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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.
| 601. |
On the top of a mountain, water boils atA. high temperatureB. same temperatureC. high pressureD. low temperature |
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Answer» Correct Answer - D On the top of a mountain , water boils at low temperature. Pressure is directly proportional to temperature.At low pressure , the boiling point of water comes down.As we go higher in altitude, the atmospheric pressure decreases. This results in decreasing the boiling point at higher altitude and increase in boiling of water. |
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| 602. |
Which of the following figures does not represent 1 mole of dioxygen gas at STP ?A. 16 grams of gasB. 22.7 litres of gasC. `6.022 xx 10^(23)` dioxygen moleculesD. 11.2 litres of gas |
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Answer» Correct Answer - A::D (a) 16 grams of dioxygen gas gas represent `0.5` mole. (b)` 11.2` lires of gas repesent `0.5` mole. |
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| 603. |
Which of the following figures does not represent 1 mole of dioxygen gas at STP ?A. 16 grams of gasB. 22.7 litres of gasC. `6.022xx10^(23)` dioxygen moleculesD. `11.2 litres of gas |
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Answer» Correct Answer - A::D 1 mole of `O_(2)` at STP =32 g of the gas=22.4L of the gas. |
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| 604. |
Why are falling liquid drops spherical? |
| Answer» This is due to the property of surface tension possessed by liquids. This makes the surface area minimum. For a given volume, sphere has the minimum surface area. | |
| 605. |
At what temperature, the r.m.s. velocity of a gas measured at `50^(@)C` will become double ?A. `626 K`B. `1019 K`C. `200^(@)C`D. `1019^(@)C` |
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Answer» Correct Answer - D `U_(rms) = sqrt((3RT)/(M))` According to available information, `sqrt((3RT)/(M)) = 2 xx sqrt((3R xx 323 K)/(M))` squaring both sides, `T = 4 xx 323 K = 1292 K` `T = 1292 - 273 = 1019^(@)C` |
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| 606. |
On heating a gas, pressure and volume both become double. By lowering temperature, one fourth of initial number of moles of air are taken in to maintain the double pressure and volume. Calculate by what fraction, the temperature must have been raised finally. |
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Answer» Initially, `P_(1)=P,V_(1)=V`, Then after heating, `P_(2)=2P, V_(2)=2V` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) :. (PV)/(T_(1))=(2Pxx2V)/(T_(2)) " or " T_(2)=4T_(1)` New P and V are kept constant (double values) when air is taken in. If initialy, npo of moles of air `(n_(1))=n` Then now no. of moles `=n+(n)/(5)=(5)/(4)n` Now temperature is raised from 4 `T_(1)` to `T_(f)` (final temperature) Applying PV=nRT As P and V are kept constant, `n_(i)T_(i)=n_(f)T_(f)` `:. " "nxx4T_(1)=(5)/(4)nxxT_(f)" or " T_(f)=(16)/(5)T_(1)` |
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| 607. |
For one mole of an ideal gas, increasing the temperature from `10^(@)C` to `20^(@)C`.A. increases the average kinetic energy by two timesB. increases the rms by `sqrt(2)` timesC. increased the rms by 2 timesD. increases both the average kinetic energy and rms velocity but not significantly |
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Answer» Correct Answer - D `bar(KE)=(3)/(2)RT :. (bar(KE_(2)))/(bar(KE_(1)))=(T_(2))/(T_(1))=(293)/(283)=1.04` `:. bar(KE_(2))=1.04bar(KE_(1))` `u_(rms)=sqrt((3RT)/(M))` `:.(u_(2))/(u_(1))=sqrt((293)/(283))=sqrt(1.04)=1.02` `:. u_(2)=1.02" u"_(1)` Thus, both increase but not significantly. |
