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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.
| 201. |
Assuming `CO_(2)` to be van der Waals gas, calculate its Boyle temperature. Given `a=3.59" L"^(2)" atm " mol^(-2)` and `b=0.0427" L "mol^(-1)`. |
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Answer» Boyle temperature in terms of van der Waals constant in given by `T_(b)=(a)/(Rb)" "` `:. " " T_(b)=(3.59" L"^(2)" atm "mol^(-1))/((0.0821" L atm "K^(-1)mol^(-1))(0.0427" L "mol^(-1)))=1024" K"` |
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| 202. |
Calculate the pressure exerted by 110 g of carbon dioxide in a vessel of 2 L capacity at `37^(@)C`. Given that the van der Waals constants are a=3.59 `L^(2) atm mol^(-2)` and b=0.0427 L `mol^(-1)`. Compare the value with the calculated value if the gas were considered as ideal. |
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Answer» According to van der Waals equation, `(p+(an^(2))/(V^(2)))(v-nb)=nRT "or "P=(nRT)/(V-nb)-(an^(2))/(V^(2))` Here, `n=(110)/(44)=2.5` moles. Substituting the given values, we get `P=((2.5 mol)(0.0821" L atm "K^(-1) mol^(-1)))/((2 L-2.5 molxx0.0427 L mol^(-1)))-((3.59 L^(2) atm" mol"^(-2))(2.5 mol^(-2)))/((2 L)^(2))` `33.61 atm-5.61 atm=28.0` If the gas were considered is ideal gas, applying ideal gas equation, PV=nRT, we get `P=(nRT)/(V)=((2.5 mol)(0.0821" L atm mol"^(-1))(310 K))/(2 L)=31.8 atm` |
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| 203. |
Calculate the pressure of `10^(23)` molecules of sulphur dioxide `(SO_(2))` when enclosed in a vessel of `2.5 L` capacity at a temperature of `27^(2)C`. |
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Answer» Correct Answer - `1.653 xx 10^(-1) atm` According to ideal gas equation : `PV = nRT` , or `P = (nRT)/(V)` `n = (1 xx 10^(22))/(N_(O)) = (1 xx 10^(22))/(6.022 xx 10^(23)) = 1.66 xx 10^(-2) mol, V = 2.5 L` `R = 0.083 L "bar" K^(-1) mol^(-1), T = 27 + 273 = 300 K` `P = ((1.66 xx 10^(-2) mol) xx (0.083 L "bar" K^(-1) mol^(-1)) xx (300 K))/((2.5 L)) = 1.653 xx 10^(-1)` atm |
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| 204. |
Assertion : Pressure of a dry gas at any temperature can be calculated by substracting aqueous tension at that temperature from the pressure of the moist gas. Reason : The pressure of water vapours is called aqueous tension.A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct and reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If assertion and reason are both incorreect. |
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Answer» Correct Answer - B Correct explanation : The molecules of the gas and that of water do not react chemically. |
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| 205. |
The value of gas constant in the units of bar `dm^(3)K^(-1)mol^(-1)` is ………………….. |
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Answer» Correct Answer - 0.083 |
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| 206. |
The critical temperature and reduced temperature of a gas are 150 K and 3 K respectively. What is the temperature of the gas ?A. 50 KB. 147 KC. 153 KD. 450 K |
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Answer» Correct Answer - D `T=thetaT_(c )` `=3xx150" K"=450" K"` |
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| 207. |
2.802 g of `N_(2)` gas is kept in one litre flask at `0^(@)C`. Calculate the pressure exerted by the gas. |
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Answer» Correct Answer - A::B::D 2.802 g of `N_(2)=2.0802//28` mol . Then apply PV=nRT |
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| 208. |
