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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.
351. |
A small quantity of gaseous `NH_(3)` and HBr are introduced simultaneously into the opposite ends of an open tube which is one metre long. Calculate the distance of the white solid `NH_(4)Br` formed from the end which was used to introduce `NH_(3)`. |
Answer» Correct Answer - C Similar to Solved Problem 5. |
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352. |
The reaction between gaseous `NH_(3)` and `HBr` produces a white solide `NH_(4)Br`. Suppose a small quantity of gaseous `NH_(3)` and gaseous `HBr` are introduced simultaneously into opposite ends of an open tube which is one metre long. Calculate the distance of white solid formed from the end which was used to introduce `NH_(3)` |
Answer» Let the distance of white solid from `NH_(3) " end" x = cm` The distance of which solid from `HBr " end " = (100 - x) cm`. Rates of diffusion shall be proportional to these distances. `(r_(1))/(r_(2)) = (x)/((100 - x)) = sqrt((M_(HBr))/(M_(NH_(3))))` Mol. mass of `HBr = 1 + 80 = 81` Mol. mass of `NH_(3) = 14 + 3 = 17` So, `(x)/((100 - x)) = sqrt((81)/(17))` or, `(x)/((100 - x)) = 2.18` So, `x = 100 xx 2.18 - 2.18 x` So, `x = (100 xx 2.18)/(3.18) = 68.55 cm` |
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353. |
Mass of `112 cm^(3)` of `NH_(3)` gas at S.T.P. isA. `0.085 g`B. `0.850 g`C. `8.5 g`D. `80.5 g` |
Answer» Correct Answer - A Mass of `112 cm^(3)` of `NH_(3)` at S.T.P. `= ((112 cm^(3)))/((22400 cm^(3))) xx (17g) = 0.085 g`. |
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354. |
If the four tubes of a car are filled to the same pressure with `N_(2),O_(2) , H_(2)`, and helium separately , then which one will be filled first ?A. `N_(2)`B. `O_(2)`C. `H_(2)`D. `He` |
Answer» Correct Answer - C Under identical conditiosn of `T` and `p`, the rates of effusion for gaseous substances are inversely proportional to the square roots of their molar masses: `r prop (1)/(sqrt(M))` Lower the molecular mass , faster it wil effuse out and will be filled first. Thus , `H_(2)` gas with the lowest molecular mass is filled first. |
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355. |
If the four tubes of a car are filled to the same pressure with `N_(2),O_(2)H_(2)` and Ne separately, then which one will be filled first ?A. `N_(2)`B. `O_(2)`C. `H_(2)`D. Ne |
Answer» Correct Answer - A::B::C::D |
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356. |
The rate of diffusion of methane at a given temperature is twice that of a gas `X`. The molecular weight of `X` isA. 64B. 32C. 4D. 8 |
Answer» Correct Answer - A |
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357. |
Ordinary rubber filled with hydrogen get deflated after sometimes. Assign reason. |
Answer» Rubber is an elastic material which is of porous nature. However, the pores are so small in size tha they are not visible. In an inflated rubber balloon, the presence inside the balloon is gerater than the atmospheric pressure prevailing outside. As a result, `H_(2)` molecular will slowly effuse of the balloon leading to contraction in volume or size of the balloon. | |
358. |
Which of the following gases has the least value of critical temperature `(T_(c ))` and critical pressure `(p_(c ))` ?A. `He`B. `H_(2)`C. `N_(2)`D. `CO_(2)` |
Answer» Correct Answer - A Because `He` atoms poses the smallest value of `a` which measure the strngth of intermolecular interaction. |
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359. |
An element with molar mass 27 g `mol^(-1)` forms a cubic unit cell with edge length `4.05 xx 10^(-8)` cm. If its density is `2.7 g cm^(-3)`, what is nature of cubic unit cell ? |
Answer» Correct Answer - FCC `rho=(ZxxM)/(a^3xxN_A)` or `Z=(rhoxxa^3xxN_A)/M=((2.7 g cm^(-3))(4.05xx10^(-8)cm)^3(6.02xx10^23 mol^(-1)))/(27 g mol^(-1))=4` Hence, it has face-centred cubic unit cell. |