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| 608. |
The average free path at `1 atm` pressure is `L`. What should be its value at `5 atm` pressure? |
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Answer» Since average free path `l=(kT)/(sqrt(2)sigma^(2)P)` `:. L prop (1)/(P)` Hence, at `5 atm` pressure, the average free path is `l=(L)/(5)` |
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| 609. |
Which of the following has the maximum value of mean free path?A. `H_(2)`B. `N_(2)`C. `O_(2)`D. `Cl_(2)` |
| Answer» Mean free path is the average distance travelled by a molecule between two successive collisions. Obivously, the molecule with smaller molecular size will have greater mean free path. Hence, the correct choice is (`a`), `H_(2)`. | |
| 610. |
The r.m.s. velocity of hydrogen is `sqrt(7)` times the r.m.s. velocity of nitrogen. If T is the temperature of the gas,A. `T(H_(2))=T(N_(2))`B. `T(H_(2))=sqrt(7)T(N_(2))`C. `T(N_(2))=2T(H_(2))`D. `T(N_(2))=sqrt(7)T(H_(2))` |
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Answer» Correct Answer - C `u=sqrt((3RT)/(M)),(u(H_(2)))/(u(N_(2)))=sqrt((T(H_(2)))/(M(N_(2)))xx(M(N_(2)))/(T(N_(2))))`, `sqrt(7)=sqrt((T(H_(2)))/(T(N_(2)))xx(28)/(2)),7=(T(H_(2)))/(T(N_(2)))xx14` or `T(N_(2))=2T (H_(2))` i.e., `T(N_(2)) gt T(H_(2))" or "T(H_(2)) lt T(N_(2))`. |
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| 611. |
The temperature and pressure in Chnadigarh are `35^(@)C` and 740 mm, respectively, whereas at shimla these are `10^(@)C` and 710 mm, respectively. Calculate the ratio of the densities, `d_(1)` and `d_(2)` of air at chandigarh and at shimla. |
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Answer» 0.957:1, `dprop1//V` `therefore(d_(1))/(d_(2))=(V_(2))/(V_(1))=(P_(1)T_(2))/(P_(2)T_(1))` |
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| 612. |
A graph is plotted between `log V` and `log T` for `2 mol` of gas at constant pressure of `0.0821 atm`. `V` and `T` are in litre and `K`. Which of the following statements are not correct? (`I`) The curve is straight line with slope `-1`. (`II`) The curve is straight line with slope `+1`. (`III`) The intercepet on `Y-`axis is equal to `2`. (`IV`) The intercepet on `Y-`axis is equal to `0.3010`.A. `I`, `II`B. `III`, `IV`C. `II,IV`D. `I`, `III` |
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Answer» `PV=nRT` or `logV=logT+log(nR)/(P)` Slope `=tan theta= tan 45^(@)=1` Intercept `= log(nR)/(P)=log[(2xx0.0821)/(0.0821)]=0.3010` |
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| 613. |
A sample of gas at `35^(@)C` and `1 atm` pressure occupies a volume of `3.75` litres. At what temperature should the gas be keep if it si required should the gas be keep if it is required to reduce the volume to `3` litres a the same pressure:A. `-26.6^(@)C`B. `0^(@)C`C. `3.98^(@)C`D. `28^(@)C` |
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Answer» Correct Answer - A `V propT implies (V_(1))/(V_(2))=(T_(1))/(T_(2))` `T_(2)=(3)/(3.75)xx3.8implies 246.4Kimplies T_(2)= -26.6^(@)C` |
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| 614. |
What is the density of `N_(2)` gas at `227^(@)C and 5.00 atm` pressure? `(R= 0.0821 atm K^(-1)mol^(-1))`A. `0.29 g//ml`B. `1.40 g//ml`C. `2.81g//ml`D. `3.41 g//ml` |
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Answer» Correct Answer - D Given `P= 5.00 atm, R= 0.0821 atm K^(-1)mol^(-1)` `T=227+273= 500 K, M_(N_(2))=28` Density`=(PM)/(RT)= (5xx28)/(0.0821xx500)= 3.41 g//ml` |
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| 615. |
In a flask of volume V litres, 0.2 mol of oxygen 0.4 mol of nitrogen, 0.1 mole of `NH_(3)` and 0.3 mol of He are enclosed at `27^(@)C`. If the total pressure exerted by these non reacting gases is one atmosphere, the partial pressure exerted by nitrogen isA. 0.1 atmosphereB. 0.2 atmosphereC. 0.3 atmosphereD. 0.4 atmosphere |