1 litre of `N_(2) " and " 7//8` litre of `O_(2)` at the same temperature and pressure were mixed together. What is the relation between the masses 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)) = 16 M_(O_(2))` |
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Answer» Correct Answer - C `PV = (M)/(m) RT` `P xx 1 = (M_(N_(2)))/(28) RT`....(i) `P xx (7)/(8) = (M_(O_(2)))/(32)RT` ....(ii) Dividing eq. (i) by e. (ii), we get `M_(N_(2)) = M_(O_(2))` |
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| 209. |
Air contains `N_(2)` and `O_(2)` in the ratio of 4:1 by volume. The average conditions of temperature and pressure ?A. `12.0`B. `14.4`C. `15.6`D. `28.8` |
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Answer» Correct Answer - B As under similar conditions of temperature and pressure, equal volumes contain equal number of moles , therefore, molar ratio of `N_(2)=4//5=0.8` and mol fraction of `O_(2)=0.2` `:.` Average vapour density of air `=(28.8)/(2)=14.4` |
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| 210. |
Which of the following has longest mean free path under identical conditions of temperature and pressure ?A. `H_(2)`B. `N_(2)`C. `O_(2)`D. `CO_(2)` |
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Answer» Correct Answer - A Mean free path (l) `(1)/(sqrt(2)pisigma^(2)n), i.e., l prop (1)/(sigma^(2))` where `sigma`=molecular diameter Under identical conditions of T and P,n is same, As `sigma` is lower for `H_(2)`, l will be largest. |
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| 211. |
How do you convert (a) pressure in atmospheres into SI units ? (b) temperature in `""^(@)C` to temperature in `""^(@)F`? |
| Answer» (a) 1 atm.=101,325 Pa or N `m^(-2)` or 1 abr `=10^(5)`Pa (b) `""^(@)C=(5)/(9)(""^(@)F-32)` | |
| 212. |
If volume of the gas is very large, then the second virial coefficient B in virial equation isA. `(b+(a)/(RT))`B. `(b-(a)/(RT))`C. `(b+(a)/(RTV))`D. `(b-(a)/(RTV))` |
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Answer» Correct Answer - B `(P+(a)/(V^(2)))(V-b)=RT` `PV+(a)/(V)-(ab)/(V^(2))-Pb=RT` If V is very large, `(ab)/(V^(2))` can be neglected. Hence, `PV+(a)/(V)-Pb=RT` or `PV=RT+Pb=(a)/(V)=RT+(RT)/(V)b-(a)/(V)` `=RT[1+(1)/(V)(b-(a)/(RT))]` Comparing with the virial equation `PV=RT[1+(1)/(B)+...]` `B=b-(a)/(RT)` |
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| 213. |
The temperature at which the volume of a gas is zero isA. `0^(@)C`B. 0 KC. 0 FD. None of these. |
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Answer» Correct Answer - B |
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| 214. |
The temperature at which the volume of a gas is zeroA. `0^(@)C`B. `0K`C. `0^(@)C`D. None of these. |
| Answer» Correct Answer - B | |
| 215. |
180 g of steam is contained in a vessel of 25 L capacity under a pressure of 50 atm. Calculate the temperature of the steam. Given that for water vapour, `a=5.46" bar" L^(2)mol^(-2)" and "b=0.031 L mol^(-1)`. |
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Answer» Correct Answer - 1531.7 K |
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| 216. |
The critical temperature `(T_(c))` and pressure `(P_(c))` of `NO` are `177 K` and `6.48MPa`, respectively, and that of `CCl_(4)` are `550 K` and `4.56MPa`, respectively. Which gas `(a)` has the smaller value for the van der Waals constant `b`, `(b)` has the smaller value for constant `a`, `(c)` has the larger critical volume, and `(d)` is most nearly ideal in behaviour at `300 K` and `1.013MPa`. |
