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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.
| 951. |
A ballon of diameter 20 metre weighs `100 kg` Calculate its pay-load if its is filled with He at 1.0 atm and `27^(@)C` Density of air is `1.2 kg,^(-3)` `[ R =0.082 dm^(3)` atm `K^(-1) mo1^(-1)]` . |
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Answer» Radius of ballon `(r) = (20 m)/(2) = 10 m` Volume of balloon `(V) = 4/3p pir^(3) = 4/3 xx 22/7 xx (10 m)^(3) = 4190.5 m^(3)` Mass of air displaced `= "Volume of air" xx "Density of air"` `= (4190.5 m^(2)) xx (1.2 kg m^(-3)) = 5028.6 kg` Moles of He present `(n) = (PV)/(RT) = ((1.0 atm) xx (4190.5 m^(3)))/((0.082 xx 10^(-3) m^(3) atm K^(-1) mol^(-1)) xx (300 K))` `= 170.346 xx 10^(3)` mol Mass of He present `= "Moles of He" xx "Molar mass of He"` `= (170.346 xx 10^(3) mol) xx (4g mol^(-1))` `= 681.38 xx 10^(3) g = 681.38 kg` Mass of filled ballon `= 100 + 681.38 = 781.38 kg` Pay load = Mass of air displaced - Mass of filled balloon `= 5028.6 -781.38 = 4247.22` kg. |
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| 952. |
A dry gas occupies `127 mL` at `N.T.P.` If the same mass of the gas is collected over water at `23^(@)C` and a total pressure of `0.98 "bar"`, what volume will it occupy ? The vapour pressure of water at `23^(@)C` is `0.028 "bar"`. |
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Answer» Correct Answer - `146.52 mL` `{:("N.T.P. conditions",,"Experimental conditions"),(V_(1)=127mL,,V_(2)=?),(P_(1)=1.013"bar",,P_(2)=(0.98-0.028) = 0.952 "bar"),(T_(1)=273K,,T_(2)=23+273=296K):}` Applying gas equation: `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2)) = ((1.013 "bar") xx (127 mL) xx (296 K))/((273 K) xx (0.952 "bar")) = 146.52 mL` |
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| 953. |
Calculate the number of moles of hydrogen contained in 18 litres of the gas at `27^(@)C` and 70 cm pressure. Given that R=0.0821 litre atm `K^(-1)`. Further, if the mass of hydrogen taken as above is found to be 1.350 g, calculate the molecular mass of hydrogen. |
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Answer» Here, we are given : V=18 litres, `T=27^(@)C`=27+273 K=300K `P=70 cm=70//76" atm", "R"=0.0821" litre atm "K^(-1)mol^(-1)` Using the ideal gas equation, PV=nRT we have `" " n=(PV)/(RT)=((70//76" atm")(18 L))/((0.0821" L atm mol"^(-1))(300" K"))=0.67" mole"` Further, we know that `n=("Mass)/("Molecular mass")=(m)/(M)" ":. M=(m)/(n)=(1.350)/(0.67)=2.015" u"`(Molar mass=mass `"mol"^(-1)`) |
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| 954. |
A given mass of a gas occuples 919.0 mL in dry state at STP. The same main when collected over water at `15^(@)C` and 750 mm pressure occupies one litre volume. Calculate the vapour pressure of water at `15^(@)C`. |
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Answer» Step 1. To calculate the pressure of the dry gas at `15^(@)C` and 750 mm pressure `{:("Given Conditions at STP","Final Conditions"),(V_(1)=919 mL,V_(2)=1000 mL),(P_(1)=760 mm,P_(2)=?("Dry state")),(T_(1)=273 K",",T_(2)=273+15=288 K):}` By applying gas equation, we have `(760xx919)/(273)=(P_(2)xx1000)/(288)" or " P_(2)=(760xx919xx288)/(1000xx273)=736.7mm` Step 2. To Calculate the vapour pressure of water at `15^(@)C`. Vapour pressure of water =Pressure of the moist gas-Pressure of the dry gas, =750-736.7=13.3 mm Alternatively, if p is the vapour pressure of water at `15^(@)C`, then take `P_(2)`=(750-p)mm.Substituting in the equation, `(P_(1)V_(1))/(T_(2))=(P_(2)V_(2))/(T_(2))`, we get `(760xx919)/(273)=((750-p)xx1000)/(288)`. Solve for p. |