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Answer» Correct Answer - D `P_(N_(2))=(0.4)/(0.2+0.4+0.1+0.3)xx1=0.4` atm (`because` Partial pressure = mole fraction `xx` total pressure) |
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| 616. |
Which of the following statement is false?A. The product of pressure and volume of fixed amount of a gas is independent of temperatureB. Molecules of different gases have the same `KE` at a given temperatueC. The gas equation is not valid at high pressure and low temperatureD. The gas constant per molecule is known as Boltzmann constant |
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Answer» Correct Answer - A `PV=` constant at constant temperature. As temperature changes, the value of constant also changes. |
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| 617. |
Van der Waals real gas acts an ideal gas at which conditions?A. High temperature, low pressureB. Low temperature, high pressureC. High temperature, high pressureD. Low temperature, low pressure |
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Answer» Correct Answer - A At higher temperature and low pressure real gas acts as an ideal gas and obey pV = nRT relation. |
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| 618. |
Van der Waals real gas acts an ideal gas at which conditions?A. High temperature, low pressureB. Low temperature, high pressureC. High temperature, high pressureD. Low, temperature, low pressure |
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Answer» Correct Answer - A At higher temperature and low pressure real gas acts as an ideal gas and obey `PV= nRT` relation. |
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| 619. |
If `v` is the volume of one molecule of a gas under given conditions, then van der Waals constant `b` isA. `4v`B. `4v//N_(0)`C. `N_(0)//4v`D. `4vN_(0)` |
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Answer» `b` is equal to `4 times` the volume of molecules in one mole of a gas (`N_(0)` molecules) Volumes of one molecule `=v` Volume of `N_(0)` molecule `=vN_(0)` Hence, `b=4vN_(0)` |
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| 620. |
A cylinder containing nitrogen gas and same liquid water at a temperature of `25^(@)C`. The total pressure n the cylinder is 600 mm. The piston is moved into the cylinder til the volume is halved keeping the temperature constant. If the aqueous tension at `25^(@)C` is 23.8 mm, calculate the final total pressure in the cylinder. |
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Answer» `p_(N_(2))=600-23.8=576.2mm` `p_(N_(2))` (after piston is moved) `=2xx576.2=1152.4mm` total pressure `=1152.4+23.8=1176.8` mm |
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| 621. |
Which of the following characterstics the critical point?A. At the critical point both liquid and solid phase coexistB. At the critical point, solid, liquid and gas phase coexistC. At the critical point liquid and gas phase coexist togetherD. At the critical point liquid and gas phase have unequal density. |
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Answer» Correct Answer - C At the critical point liquid and gas phase coexist together with equal density. |
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| 622. |
For hydrogen gas, Z is ......unity at all pressure. |
| Answer» Correct Answer - Greater than | |
| 623. |
The value of compressibility factor at the critical state the gas matches with the `Z_(c )` isA. `CH_(4), Z_(c )=0.29`B. `CF_(4), Z_(c )= 0.375`C. `CH_(3)CN, Z_(c )=0.29`D. `H_(2)O, Z_(c )= 0.35` |
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Answer» Correct Answer - A For the symmetrical and non-polar molecular molecule the experimental value of `Z_(c )` is approximately equal to `0.29`. |