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Answer» The given values are `{:(T_(c)(NO)=177K, T_(c)(C Cl_(4))=550 K), (P_(c)(NO)=6.485MPa, P_(c)(C Cl_(4))=4.56MPa):}` `a` Since `(P_(c))/(T_(c))=(a//27b^(2))/(8a//27Rb)=(R )/(8b)`, `b=(T_(c)R)/(8P_(c))` Thus, `b(NO)=((177K)(8.314MPadm^(3)K^(-1)mol^(-1)))/((8)(6.485MPa))` `=28.36 cm^(3)mol^(-1)` and `c(C Cl_(4))=((550 K)(8.314MPa cm^(3)K^(-1)mol^(-1)))/((8)(4.56MPa))` `=125.35 cm^(3)mol^(-1)` Hence, `b(NO)ltb(C Cl_(4))`. `b` Since `a=27P_(c)b^(2)`, `a(NO)=(27)(6.485MPa)(28.36 cm^(3)mol^(-1))^(2)` `=140.827 kPa dm^(6) mol^(-2)` `a(C Cl_(4))=(27)(4.56MPa)(125.35 cm^(3)mol^(-1))^(2)` `=1934.538kPa dm^(6) mol^(-2)` Hence, `a(NO)lta(C Cl_(4))`. `c` Since `V_(C)=3b`, `V_(C)(NO)=3xx(28.36 cm^(3)mol^(-1))` `=85.08 cm^(3) mol^(-1)` `V_(C)(C Cl_(4))=3xx(125.35 cm^(3)mol^(-1))` `=376.05 cm^(3) mol^(-1)` Hence, `V_(C)(NO)ltV_(C)(C Cl_(4))`. `d` `NO` is more ideal in behaviour at `300 K` and `1.013 MPa` because its critical temperature is less than `300 K`, whereas the corresponding critical temperature of `C Cl_(4)` is greater than `300 K`. |
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| 217. |
The CORRECT statement for cubic close packed (ccp) three-dimensional structure is (are)A. The number of neighbours of an atom present in the topmost layer is 12B. The efficiency of the atom packing is 74%C. The number of octahedral and tetrahedral voids per atom are 1 and 2 respectivelyD. The unit cell edge length is `2sqrt2` times the radius of the atom |
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Answer» Correct Answer - B,C,D (a) is incorrect because for any atom in the top most layer, coordination number is not 12 as there is no layer above the topmost layer. (b ) is a known fact. (c ) is correct because in ccp (fcc), number of atoms per unit cell is 4. hence , octahedral voids =4 and tetrahedral voids=8. Therefore , number of octahedral voids per atom =1 and number of tetrahedral voids per atom =2. (d)For ccp (fcc), `r=a/(2sqrt2)` or `a=2sqrt2r` |
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| 218. |
A uni-univalent ionic crystal AX is composed of the following radii (arbitrary units) : `{:(A^+,A^-),(1.0,2.0):}` Assuming that ions are hard spheres , predict giving reasons whether the crystal will have sodium chloride cesium chloride structure. Calculate the volume of the unit cell. |
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Answer» Correct Answer - NaCl, `216 (au)^3` `r_(+)//r_(-)`=2/1=0.5 which lies in the range 0.414-0.732 , Hence , it has sodium chloride structure. Edge=2 `(r_+ + r_-)` =6 au . Volume =`(6 au)^3=216(au)^3` |
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| 219. |
In an ionic compound `A^+ X^-` , the radii of `A^+` and `X^-` ions are 1.0 pm and 2.0 pm respectively. The volume of the unit cell of the crystal AX will beA. `"27 pm"^3`B. `"64 pm"^3`C. `"125 pm"^3`D. `"216 pm"^3` |
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Answer» Correct Answer - D `r_(A^+)/r_(X^-)=1/2`=0.5 . As it lies in the range 0.414-0.732, AX has octahedral structure like that of NaCl. Hence, edge length (a)=`2(r_+ +r_(-))` =2(1+2)pm =6 pm Volume of the unit cell=`a^3`=`"(6 pm)"^3="216 pm"^3` |
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| 220. |
A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion `(Y^-)` will beA. 275.1B. 322.5 pmC. 241.5 pmD. 165.7 pm |
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Answer» Correct Answer - C NaCl has face-centred cubic arrangement of `Cl^-` ions and `Na^+` ions are present in the octahedral voids. Hence, for such a solid radius of cation =0.414 x radius of the anion ( or `r_+/r_(-)`=0.414) i.e., `r_+= 0.414xxr_(-)` or `r_(-)=r_+/0.414=100/0.414`=241.5 pm |
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| 221. |
At what temperature is the average velocity of `O_(2)` molecule equal to the root mean square velocity at `27^(@)C`? |
