at the same temperature and pressure were mixed together. What is the relation between the mases of the two gases in the mixture?A. `M_(N_(2))=3M_(O_(2))`B. `M_(N_(2))=8M_(O_(2))`C. `M_(N_(2))=M_(O_(2))`D. `M_(N_(2))=16M_(O_(2))` |
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Answer» `PV=(M)/(m)RT` `Pxx1=(M_(N_(2)))/(28)RT` `Pxx(7)/(8)=(M_(O_(2)))/(32)RT` Dividing equation (`i`) by (`ii`), we get `M_(N_(2))=M_(O_(2))` |
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| 593. |
`120 g` of an ideal gas of molecular weight `40` is confirmed to a volume of `20 litre at 400 K`, then the pressure of is:A. `490 atm`B. `4.92 atm`C. `2236 atm`D. `22.4 atm` |
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Answer» Correct Answer - B `P=(3xx0.082xx400)/(20)implies P= 4.92 atm` |
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| 594. |
A student forgot to add the reaction mixture to the round bottomed flask at `27^(@)C` but instead, he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer, he found the temperature of the flask was `477^(@)C`. What fraction of air would have been expelled out ? |
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Answer» Suppose volume of vessel =V `cm^(3)`, i.e., volume of air in the flask at `27^(@)C=V cm^(3)`. `(V_(1))/(T_(1))=(V_(2))/(T_(2))," "i.e.," "(v)/(300)=(V_(2))/(750)" or "V_(2)=2.5" V"` `:. " " "Volume expelled"=2.5" V"-V=1.5" V"` `:. " " "Fraction of air expelled" `=(1.5" V")/(2.5" V")=(3)/(5)` |
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| 595. |
Rate of diffustion of a gas is :A. directly proportional to its densityB. directly proportional to its molecular massC. diectly proportional to th squre of its molecular massD. inversely proportional to the square root of its molecular mass. |
| Answer» Correct Answer - D | |
| 596. |
Point out the correct statement for the set of characteristics of ZnS crystal.A. Coordination number (4:4), ccp,`Zn^(2+)` ion in all the alternate tetrahedral voidsB. Coordination number (6:6), hcp,`Zn^(2+)` ion in all the alternate tetrahedral voidsC. Coordination number (6:4), hcp,`Zn^(2+)` ion in all the alternate octahedral voidsD. Coordination number (4:4), ccp,`Zn^(2+)` ion in all the tetrahedral voids |
| Answer» (a) ZnS has zinz blende type structure . The `S^(2-)` ions are present at the corers of the cube and at the centre of each face. Zinc ions occupy half of the tetrahedral sites. Each zinc ion is surrounded by four sulphate sites. Each zinc ion is surrounded by four sulphide ions which are disopsed towards the corner of regular tetrahedruon. Similarly, `S^(2-)` ion is surrounded by four `Zn^(2+)` ions. | |
| 597. |
The temperature `30.98^(@)C` is called critical temperature `(T_(C))` of carbon dioxide. The critical temperature is theA. lowest tempertaure at which liquid carbon dioxid eis observedB. highest temperature at which gas carbon dioxide is observedC. highest temperature at which solid carbon dioxide is observedD. highest temperature at which liquid carbon dioxide is observed |
| Answer» Correct Answer - (d) | |
| 598. |
Which of the following liquid will exhibit highest vapour pressure?A. `C_(2)H_(5)OH`B. `NH_(3)`C. HFD. `H_(2)O` |
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Answer» Correct Answer - B H-bonds in liquid `NH_(3)` are weaker hence, its escaping tendency is highest and as a result its vapour pressure will be highest. |
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| 599. |
A liquid boils, whenA. vapour pressure becomes equal to the atmospheric pressureB. atmospheric pressure is lower than vapour pressureC. vapour pressure becomes higherD. temperature becomes very high at constant pressure |
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Answer» Correct Answer - A A liquid boils when vapour pressure becomes equal to the atmospheric pressure. |
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| 600. |
During the evaporation of liquidA. the temperature of the liquid will riseB. the temperature of the liquid will fallC. may arise or fall depending on the natureD. the temperature remains unaffected |
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Answer» Correct Answer - B During evaporation, molecules having high energy leave the surface of the liquid. As a result, average kinetic energy of liquid decreases. `KEpropT` `:.` Temperature of liquid falls. |
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