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| 624. |
Volume occupied by gas at `T_(c )` and `P_(c )` is called........ |
| Answer» Critical volume `V_(c )` | |
| 625. |
If `bar(V)` is the observed molor volume of real gas and `bar(V)_(id)` is the molar volume of an ideal gas, then `Z` isA. `bar(V)bar(V)_(id)`B. `(barV)/(bar(V)_(id))`C. `(barV)/(bar(V)_(id))`D. `(barV)/(bar(V)_(id))` |
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Answer» Correct Answer - B For ideal gas `(PbarV_(id))/(RT)=1` `:. bar(V)_(id)=(RT)/(P)` For real gas, `Pbar(V)=RT` `(Pbar(V))/(RT)=Z " " (bar(V))/(bar(V)_(id))=Z` |
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| 626. |
Regarding the van der Waals constant which of the following is/are correct?A. `'a'` depends on the intermolecular interactionsB. `'b'` depends on the size of the gas moleculesC. `'a'` and `'b'` are the characterstic constant not the universal gas constantD. All of the above are correct |
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Answer» Correct Answer - D In van der Waals equation `(P+(am^(2))/(V^(2)))` `(V-nb)=nRT` `'a' & 'b'` are the characteristic constant not the universal gas constant and depends on the intramolecular interactions and size of the gas molecule respectively |
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| 627. |
The van der Waals equation for one mol of `CO_(2)` gas at low pressure will beA. `(P+(a)/(V^(2)))V=RT`B. `P(V-b)=RT-(a)/(V^(2))`C. `P=(RT)/(V-b)`D. `P=((RT)/(V-b)-(a)/(V^(2)))` |
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Answer» `(P+(a)/(V^(2)))(V-b)=RT` At low pressure, volume is high and `b` may be ignored. So van der Waals equation becomes `(P+(a)/(V^(2)))V=RT` Hence, the answer is (`a`). |
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| 628. |
The gas molecule can be liquefied and solidified due to the pressure of ...... Force of attraction. |
| Answer» Correct Answer - Intermolecular | |
| 629. |
Which of the following is the correct set of volume calculated by ideal gas equation and van der Waals equation respectively for `1` mole `CO_(2)` gas at `300 K` and `10 atm` pressure. `(R= 0.0821 L atm K^(-1)mol^(-1))`A. `2.463L, 2.56L`B. `2.463L, 2.38L`C. `2.463L, 2.463L`,D. `2.463L, 2.5L` |
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Answer» Correct Answer - B `V_(I)=(nRT)/(p)=(1xx0.0821xx300)/(10)= 2.463L` For `CO_(2)`, at `10 atm Zlt1` `implies (V_(R))/(V_(I))lt1 implies V_(R) lt V_(I)` |
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| 630. |
The ratio of molar volume to ideal molar volume is called ....... |
| Answer» compressibility factor, `Z = (V_(m))/(V_("ideal"))` | |
| 631. |
The volume of 0.0168 mol of `O_(2)` obtained by decomposition of `KClO_(3)` and collected by displacement of water is 428 ml at a pressure of 754 mm Hg at `25^(@)C`. The pressure of water vapour at `25^(@)C` isA. 18 mm HgB. 20 mm HgC. 22 mm HgD. 245 mm Hg. |
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Answer» Correct Answer - D Pressure of dry gas, P=nRT/V `=0.0168xx(0.0821xx1000xx760" ml "mm)xx(298)/(428)` 730 mm Pressure of moist gas=754 mm. Hence, pressure of water vapour =(754-730)mm=24 mm. Alternatively, volume of `0.0168" mol"` of `O_(2)` at STP `=0.0168xx22400=376.3" ml"` Thus, `V_(1)=376.3" ml", P_(1)=760" mm", T_(1)=273" K"` `V_(2)=428" ml",P_(2)=?,T_(2)=298" K"`. Calculate `P_(2)`. |
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| 632. |
What is the pressure of `2 "moles"` of `NH_(3)` at `27^(@)C` when its voulume is `5 "litre"` in van der waals equation `(a= 4,17,b= 0.03711)`?A. `10.33 atm`B. `9.33 atm`C. `9.74 atm`D. `9.2 atm` |