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Answer» `v_(av) = v_(rms)` `sqrt((8RT)/(pi M)) = sqrt((3RT)/(M))` `(8RT)/(pi M) = (3R xx 300)/(M)` `T = 353.57 K` `t = 80.57^(@)C` |
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| 222. |
The root mean square velocity of an ideal gas to constant pressure varies with density (`d`) asA. `1//sqrt(d)`B. `sqrt(d)`C. `d`D. `d^(2)` |
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Answer» Correct Answer - A From Eq.(5.35), the root mean square velocity is given as `u_(rms) = sqrt((3p)/(d))` Hence, at constant pressure, `u_(rms) prop sqrt((1)/(d))` |
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| 223. |
Which of the following molecule is polar and non-planarA. dipole - dipole forcesB. hydrogen bondsC. dipole - induced dipole forcesD. dispersion forces. |
| Answer» Correct Answer - C | |
| 224. |
According to kinetic theory of gases:A. collisions are always elasticB. heavier molecules transfer more momentum to the wall of the containerC. only a small number of molecules have very high velocityD. between collisions, the molecules move in straight lines with constant velocities |
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Answer» Correct Answer - A |
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| 225. |
If a gas expended at constant temperatureA. the pressure decreasesB. the kinetic energy of the molecules remains the sameC. the kinetic energy of the molecules decreasesD. the number of molecules of the gas increases |
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Answer» Correct Answer - A::B |
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| 226. |
At what temperature root mean sqaure of `N_(2)` gas is equal to that of propance gas at `STP` conditions.A. `173.7^(@)C`B. `173.7K`C. `273 K`D. `- 40^(@)C` |
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Answer» Correct Answer - B `v_(rms) (N_(2)) = v_(rms) (C_(3) H_(8))` `sqrt((3RT)/(28)) = sqrt((3R 273)/(44))` `T = 173.7 K` |
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| 227. |
With increase in temperature, the fraction of molecules possessing most probable speed ………………. |
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Answer» Correct Answer - decreases |
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| 228. |
Assertion. Different gases at the same conditions of temperature and pressure have same root mean velocity. Reason. Root mean square velocity lies between average velocity and most probable velocity. |
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Answer» Correct Answer - D Correct A. RMS also depends upon the nature (molecular mass) of the gas (u`=sqrt(3RT//M)` Correct R. `alpha:v:u=1:1.128:1.224(u gt u gtalpha)` |
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| 229. |
Assertion. At same temperature, most probable speed of `N_(2)` is greater than that of `Cl_(2)`. Reason. At the same temperature, fraction of `N_(2)` molecules possessing the most probable slope speed is greater than that of `Cl_(2)` molecules. |
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Answer» Correct Answer - C Correct R. At the same temperature, fraction of `N_(2)` molecules possessing the most probable speed is less than that of `Cl_(2)` molecules |
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| 230. |
Which of the following statements is not true about NaCl structure?A. `Cl^(-)` ions ar fcc arrangementB. `Na^(+)` ions have corrdination number fourC. `Cl^(-)` ions have coordination number sixD. Each unit cell contains 4 NaCl molecules |
| Answer» (b) NaCl has a fcc structure have CN 6:6 | |
| 231. |