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360. |
A volume of hydrogen measures one cubic decimetre at `20^(@)C` and at a pressure of half an atmosphere. What will be its volume at `10^(@)C` and at 700 mm pressure ? |
Answer» Correct Answer - B::C::D | |
361. |
In `A^(+)B^(-)` ionic compound, radii of `A^(+) and B^(-)` ions are 180 and 187pm respectively. The crystal structure of the compound will beA. NaCl typeB. CsCl typeC. ZnS typeD. similar to diamond |
Answer» (b) `(r^(+))/(r^(-))=(180)/(187)=0.96`, this value lies in the range of 0.732-1.000 | |
362. |
A compound having bcc geometry has atomic mass 50. Calculate the density of the unit cell, if the edge length is 290pm.A. `6.81gcm^(-3)`B. `3.40gcm^(-3)`C. `13.62gcm^(-3)`D. none of these |
Answer» (a) density of unit cell, `d=(ZxxM)/(N_(A)xxa^(3))` `d=(2xx50)/((290xx10^(10))^(3)xx6.023xx10^(23))=6.81g//cm^(3)` |
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363. |
Calculate the approxiimate number of unit cells present in 1g of gold. It is well known that gold crystallises in the face cubic lattice (atomic mass of gold is 197u).A. `7.64xx10^(20)`B. `6.02xx10^(23)`C. 197D. 4 |
Answer» (a) 1 mole of gold=197g=`6.02xx10^(23)` atoms. Number of atoms available in 1g of gold =`(6.02xx10^(23))/(197)` As fcc unit cell contains 4 atoms. Number of unit cells present=`(6.02xx10^(23))/(197xx4)=7.64xx10^(20)` |
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364. |
Calculate the volume of hydrogen librated at N.T.P. when `500 cm^(3)` of `0.5 N` sulphuric acid react with excess of zinc. |
Answer» Correct Answer - `2800 cm^(3)` Step I. Mass of sulphuric acid in solution `"Normality" (N) = (("Mass of" H_(2)SO_(4))/("Equivalent mass"))/(("Volume of solution in" cm^(3))/(100))` or `(0.5 "equiv" dm^(-3)) = (W)/(((49 "equiv"^(-1)))/((0.5 dm^(3))))` `W = (0.5 "equiv" dm^(-3)) xx (49 g "equiv"^(-1)) xx (0.5 dm^(3)) = 12.25 g`. Step II. Volume of hydrogen liberarted at N.T.P. The chemical equation for the reaction is : `{:(Zn+H_(2)SO_(4),rarrZnSO_(4)+,H_(2)),(2+32+64,,22400cm^(3)),(=98g,,):}` `98g` of `H_(2)SO_(4)` evolve `H_(2)` at N.T.P. `22400 cm^(3)` `12.25 g` of `H_(2)SO_(4)` evolve `H_(2)` at N.T.P. `= 22400 (cm^(3)) xx ((12.25 g))/((98g)) = 2800 cm^(3)` |
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365. |
Which of the following statements are correct ?A. Ferrimagnetic substances lose ferrimagnetism on heating and become paramagneticB. Ferrimagnetic substances do not lose ferrimagnetism on heating and remain ferrimagneticC. Antiferromagnetic substances have domain structures similar to ferromagnetic substances and their magnetic moments are not cancelled by each otherD. In ferromagnetic substances , all the domains get oriented in the direction of magnetic field and remain as such even after removing magnetic field |
Answer» Correct Answer - a,d | |
366. |
Under what conditions does the behavior of real gases deviate most from that predicted by the ideal gas law? |
Answer» Real gases show maximum deviation from the ideal gas behaviour when the pressure of the gas is very high and its temperature is very low. | |
367. |
The ratio of the root mean square velocity of `H_(2)` at `50 K` to that of `O_(2)` at `800K` isA. `4`B. `2`C. `1`D. `(1)/(4)` |
Answer» `mu_(ms)=sqrt((3RT)/(M)), mu_(ms) prop sqrt((T)/(M))` `:. (mu_(ms)(H_(2) at 50 K))/(mu_(ms)(O_(2) at 800 K))=sqrt((50)/(2)xx(32)/(800))=1` So the correct choice is (`c`). |
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368. |
An element (with atomic mass=250 g) crystallises in a simple cube. If the density of unit cell is 7.2 g `cm^(-3)` , what is the radius of the element ?A. `1.93xx10^(-6)` cmB. `1.93xx10^(-8)` cmC. `1.93xx10^(-8) Ã…`D. `1.93 xx 10^(-8)`m |