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Answer» Correct Answer - B `(P-(n^(2)a)/(V^(2)))(V-nb)=nRT` `(P-((2)^(2)xx4.17)/((5)^(2)))(5-2xx.03711)` `=2xx.0821xx300` `P=(2xx.0821xx300)/(5-2xx0.3711)-(4.7xx2^(2))/(5^(2))` `implies 10-0.66= 9.3 atm` |
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| 633. |
Carbon dioxide is heavier than oxygen and nitrogen but it does not form the lower layer of the atmosophere. Exoplain. |
| Answer» The diffusion of the gases is quite independent of the gravitational pull by the earth. The molecules of carbon dioxide remain distributed throughout the amosphere. Therefore, carbon dioxides does not form the lower layer of the atmosphere. | |
| 634. |
What is the pressure of 2 mole of `NH_(3)` at `27^(@)C` when its volume is 5 litre in van der Waals equation ? (a=4.17, b=0.03711)A. 10.33 atmB. 9.33 atmC. 9.74 atmD. 9.2 atm |
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Answer» Correct Answer - B `(P+(an^(2))/(V))(V-nb)=nRT` or `P=(nRT)/(V-nb)-(an^(2))/(V^(2))` `=(2xx0.0821xx300)/(5-2xx0.03711)-(4.17xx2^(2))/(5^(2))` `10-0.66=9.33" atm"` |
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| 635. |
`CO_(2)` is heavler than `n_(2)` and `O_(2)` gases present in the atmosphere, yet it does not form the lower layer of the atmosphere. Why ? |
| Answer» Gases prossess the property of diffusion which is independent of the force of gravitation. As a result of diffusion, gases mix with each other and remain almost uniformly distributed in the atmosphere. | |
| 636. |
The circulation of blood in human body supplies `O_(2)` and releases `CO_(2)`. The concentration of `O_(2)` and `CO_(2)` is variable but on the average, 100 mL blood contains 0.02 g of `O_(2)` and 0.08 g of `CO_(2)`. Calculate the volume of `O_(2)` and `CO_(2)` at 1 atmosphere and body temperature of `37^(@)C` assuming 10 L of blood in human body. |
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Answer» 100 mL blood contains `0.02 g O_(2)` and `0.08 g CO_(2)` `:.` 10 L=10,000 mL blood will contain `O_(2)=(0.02)/(100)xx10000=2g" and " CO_(2)=(0.08)/(100)xx10000=8g` `PV=nRT=(w)/(M)RT` `:. " For "O_(2)," "1xxV=(2)/(32)xx0.0821xx310 " or "V_(o_(2))=1.59 L` For `CO_(2), " " 1xxV=(8)/(44)xx0.0821xx310 " or "V_(CO_(2))=4.62 L` |
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| 637. |
The circulation of blood in human body supplies `O_(2)` and releases `CO_(2)` The concentration of `O_(2)` and `CO_(2)` is variable but on the average `100mL` blood contains `0.02 g` of `O_(2)` and `0.08 g` of `CO_(2)` Calcultate the volume of `O_(2)` and `CO_(2)` at 1 atm and body temperature `37^(@)C` assuming 10 litre blood in human body .A. `2 L, 4 L`B. `1.5L, 4.5 L`C. `1.59 L, 4.62 L`D. `3.82 L`, `4.62 L` |
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Answer» Correct Answer - C `:. 100 ml` blood has `0.02gO_(2)` and `8gCO_(2)` `:. 10,000 ml` blood has `2gO_(2)` and `8gCO_(2)` Using `PV= nRT` for `O_(2), 1xxV_(O_(2))=(2)/(32)xx0.0821xx310` `implies V_(o_(2))= 1.59 Litre` for `CO_(2), 1xxV_(CO_(2))=(8)/(44)xx0.0821xx310` `implies V_(O_(2))= 4.62 Litre` |
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| 638. |
What is the relationship between the average velocity `(v)`, root mean square velocity `(u)` and most probable velocityA. `alpha:v:u::1:1.128:1.224`B. `alpha:v:u::1.128:1:1:1.224`C. `alpha:v:u::1.128:1.224:1`D. `alpha:v:u::1.124:1.228:1` |
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Answer» Correct Answer - A `V_(av):V_(rms):V_(most)` probable `= V:U:alpha` `sqrt((8RT)/(pi M)):sqrt((3RT)/(M)):sqrt((2RT)/(M))` `alpha:V:U = sqrt(2):sqrt((8)/(pi)):sqrt(3) = 1:1.128:1.224` |
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| 639. |
`3.2 g` of S is heated to occupy a volume of `780 ml` at `450^(@)C` and `723 mm` pressure. Formula of sulphure isA. `S_(2)`B. `S`C. `S_(4)`D. `S_(8)` |