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| 955. |
At `25^(@)C` and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is `10^(@)C `and volume of the gas is 640 mL. |
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Answer» Correct Answer - `676.64 mm` `{:("Initial Conditions",,"Final Conditions"),(V_(1)=600mL,,V_(2)=640 mL),(P_(1)=760 mm,,P_(2)=?),(T_(1)=25^(@)C=298K,,T_(2)=10^(@)C=283K):}` Applying gas equation : `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `P_(2) = (P_(1)V_(1)T_(2))/(T_(1)V_(2))` `P_(2) = ((760 mm) xx (600 mL) xx (283 K))/((298 K) xx (640 mL)) = 676.64 atm` |
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| 956. |
Calculate the total pressure in a mixture of `8 g` of oxygen and `4 g` hydrogen confined in a vessel of `1 dm^(3)` at `27^(@)C`. `(R = 0.083 "bar" dm^(3) K^(-1) mol^(-1))` |
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Answer» Correct Answer - 65.025 bar Given, mass of dioxygen `(O_(2))` = 8 g Thus, numbe of moles of `O_(2)=8/32=0.25`mole Thus number of moles of `H_(2)=4/2=2` mole Therefore, total number of moles in the mixture =0.25+2=2.25 mole Given, V= 1 `dm^(3)` n= 2.25 mol R= 0.083 bar `dm^(3)K^(-1)mol^(-1)` T= `27^(@)C`=300K Total pressure (p) can be calculate as : pV= nRT `rArr=(nRT)/V` `=(225xx0.083xx300)/1` =56.025 bar Hence, the total pressure of the mixture is 56.025 bar. |
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| 957. |
Pay load is defined as the difference between the mass of the displaced air and the mass of the balloon. Calculate the say load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at `27^(@)C` (Density of air `=1.2" kg m"^(-3)` and R=0.0833 bar `dm^(3) K^(-1)mol"^(-1)` ) |
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Answer» Radius of the balloon =10 m `:.` Volume of the balloon`=(4)/(3)pir^(3)=(4)/(3)xx(22)/(7)xx(10 m)^(3)=4190.5" m"^(3)` Volume of He filled at 1.66 bar and `27^(@)C=4190.5" m"^(3)` Calculation of mass of He : PV=nRT`=(w)/(M)RT` or `" " w=("MPV")/("RT")=((4xx10^(-3)" kg mol"^(-1))(1.66"bar")(4190.5xx10^(3)"dm"^(3)))/((0.083" bar dm"^(3)" K "^(-1)"mol"^(-1))(300K))=1117.5" kg"` Total mass of the balloon along with He `=100+1117.5=1217.5" kg"` Maximum mass of the air that can be displaced by balloon to go up =Volume xxDensity `=1490.5" m"^(3)xx1.2" kg "m^(-3)=5028.6" kg"` `:. " Pay load" =5028.6-1217.5" kg" =3811.1" kg"` |
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| 958. |
A sample of nitrogen gas occupies a volume of `1.0 dm^(3)` at a pressure of `0.5` bar and at `40^(@)C`. Calculate the pressure the gas if compressed to `0.225 dm^(3)` at `-6^(@)C`. |
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Answer» From the available data : `V_(1) = 1.0 dm^(3) , V_(2) = 0.225 dm^(3)` `P_(1) = 0.5` bar , `P_(2) = ?` `T_(1) = 40 + 273 = 313 K , T_(2) = -6 + 273 = 267 K` According to Gas equation, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `P_(2) = (P_(1)V_(1)T_(2))/(T_(1)V_(2))` By substituting the values, `P_(2) = ((0.5 "bar") xx (1.0 dm^(3)) xx (267 K))/((313 K) xx (0.225 dm^(3))) = 1.896` bar |
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| 959. |