At the same pressure, the rate of diffusion of a gas at `927^(@)C` will be ……………………. Times that at `27^(@)C`. |
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Answer» Correct Answer - 2 `r prop sqrt(T)` `:. (r_(1))/(r_(2))=sqrt((T_(1))/(T_(2)))=sqrt((300K)/(1200K))=sqrt((1)/(4))=(1)/(2) :. r_(2)=2r_(1)` |
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| 232. |
If rate of diffusion of `A` is `5` times that of `B`, what will be the density ratio of `A` and `B` ? |
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Answer» Correct Answer - a The rate of diffusion a gas is inversely proportional to the square root of its density `sqrt((dB)/(dA)) = (r_(A))/(r_(B)) = 5` or `(dB)/(dA) = ((r_(A))^(2))/((r_(B))^(2)) = 25` `(d_(A))/(d_(B)) = (d_(A))/(d_(B)) = 1/25` |
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| 233. |
Two separate bulbs contain ideal gas `A` and `B`. The density of a gas `A` is twice that of a gas `B`. The molecular mass of `A` is half that of gas `B`. The two gases are at the same temperature. The ratio of the pressure of `A` to that gas `B` isA. `2`B. `1//2`C. `4`D. `1//4` |
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Answer» Correct Answer - C `d_(a)= 2d_(b), 2M_(a)=M_(b)` `PV= nRT=(m)/(M)RT, P=(m)/(V).(RT)/(M)=(dRT)/(M)` `(P_(a))/(P_(b))=(d_(a))/(d_(b))(M_(b))/(M_(a))=(2d_(b))/(d_(b))xx(2M_(A))/(M_(a))=4` |
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| 234. |
Equal masses of the two gases A and B are kept in two separate vessels at the same temperature and pressure. If the ratio of the molecular masses of A and B is `2 : 3`, find the ratio of the volumes of the two vessels. |
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Answer» Let the molecular masses of the gases A and B be 2M and 3M respectively. Let the volume of the two vessels in which the gases are kept be `V_(1)` and `V_(2)` respectively. According to ideal gas equation , `PV = nRT = (W)/(M) RT` For gas A : `PV_(1) = (W)/(2M) RT` For gas B : `PV_(2) = (W)/(3M) RT` Divide eqn. (i) by eqn (ii), `(PV_(1))/(PV_(2)) = (W)/(2M) xx (3M)/(W) = 3/2` or `(V_(1))/(V_(2)) = 3/2` The volume of the two gases are in the ratio `3 : 2`. |
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| 235. |
The density of a gas `A` is twice that of a gas `B` at the same temperature. The molecular mass of gas `B` is thrice that of `A`. The ratio of the pressure acting on `A` and `B` will beA. `6 : 1`B. `7 : 8`C. `2 : 5`D. `1 : 4` |
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Answer» Correct Answer - A `Pm = dRT` `(P_(A) m_(A))/(P_(B) m_(B)) = (d_(A) RT)/(d_(B) RT)` `(P_(A))/(P_(B)) xx (1)/(3) = 2` `P_(A): P_(B) = 6 : 1` |
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| 236. |
The density of a gas a is twice that of gas B. Molecular mass of A is half of the molecular of B. The ratio of the partial pressures of A and B is :A. `1//4`B. `1//2`C. `4//1`D. `2//1` |
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Answer» Correct Answer - C `PV = nRT` `P = (wRT)/(Mv) = (dRT)/(M)` `(P_(A))/(P_(B)) = (d_(A))/(d_(B)) xx (M_(B))/(M_(A)) = 2/1 xx 2/1` `= 4//1` |
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| 237. |
Two separate bulbs contain ideal gas `A` and `B`. The density of a gas `A` is twice that of a gas `B`. The molecular mass of `A` is half that of gas `B`. The two gases are at the same temperature. The ratio of the pressure of `A` to that gas `B` isA. 2B. `1//2`C. 4D. `1//4` |
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Answer» Correct Answer - c `d_(A) = (P_(A)M_(A))/(RT),d_(B) = (P_(B)M_(B))/(RT)` `(d_(A))/(d_(B)) = (P_(A)M_(A))/(P_(B)M_(B))` `2 = (P_(A))/(P_(B)) xx 1/2` or `(P_(A))/(P_(B)) = 4` |