Answer» Correct Answer - B `Z=(l^(3)xxrhoxxN_(A))/(M)` `1=(l^(2)xx7.2xx6.023xx10^(23))/(250)` `l=3.86xx10^(-8)` cm `l=2r` (for simple cubic unit cell) `r=(l)/(2)=(3.86xx10^(-8))/(2)=1.93xx10^(-8)` cm |
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369. |
The correct order to the packing efficiency in different types of unit cells is _______A. fcc lt bcc lt simple cubicB. fcc gt bcc gt simple cubicC. fcc lt bcc gt simple cubicD. bcc lt fcc gt simple cubic |
Answer» Correct Answer - b fcc =0.74 , bcc=0.68 , simple =0.524 . Hence , fcc gt bcc gt simple |
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370. |
A solid `A^+B^-` has NaCl type close packed structure .If the anion has a radius of 241.5 pm , what should be the ideal radius of the cation ? Can a cation `C^+` having radius of 50 pm be fitted into the tetrahedral hole of the crystal `A^+ B^-` ? |
Answer» As `A^+ B^-` has NaCl structure, `A^+` ions will be present in the octahedral voids. Ideal radius of the cation will be equal to the radius of the octahedral void because in that case. It will touch the anions and the arrangement will be close packed. Hence, Radius of the octahedral void =`r_(A^+)=0.414 xxr_(B^-)=0.414xx241.5` pm =100.0 pm Radius of the tetrahedral void =`0.225 xxr_(B^-)` =0.225 x 241.5 pm =54.3 pm As the radius of the cation `C^+` (50 pm) is smaller than the size of the tetrahedral void, it can be placed into the tetrahedral void (but not exactly fitted into it) |
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371. |
A solid AB has NaCl structure. If the radius of cation `A^+` is 170 pm, calculate the maximum possible radius of the anion `B^-` |
Answer» Correct Answer - 410.6 pm The formula `r_+=0.414xxr_(-)` gives maximum possible radius of anion for a given cation for close packing. |
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372. |
The ionic radii of `A^+` and `B^-` ions are `0.92xx10^(-10)` m and `1.81xx10^(-10)` m . The coordination number of each ion in AB isA. 2B. 6C. 4D. 8 |
Answer» Correct Answer - B Radius ratio , `r_(+)/r_(-)=(0.98xx10^(-10))/(1.81xx10^(-10))=0.541` It lies in the range 0.414-0.732 . Hence, coordination number of each ion will be 6, as in the case of NaCl |
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373. |
Which kind of defect are introduced by doping ?A. Dislocation defectB. Schottky defectC. Frenkel defectD. Electronic defects |
Answer» Correct Answer - d Doping is done with electron rich or electron deficit impurities. Hence , it introduces electronic defects. |
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374. |
By X-ray studies, the packing of atoms in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly identical . The density of gold is found to be `19.4 g cm^(-3)` and its atomic mass is 197 a.m.u. The coordination number of gold atom in the crystal isA. 4B. 6C. 8D. 12 |
Answer» Correct Answer - D The arrangement of layers is ABC ABC…. Type, i.e., it has cubic close packing arrangement . Hence, the coordination number is 12. |
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375. |
Which of the following has highest surface tension ?A. WaterB. Soap solution in waterC. Detergent solution in waterD. Glycerol in water |
Answer» Correct Answer - D Glycerol in water has highest surface tension due to H-bonding. |
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376. |
Which of the following phenomena are caused due to the characteristic property of liquids called surface tension ?A. Small drops of mercury form spherical beads instead of spreading on the surface.B. Particles of soil at the bottom of river remain separated but they stick together when taken out.C. A liquid rise (or fall) in a thin capillary as soon as the capillary touches the surface of the liquid.D. All of these. |