A sample of a gas occupies a volume of 512 mL at `20^(@)C` and `74 cm` of Hg as pressure. What volume would this gas occupy at STP ? |
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Answer» `{:("Initial conditions","Final conditions" (STP),),(P_(1) = 74 cm,P_(2) = 76 cm,),(T_(1) = 20^(@)C = (20 + 273) = 293 K,T_(2) = 0^(@)C = 273 K,),(V_(1) = 512 mL,V_(2) = ?,):}` We know that, `(P_(1)V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2))` `(74 xx 512)/(293) = (76 xx V_(2))/(273)` So, `V_(2) = (74 xx 512 xx 273)/(293 xx 76)` `= 464.5 mL` |
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| 960. |
`34.05 mL` of phosphorus vapours weighs `0.0625 g` at `546^(@)C` and `0.1` bar pressure. What is the molar mass of phossphorus ? |
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Answer» According to ideal gas equation : `PV = nRT , PV = (WRT)/(M)` or `M = (WRT)/(PV)` According to available data , Mass of phosphrous vapours `(W) = 0.0625 g` Volume of vapours `(V) = 34.05 mL = 34.05 xx 10^(-3) L` Pressure of vapours `(P) = 1.0 "bar"` Gas constant `(R) = 0.083 "bar" L K^(-1) mol^(-1)` Temperature `(T) = 546+273 = 819 K` `:. M = ((0.0625 g) xx (0.083 "bar" LK^(-1) mol^(-1)) xx (819 K))/((1.0 "bar") xx (34.05 xx 10^(-3) L)) = 125 g mol^(-1)`. |
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| 961. |
Calculate the total pressure in a mixture of 8 g of oxygen and 4 g of hydrogen confined in a vessel of 1 `dm^(3)` at `27^(@)C`. R=0.083 bar `dm^(3)K^(-1)mol^(-1)`. |
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Answer» Molar mass of `O_(2)`=32 g `mol^(-1)" ":. 8 g O_(2)=(8)/(32)mol=0.25 mol` Molar mass of `H_(2)=2 g mol^(-1)" " :. 4 g O_(2)=(4)/(2)=2 mol` `:.` Total number of moles (n)=2+0.25=2.25 V=1 `dm^(3)`, T=`27^(@)C=300 K,R=0.083 bar dm^(3)K^(-1)mol^(-1)` PV=nRT or P`=(nRT)/(V)=((2.25 mol)(0.083 bar dm^(3)K^(-1)mol^(-1))(300 K))/(1 dm^(3))=56.025 bar` |
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| 962. |
Density of a gas is found to be `5.46 g//dm^(3)` at `27^(@)C` and at 2 bar pressure. What will be its density at STP ? |
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Answer» `d=(MP)/(RT)`. For the same gas at different temperatures and pressure, `(d_(1))/(d_(2))=(P_(1))/(T_(1))xx(T_(2))/(P_(2))`. Here, `" " d_(1)=5.46 g dm^(-3), T_(1)=27^(@)C=300 K, P_(1)=2" bar"`. At STP, `" " d_(2)=? " " T_(2)=0^(@)C=273 K, P_(2)=1" bar"` `:. (5.46 g dm^(-3))/(d_(2))=(2" bar" )/(300 K)xx(273 K)/(1" bar" )` or `" " d_(2)=3 g dm^(-3)` |
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| 963. |
A sample of gas occupies volume of `2.74 dm^(3)` at `0.9` bar and `27^(@)C`. What will be the volume at `0.75` bar and `15^(@)C`? |
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Answer» Form the available data : `V_(1) = 2.74 dm^(3)` , `V_(2) = ?` `P_(1) = 0.9` bar , `P_(2) = 0.75` bar `T_(1) = 27 + 273 = 300 K` , `T_(2) = 15 + 273 = 288 K` According to gas equation, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))` By substituting the values, `V_(2) = ((0.9 "bar") xx (2.74 dm^(3)) xx (288 K))/((300 K) xx (0.75 "bar" )) = 3.16 dm^(3)` |
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| 964. |
The density of `CO_(2)` is `0.326 g dm^(-3)` at `27^(@)C` and `0.25` bar pressure. What is the density of the gas at `47^(@)C` keeping the pressure constant? |