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Answer» Correct Answer - D `PV=(w)/(M)RT` `(723)/(760)xx(780)/(1000)=(3.2)/(M)xx0.0821xx683` `M=256` `:.` Atomicity `=(256)/(32)=8` |
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| 640. |
A gas can be liquefied by pressure alone when its temperatureA. above its critical temepratureB. at its critical temperatureC. below its critical temperatureD. at any temperature |
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Answer» Correct Answer - C The temperature below which the gas can be liquefied by the application of pressure alone is called critical temperature. |
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| 641. |
A gas can be liquefied by pressure alone when its temperatureA. higher than its critical temperatureB. lower than its critical temperatureC. either of theseD. none |
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Answer» Correct Answer - B A gas cab be liquefied only if its temperature is lower than its critical temperature. |
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| 642. |
Calculate the volume occupied by 4 mole of an ideal gas at `2.5 xx 10^(5) Nm^(-2)` pressure and 300 K temperature |
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Answer» Correct Answer - `39.9 dm^(3)` `V = (nRT)/(P)`, Given, `n = 4, R = 8.314 Nm K^(-1) mol^(-1)` `T = 300, P = 2.5 xx 10^(5) Nm^(-2)` |
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| 643. |
The numerical value of b is ......times the actual volume occupied by one mole of gas molecule. |
| Answer» Correct Answer - Four | |
| 644. |
Which of the following is not a crystalline solid ?A. KClB. CsClC. GlassD. Rhombic sulphur |
| Answer» (c ) Glass is an amorphous solid. | |
| 645. |
Assertion: Gases are easily absorbed on the surface of metals, especially transition metals. Reason: Transition metals have free valenciesA. 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 Because of free valencies of transition metal, gases easily gets abosorbed on the surface of metal. |
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| 646. |
An excess of potassium ions makes KCl crystals appear violet or lilac in colour sinceA. some of the anionic sites are occupied by an unpaired electronB. some of the anionic sites are occupied by a pair of electronsC. there are vacancies at some anionic sitesD. F-centres are created which impart colour to the crystals |
| Answer» Correct Answer - a,d | |
| 647. |
Lithium forms body centred cube structure .The length of the side of its unirt cell is 351 pm Atomic radius of the lithium will beA. 240.8 pmB. 151.8 pmC. 75.5 pmD. 300.5 pm |
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Answer» Correct Answer - B In case of body centred cubic (bcc) crystal, `asqrt3=4r` Given, edge length, a=351 pm Hence, atomic radius of lithium, `r=(asqrt3)/(4)=(351xx1.732)/(4)` `=151.98" pm"` |
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| 648. |
If `20 cm^(3)` gas at `1 atm` is expanded to `50 cm^(3)` at constant `T`, then what is the final pressureA. `20xx(1)/(50)`B. `50xx(1)/(20)`C. `1xx(1)/(20)xx50`D. None of these |
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Answer» Correct Answer - A At constant `T, P_(1)V_(1)=P_(2)V_(2)` `1xx20= P_(2)xx50, P_(2)=(20)/(50)xx1` |
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| 649. |
Virial equation is: `PV_(M)=RT[A+(B)/(V_(M))+(C )/(V_(M^(2)))+…]`, where `A`, `B`, `C`, …. are first second,third, … virial coefficent, respectively, For an ideal gasA. `A=` unity and `B`, `C` are zero.B. `A`, `B`, `C` are all equal to unity.C. `A` is dependent of temperature.D. All `A`, `B`, `C` depend on temperature. |
| Answer» `PV=RT` for ideal gases. | |
| 650. |
Rate of diffusion is proportional to......... |
| Answer» `r prop sqrt((1)/(d_(1)))` or `r_(1) prop sqrt((1)/(M_(1)))` | |