Answer» Correct Answer - D Liquids tend to minimize their surface area , i.e., liquids tend to thave minimum number of molecules at their surface. This is because the molecules on the surface experience a net downward force and have more energy than the molecules in the bulk, which do not experience any net force. Moist soil grains are pulled together because the surface area of thin film of water is reduced. |
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377. |
Which of the following phenomena does not involve surface tension ?A. Mercury drops acquire spherical shape .B. Liquids tend to rise in the capillary .C. A liquids flows over a fixed surface .D. Moist soil grains are pulled together . |
Answer» Correct Answer - C Liquids tend to minimise their surface due to surface tension. |
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378. |
Which of the following is not physical characteristic of gases ?A. Densities of gases can be increases by applying increased pressure.B. Pressure must be xerted to confine gases.C. Different gases in a mixture do not separate on standing .D. A sample of a gas contains more molecules when hot than it contains when cold at the same pressure. |
Answer» Correct Answer - D A sample of a gas occupies a greater volume when hot than it does when cold at the same pressure, but the number of molecules does not change. |
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379. |
The r.m.s. velocity of molecules of a gas of density 4 kg `m^(-3)` and pressure `1.2xx10^(5)Nm^(-2)` isA. `900 ms^(-1)`B. `120 m s^(-1)`C. `600 m s^(-1)`D. `300 m s^(-1)` |
Answer» Correct Answer - A::C::D |
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380. |
The `R.M.S`. Speed of the molecules of a gas of density `kg m^(-3)` and pressure `1.2xx10^(5)Nm^(-2)` is:A. `120 ms^(-1)`B. `300 ms^(-1)`C. `600 ms^(-1)`D. `900 ms^(-1)` |
Answer» Correct Answer - B `U_(rms)=sqrt((3PV)/(M))=sqrt((3P)/(d))` `=sqrt((3xx1.2xx10^(5))/(4))= 300 m//sec` |
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381. |
The pressure of a gas can be expressed in many different units. The `SI` unit of pressure isA. pascalB. torrC. barD. atm |
Answer» Correct Answer - A It is defined as the pressure exerted by a force of one newton acting on an area of one square meter. |
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382. |
When a gas is compressed as constant temperature:A. the speeds of the molecule increaseB. the collisions between the molecules increaseC. the speeds of the molecules decreaseD. the collisions between the molecule decrease |
Answer» Correct Answer - B Frequency of collision will increase. |
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383. |
At sea leval, at a latitude of `45^(@)` , the average atmospheric pressure supports a column of mercury ……… mm high in a single mercury barometter when the mercury is at `0^(@)C`.A. `700`B. `760`C. `800`D. `786` |
Answer» Correct Answer - B This average sea leval pressure of `760 mm Hg` is called one atmosphere of pressure. |
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384. |
What is the pressure in atmosphers if the barometer reading is `688 mm Hg`?A. `0.905 atm`B. `0.986 atm`C. `1 atm`D. `0.876 atm` |
Answer» Correct Answer - A Since `1 atm = 760 mm Hg` , the pressure in atmospheres is calculated as follows: `Pressure = 688 mm Hg xx (1 atm)/(760 mm Hg)` ` = 0.905 atm` |
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385. |
Cations are present in the interstitial sites in …………… .A. Vacancy defectB. Frenkel defectC. Metal deficiency defectD. Schottky defect |
Answer» Correct Answer - B In Frenkel defect, few cations are displaced from their normal lattice points to occupy interstitial site. |
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386. |
Cations are present in the interstitial sites in ______A. Frenkel defectB. Schottky defectC. Vacancy defectD. Metal deficiency defect |