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Answer» From the available data : `rho_(1) = 0.326 g dm^(-3) , T_(1) = 27+273 = 300 K` `rho_(2) = ? , T_(2) = 47+273 = 320 K`. We know that, `rho_(1)T_(1) = rho_(2)T_(2)` or `rho_(2) = (rho_(1)T_(1))/(T_(2)) = ((0.326 g dm^(-3)) xx (300 K))/((320 K)) = 0.31 g dm^(-3)` |
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| 965. |
`1.22g` of a gas mesured over water at `15^(@)C` and undr a pressure of `1.02` bar of mercury occupied `0.9 dm^(3)`. Calculate the volume of the volume of the dry gas at `N.T.P.` Vapour pressure of water at `15^(@)C` is `0.018` bar. |
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Answer» Pressure of dry gas = Pressure of moist gas - Aqueous tension `= 1.02 - 0.018 = 1.002` bar From the available data: `V_(1) = 0.9 dm^(3) , V_(2) = ?` `P_(1) = 1.002` bar , `P_(2) = 1.013` bar `T_(1) = 15+273 = 288 K , T_(2) = 273 K` According the Gas equation , `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))` Substituting the values, `V_(2) = ((1.022 "bar")xx (0.9 dm^(3)) xx (273 K))/((288 K) xx (1.013 "bar" )) = 0.844 dm^(3)` |
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| 966. |
The vapour of phosphine gas at `27^(@)C` and 3 bar pressure has density:A. `4.09 g mL^(-1)`B. `4.14 g L^(-1)`C. `2.04 kg L^(-1)`D. `2.04 g L^(-1)` |
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Answer» Correct Answer - B `d = (Pm)/(RT) = (3 xx 34)/(0.0821 xx 300) = 4.14 g L^(-1)` |
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| 967. |
1.22 g of a gas measured over water at `150^(@)C` and a pressure of 775 mm of mercury occupied 900 mL. Calculate the volume of dry gas at NTP. Vapour pressure of water at `15^(@)C` is 14 mm. |
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Answer» Pressure of dry gas = Pressure of moist gas `-` Aqueous tension `= 775 - 14` `= 761 mm` `{:("Initial conditions","NTP Conditions",),(V_(1) = 900 mL,V_(2) = ?,),(P_(1) = 761 mm,P_(2) = 760 mm,),(T_(1) = (273 + 15) = 288 K,T_(2) = 273 K,):}` Since, `(P_(1)V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2))` So, `V_(2) = (P_(1) V_(1) T_(2))/(T_(1) P_(2))` `= (761 xx 900 xx 273)/(288 xx 760)` `= 854.2 mL` |
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| 968. |
What is the volume occupied by 11 g of carbon dioxide at `27^(@)C` and 780 mm of Hg pressure ? |
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Answer» Correct Answer - 6 litre `780 mm " of " Hg = (780)/(760) atm, w = 11 g, M = 44 g mol^(-1)` Now apply `V = (w)/(M). (RT)/(P)` |
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| 969. |
Calculate the volume occupied at `27^(@)C` and 2 bar pressure of a gas evolved from 2 mL of solid carbon dioxide. Given the density of solid carbon dioxide is `1.53 g mL^(-1)`. |
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Answer» Correct Answer - `0.865 L` Step I. No. of moles of `CO_(2)(g)` No. of moles of `CO_(2)(n) = ("Mass of " CO_(2)(g))/("Molar mass ") = ("Density" xx "Volume")/("Molar mass")` `= ((1.53 g mL^(-1)) xx (2 mL))/((44 g mol^(-1))) = 0.0695` mol. Step II. Volume of `CO_(2)(g)` According to ideal gas equation, `PV = nRT , n = 0.0695 mol, P = 2 "bar", T = 300 K`, `:. V = (nRT)/(P) = ((0.0695 mol) xx (0.083 L "bar" K^(-1) mol^(-1)) xx (300 K))/((2 "bar")) = 0.865 L` |
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| 970. |
A gas has a vapour density 11.2. The volume occupied by gram of the gas at STP will be:A. 11.2 LB. 22.4 LC. 1 LD. 10 L |