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| 238. |
A sample of gas is at `0^(@)C`. The temperature at which its rms speed of the molecule will be doubled isA. `103^(@)C`B. `273^(@)C`C. `723^(@)C`D. `1092 K` |
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Answer» Correct Answer - D `(r_(1))/(r_(2))=sqrt((T_(2))/(T_(2)))` `2=sqrt((T_(2))/(273))` `T_(2)= 273xx4= 1092 K` |
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| 239. |
Under identical conditions of temperature, the density of a gas `X` is three times that of gas `Y` while molecular mass of gas `Y` is twice that of `X`. The ratio of pressure of `X` and `Y` will beA. 6B. `1//6`C. `2//3`D. `3//2` |
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Answer» Correct Answer - A `rho=(PM)/(RT)implies P=(rhoRT)/(M)` `P_(x)=(rhoxxRT)/(M_(x)) " " P_(y)=((rho_(x)RT)/(3))/(2Mx)` `6P_(y)=(rho_(x)RT)/(M_(X))` `:. 6P_(Y)=P_(X) " " (P_(X))/(P_(Y))=6` |
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| 240. |
An effusion experiment requires `40 s` of a certain number of moles of a gas of unknown molar mass to pass through a small orifice into a vaccum. Under the same conditions, `16 s` was required for the same number of moles of `O_(2)` to effuse. What is the molar mass of the unknown gas?A. `5.1 g//mol`B. `12.8 g//mol`C. `80 g//mol`D. `200 g//mol` |
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Answer» Correct Answer - D `(t_((gas)))/(t(O_(2)))=sqrt(M_((gas))/(M_((O_(2)))))` `implies(40)/(16)=sqrt(M_((gas))/(32))` `implies (5)/(2)=sqrt(M_((gas))/(32))implies (25)/(4) =(M_((gas)))/(32)` `implies M_(gas)= 200 g// mol` |
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| 241. |
Compressibility factor for `H_(2)` behaving as real gas is:A. 1B. `(1-(a)/(RTV))`C. `(1+(pb)/(RT))`D. `(RTV)/(1-a)` |
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Answer» Correct Answer - C For `H_(2),` attractive forces are neglected, hence `(a)/(V^(2))=0` `:." " p(V-b)=RTrArrpV=RT+pb` `:." "(pV)/(RT)=(1+(pb)/(RT))=Z` |
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| 242. |
As the temperature is raised from `20^(@)C` to `40^(@)C` the average kinetic energy of neon atoms changes by a factor of which of the following ?A. `1//2`B. `sqrt(313//293)`C. 313/293D. 2 |
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Answer» Correct Answer - C For monoatomic neon atoms, `E(3)/(2)RT` `:. (E_(40.))/(E_(20.))=(T_(40.))/(T_(20.))=(40+273)/(20+273)=(313)/(293)`. |
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| 243. |
If 300 ml of a gas weigh 0.368 g at STP, what is its molecular weight ?A. 30.16B. 2.55C. 27.5D. 37.5 |
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Answer» Correct Answer - A::B::C::D |
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| 244. |
The density of neon will be highest atA. STPB. `0^(@)C,2 atm`C. `273^(@)C, 1 atm`D. `273^(@)C,2 atm` |
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Answer» Correct Answer - B::D |
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| 245. |
A pre-weighed vessel was filled with oxygen at `NTP` weighted. It was the evacuated, filled with `SO_(2)` at the same temperature and pressure, and again weighted. The weight of oxygen will beA. The same as that of `SO_(2)`B. `(1)/(2)` that of `SO_(2)`C. Twice that of `SO_(2)`D. One-fourth that of `SO_(2)` |
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Answer» Correct Answer - B `("M.wt.of".O_(2))/("M.wt. of" SO_(2))implies (M_(1))/(M_(2))implies(32)/(64)= (1)/(2)` The weight of oxygen will be `(1)/(2)` that of `SO_(2)` |
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| 246. |
A `2L` flask contains `1.6 g` of methane and `0.5 g` of hydrogen at `27^(@)C`. Calculate the partial pressure of each gas in the mixture and hence calculate the total pressure. |