Answer» Correct Answer - a | |
387. |
Which of the following statements is not correct ?A. The fraction of the total volume unoccupied by the atoms in a primitive cell is 0.48B. Molecular solids are generally volatileC. The number of carbon atoms in a unit cell of Diamond is 4D. The number of Bravais lattices in which a crystal can be categorized is 14. |
Answer» Correct Answer - C The number of C-atoms in a unit cell of diamond is 8 and not 4 |
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388. |
If the positions of `Na^+` and `Cl^-` are interchanged in NaCl , having fcc arrangement of `Cl^-` ions then in the unit cell of NaClA. `Na^+` ions will decrease by 1 while `Cl^-` ions will increase by 1B. `Na^+` ions will increase by 1 while `Cl^-` ions will decrease by 1C. Number of `Na^+` and `Cl^-` ions will remain the sameD. The crystal structure of NaCl wil change |
Answer» Correct Answer - B In NaCl with fcc arrangement of `Cl^-` ions , number of `Cl^-` ion =14 , `Na^+` ions =13. On interchanging their positions, `Cl^-` ions will be 13 and `Na^+` ions will be 14 |
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389. |
A solid AB has the NaCl structure. If radius of the radius of the cation `A^(+)` is 120 pm, calculate the maximum value of the radius of anion `B^(-)`. |
Answer» Since, NaCl has octahedral structure, The limiting ration `(r_(A)^(+))/(r_(B)^(-))=0.414` or `r_(B)^(-)=(r_(A)^(+))/(0.414)=(120)/(0.414)=290` pm |
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390. |
1500 mL flask contains `400 mg O_(2) " and " 60 mg H_(2) " at " 100^(@)C`. (a) What is the total pressure in the flask ? (b) If the mixture is permitted to react to form water vapour at `100^(@)C`, what will be left and what will be their partial pressure ? |
Answer» (a) No. of moles of `O_(2) = (400)/(1000 xx 32) = 0.0125` No. of moles of `H_(2) = (60)/(1000 xx 2) = 0.03` Partial pressure of `O_(2) = (0.0125 xx 0.0821 xx 373)/(1.5) = 0.255 atm` Partial pressure of `H_(2) = (0.03 xx 0.0821 xx 373)/(1.5) = 0.612 atm` Total pressure `= 0.255 + 0.612 = 0.867 atm` (b) `{:(,2H_(2),+O_(2),= 2H_(2) O),("Initital",0.03,0.0125," "0),("After reaction",0.005,0," "0.025):}` Partial pressure of `H_(2) = (0.005 xx 0.0821 xx 373)/(1.5) = 0.102 atm` Partial pressure of `H_(2) O = (0.025 xx 0.0821 xx 373)/(1.5) = 0.51 atm` |
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391. |
A gas occupies a volume of 2.5 L at `9xx10^(5)N m^(-2)`. Calculate the additional pressure required to decrease the volume of the gas to 1.5 L, keeping the temperature constant. |
Answer» Correct Answer - A::B `V_(1)=2.5 L, P_(1)=9xx10^(5) N m^(-2) , V_(2)=1.5 L, P_(2)`=? Apply `P_(1)V_(1)=P_(2)V_(2)`. Calculate `P_(2)`. We get `P_(2)=15xx10^(6) N m^(-2)` Additional pressure required `=15xx10^(6)N m^(-2)-9xx10^(5)N m^(-2)=6xx10^(5)N m^(-2)`. |
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392. |
3.2 g of oxygen (At. wt. = 16) and 0.2 g of hydrogen (At. wt. = 1) are placed in a `1.12 L` flask at `0^(@)C`. The total pressure of the gas mixture will beA. 1 atmB. 4 atmC. 3 atmD. 2 atm |
Answer» Correct Answer - b `P_(O_(2)) = (n_(O_(2))RT)/(V_(O_(2))) = (3.2 xx 0.0821 x 273)/(32 xx 1.12)` `= 2 atm` `P_(O_(2)) = (n_(H_(2))RT)/(V_(H_(2))) = (0.1 xx 0.0821 xx 273)/(2 xx 1.2)` `= 2` atm Total pressure `= 2 + 2 = 4` atm |
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393. |
Assertion: On compressing a gas to half the volume, the number of molecules is halved. Reason: The number of moles present decreases with decrease in volume.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. |
Answer» Correct Answer - A::B::C::D Number of molecule is independent of pressure, and moles number is independent of volume. |
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394. |