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Answer» Correct Answer - C Number of moles `= ("Mass")/("Molar mass")` `= ("Mass")/(2 xx " Vapour density")` `= (1)/(2 xx 11.2) = (1)/(22.4)` Volume of gas `= n xx 22.4 L` `= (1)/(22.4) xx 22.4 L = 1L` |
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| 971. |
The quantity `(PV//K_(B)T)` representsA. number of molecules in the gasB. mass of the gasC. number of moles of the gasD. translational energy of the gas |
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Answer» Correct Answer - A `PV = nRT` `n = (PV)/(RT)` Number of molecules `= N_(A) (PV)/(RT)` `= (PV)/((R//N_(A))T) = (PV)/(k_(B)T)` |
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| 972. |
Assertion : At constant temperature PV vs P plot for real gases is not a straight line . Reason : In the curves of dihydrogen and helium , as the pressure increases the value of PV also increases .A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false . |
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Answer» Correct Answer - B PV `vs` P plot for real gases is not a straight line because real gases do not follow ideal gas equation perfectly under all conditions. |
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| 973. |
One way of writing the equation of state for real gases is : `PV=RT[1+(B)/(V)]` where B is a constant. Derive an approximate expression for B in terms of the van der Waals constant a and b |
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Answer» According to van der Waals equation, `(P+(a)/(V_(2)))(V-b)=RT " "rArr " " P=(RT)/(V-b)-(a)/(v_(2)) rArr " " PV=(RTV)/(V-b)-(a)/(V)` or `" " PV=RT (1-(b)/(V))^(-1)-(a)/(V)=RT(1+(b)/(V))-(a)/(V)" "` (Neglecting higher powers of b/V) `rArr " " PV=RT (1+(b)/(V)-(a)/(VRT))=RT[1+(1)/(V)(b-(a)/(RT))]` Comparing with the given form of teh equation, `B=b-(a)/(RT)` |
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| 974. |
Slope of the plot between PV and P at constant temperature isA. zeroB. 1C. `1//2`D. `1//sqrt(2)` |
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Answer» Correct Answer - A::B::C::D |
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| 975. |
The value of van der Waals constant `a` for the gases `O_(2)`, `N_(2)`, `NH_(3)`, and `CH_(4)` are `1.360`, `1.390`, `4.170`, and `2.253 L^(2) atm mol^(-2)`, respectively. The gas which can most easily be liquefied isA. `CH_(4)`B. `N_(2)`C. `NH_(3)`D. `O_(2)` |
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Answer» Correct Answer - C The ease of lineequation of gases is directly related to the strength of intermolecular attractions . The stronger the attractions , the more easily the gas will be liquefied. The value of constant `a` measures the strength of molecular attractions. Thus , `NH_(3)` with the highest value of `a` is most easily liquefied. |
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| 976. |
Assertion `:-` At constant temperature `PV` vs `V` plot for real gas is not a straight line. Reason `:-` At high pressure, all gases have `Zgt1` but at low pressure most gases have `Zlt1`A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true and R is not the correct explanation of A.C. A is true but R is false.D. A is false and R is true. |
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Answer» Correct Answer - B Correct explanation. At high pressure, `Z gt 1`for real gases. Therefore, graph is not a straight line. |