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Answer» Moles of `CH_(4)=(W)/(Molar weight)=(1.6)/(16)mol=0.1 mol` Partial pressure of `CH_(4)` `P_(CH_(4))=n_(CH_(4))xx(RT)/(V)=(0.1xx0.0821xx300)/(2)=1.23 atm` Moles of `H_(2)=(W)/(Molar weight)=(0.5)/(0.2) mol=0.25 mol` Partial pressure of `H_(2)` `P_(H_(2))=n_(H_(2))xx(RT)/(V)=(0.25xx0.0821xx300)/(2)=3.079 atm` `:. ` Total pressure `=P_(CH_(4))+P_(H_(12))=1.23+3.079=4.31 atm` |
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| 247. |
A 10 litre flask contains 0.2 mole of methane, 0.3 mole of hydrogen and 0.4 mole of nitrogen at `25^(@)C`. What is the partial pressure of each component and what is the pressure inside the flask ? |
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Answer» `P = (nRT)/(V)` Partial pressure of methane `= (0.2 xx 0.00821 xx 298)/(10) = 0.489 atm` Partial pressure of hydrogen `= (0.3 xx 0.0821 xx 298)/(10) = 0.734 atm` Partial pressure of nitrogen `= (0.4 xx 0.0821 xx 298)/(10) = 0.979` atm Total pressure `= (0.489 + 0.8734 + 0.979) atm` `= 2.202 atm` Alternative solution: Total number of moles `= 0.2 + 0.3 + 0.4 = 0.9` Let the total pressure be P. We know that, `P = ("total number of moles")/(V). RT` `V = 10` litre, `R = 0.0821 L atm K^(-1), T = (25 + 273) = 298 K` `+ (0.9)/(10) xx 0.0821 xx 298 = 2.20 atm` Partial pressure of `CH_(4) =` Mole fraction of methane `xx` total pressure `= (0.2)/(0.9) xx 2.20 = 0.489 atm` Partial pressure of `H_(2) =` Mole fraction of `H_(2) xx` total pressure `= (0.3)/(0.9) xx 2.20 = 0.489 atm` Partial pressure of `H_(2) =` Mole fraction of `H_(2) xx` total pressure `= (0.3)/(0.9) xx 2.20 = 0.733` atm Partial pressure of `N_(2) =` Mole fraction of `N_(2) xx` total pressure `= (0.3)/(0.9) xx 2.20 = 0.733 atm` Partial pressure of `N_(2) =` Mole fraction of `N_(2) xx` total pressure `= (0.4)/(0.9) xx 2.20 = 0.978 atm` |
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| 248. |
A vessel of `4.00L` capacity contains `4.00 g` of methane and `1.00 g` of hydrogen at `27^(@)C`. Calculate the partial pressure of each gas and also the total pressure in the container. |
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Answer» Let the partial pressure of hydrogen be `P_(H_(12))` and the partial pressure of methane be `P_(CH_(4))`. The number of moles of hydrogen is `n_(1)=(1)/(2)=0.5 mol` The number of moles of methane is `n_(2)=(4)/(16)=0.25 mol` Now applying ideal gas equation for each gas, we get `P_(H_(2))=(n_(1)RT)/(V)=(0.5xx0.0821xx300)/(4)=3.07 atm` Similarly, `P_(CH_(4))V=n_(2)RT` `P_(CH_(4))=(n_(2)RT)/(V)=(0.25xx0.0821xx300)/(4)=15.4 atm` Total pressure of the gaseous mixture is `P_(H_(2))+P_(CH_(4))=3.07+1.54=4.61 atm`. |
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| 249. |
Why dry air is heavier than moist air? |
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Answer» Average molar weight of dry air is `=(% of N_(2)xx28+% of O_(2)xx32)/(100)` Average molar weight of moist air is `= (% of N_(2)xx28+% of O_(2)xx32+% of H_(2)Oxx18)/(100)` Evidently average molar weight of dry air is more and so is its density `(d=M//V)`. |
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| 250. |
Helium is used in balloons in place of hydrogen because it isA. lighter than hydrogenB. incombustibleC. more abundant than hydrogenD. radioactive |
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Answer» Correct Answer - B Helium gas is heavier than hydrogen gas , but it is preferred over `H_(2)` gas because `H_(2)` gas is highly inflammable while `He` is incombustible. |
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