Volume of the air that will be expelled from a vessel of `300 cm^(3)` when it is heated from `27^(@)C` to `37^(@) C` at the same pressure will beA. `310 cm^(3)`B. `290 cm^(3)`C. `10 cm^(3)`D. `37 cm^(3)` |
Answer» Correct Answer - C `(V_(2))/(V_(1))=(T_(2))/(T_(1))` `:. V_(2)=(T_(2))/(T_(1))V_(1)=(310K)/(300K)xx300 cm^(3)= 310cm^(3)` |
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395. |
A 2.0 L container at `25^(@)C` contains 1.25 mol of oxygen and 3.3 mol of carbon. (a) What is the initial in the flask ? (b) If carbon and oxygen and oxygen react as completely as possible to form CO, what will be the final pressure in the container ? |
Answer» (a) V=2.0 L, T=298 K, n=1.25 mol (because only gas will exert pressure). Hence, PV=nRT or `P=(nRT)/(V)=((1.25" mol")(0.0821" L atm "K^(-1)mol^(-1))(298 K)/(2.0" L")=15.3" atm"` (b) The reaction will be :`C(s)+(1)/(2)O_(2)(g) to CO(g)` As 1 mol of C reacts with `(1)/(2)` mol of `O_(2)`, the limiting reactant will be 1.25 mol of `O_(2)`. Thus, `(1)/(2)` mol of `O_(2)` produces CO=1 mol. `:.1.25` mol of `O_(2)` will produce CO=2.50 mol. Now, n(gaseous)=2.50 mol. Hence, `P=(nRT)/(V)=((2.50" mol")(0.0821" L atm "K^(-1)mol^(-1)(298"K"))/(2.0" L ")=30.6" atm "`. |
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396. |
What volume of air will be expelled from a vessel containing `400 cm^(3)` at `7^(@)C` when it is heated to `27^(@)C` at the same pressure? |
Answer» `(V_(1))/(T_(1))=(V_(2))/(T_(2))`, i.e., `(400 cm^(3))/((273+7)K)=(V_(2))/((273+27)K)` or `V_(2)=(400)/(280)xx300 cm^(3)=482.6 cm^(3)` This is the volume after expansion. So Volume expelled `=(428.6-400)cm^(3)=28.6 cm^(3)` |
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397. |
For `1 mol` of an ideal gas, `V_(1)gtV_(2)gtV_(3)` in fig. (`I`),`T_(1)gtT_(2)gtT_(3)` in fig. (`II`), `P_(1)gtP_(2)gtP_(3)` in fig. (`III`), and `T_(1)gtT_(2)gtT_(3)` in fig. (`IV`) , then which curves are correct. A. `I,II`B. `I,II,III`C. `II,IV`D. `I,III,IV` |
Answer» `PV=RT` if `P_(1) gt P_(2)`, `V_(1) lt V_(2)`. Also `log P= -log V+ log RT` |
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398. |
A quantity of hydrogen gas occupies a volume of `30.0 mL` at a certain temperature and pressure. What volume would half of this mass of hydrogen occupy at triple the initial temperature, if the pressure was one-ninth that of the original gas? |
Answer» `{:(,"Initial","Final"),(V, 30mL,V_(2)),(P,P_(1), P_(2)=P_(2)=1//9P_(1)),(T,T_(1),T_(2)=3T_(1)),(n,n_(1),n_(2)=1//2n_(1)):}` Using gas equation, `PV=nRT`, we get `(P_(1)V_(1))/(P_(2)V_(2))=(n_(1)RT_(1))/(n_(2)RT_(2))` `(P_(1)xx30)/((1)/(9)P_(1)xxV_(2))=(nT_(1))/(((1)/(2)n_(1))(3T_(1)))` `V_(2)=(1)/(2)xx3xx9xx30.0=405 mL` |
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399. |
A flask containing 250 mg of air at `27^(@)C` is heated till 25.5% of air by mass is expelled from it. What is the final temperatuer of the flask ? |
Answer» Let the volume of flask be V. Let final temeperature be T(K). Mass of air expelled at `T(K)=(250x25)/(100)=62.5mg` Mass of air contained in flask `=250-62.5=187.5mg` now, volume of total air (250 mg) at higher temperature `V_(2)=(Vxx250)/(187.5)mL` Now `(V_(2))/(T)=(V)/(300)` or `T=(V_(2)xx300)/(V)=(Vxx250xx300)/(187.5xxV)=400K` or `127^(@)C` |
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400. |
A quantity of hydrogen gas occupies a volume of `30.0 mL` at a certain temperature and pressure. What volume would half of this mass of hydrogen occupy at triple the initial temperature, if the pressure was one-ninth that of the original gas?A. `270 mL`B. `90 mL`C. `405 mL`D. `137 mL` |
Answer» `P_(1)V_(1n_(1)RT_(1)` `P_(2)V_(2)=n_(2)RT_(2)` `(P_(2)V_(2))/(P_(1)V_(1))=(n_(2)T_(2))/(n_(1)T_(1))` `(P_(1))/(9P_(1))(V_(2))/(30)=(1)/(2)(n_(1))/(n_(1))(3T_(1))/(T_(1))` `V_(2)=405 mL` |
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