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| 977. |
The value of van der Waals constant `a` for the gases `O_(2)`, `N_(2)`, `NH_(3)`, and `CH_(4)` are `1.360`, `1.390`, `4.170`, and `2.253 L^(2) atm mol^(-2)`, respectively. The gas which can most easily be liquefied isA. `O_(2)`B. `N_(2)`C. `NH_(3)`D. `CH_(4)` |
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Answer» `(P+(an^(2))/(V^(2)))(V-nb)=nRT` The van der Waals constant `a` is used in pressure correction, and its value depends upon the intermolarcular forces between the gas molecules The larger the value of `a` for a gas, the more easily that gas can be liquefied. |
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| 978. |
The compressibilty factor for a real gas at high pressure is:A. `1 + (RT)/(Pb)`B. `1`C. `1 + (Pb)/(RT)`D. `1 - (Pb)/(RT)` |
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Answer» Correct Answer - C For 1 mol gas `(P+ (a)/(V^(2))) (V - b) = RT` Pressure correction can be neglected at high pressure. Thus, `P (V - b) = RT` `PV = Pb + RT` `(PV)/(RT) = 1 + (Pb)/(RT)` `Z = 1 + (Pb)/(RT)` |
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| 979. |
An evacuated glass vessel weighs `50.0 g` when empty, `148.0 g` when filled with a liquid of density `0.98 g mL^(-1)`, and `50.5 g` when filled with an ideal gas at `760 mm Hg` at `300 K`. Determine the molar mass of the gas. |
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Answer» Mass of ideal gas `=50.5-50=0.5 g` Mass of liquid `=148-50=98 g` `Volume=(Mass)/(Density)=(98)/(0.98)=100 mL` or `0.1 L` For the gas, `PV=(w)/(M)RT` or `(760)/(760)xx0.1=(0.5)/(M)xx0.0821xx300` or `M=123.15` |
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| 980. |
If the most probable velocity of methane at a certain temperature is `400 ms^(-1)`, the kinetic energy of one mole of methane at the same temperature in J is:A. 1024B. 2048C. 3072D. 1920 |
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Answer» Correct Answer - D `v_(rms) = (sqrt3)/(sqrt2) xx v_(mp)` `sqrt((2E)/(m)) = (sqrt3)/(sqrt2) xx 400` `sqrt((2E)/(16 xx 10^(-3))) = (sqrt3)/(sqrt2) xx 400` On solving: `E = 1920 J mol^(-1)` |
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| 981. |
28 g of each of the following gases are taken at `27^(@)C` and 600 mm pressures. Which of these will have the least volume?A. HBrB. HClC. HFD. HI |
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Answer» Correct Answer - D PV=nRT or `V=(nRT)/(P)`. As P and T are same for all given gases, `V prop n`. As `n=(w)/(M)`, n will be least for which M is maximum viz. HI. |
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| 982. |
An evacuated vessel weighs 50 g when empty, 144 g when filled with a liquid of density 0.47 g `mL^(-1)` and 50.5 g when filled with an ideal gas at 760 mm Hg at 300 K. The molar mass of the ideal gas is (Given R=0.082 L atm `K^(-1)mol^(-1)`)A. 61.575B. 130.98C. 87.943D. 123.75 |
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Answer» Correct Answer - A Mass of liquid filling the vessel =144-50=94 g Volume of liquid=Volume of vessel`=(94 g)/(0.47" g "mL^(-1))` `:.` Volume of gas(V)=200mL=0.02L Pressure (P)=760 mm=1 atm Temperature (T)=300 K Mass of the gas=50.5-50=0.5 g PV=nRT`=(w)/(M)RT` `:. M=(wRT)/(PV)=(0.5xx0.0821xx300)/(1xx0.2)=61.575` |
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| 983. |
For gaseous state, if most probable speed is denoted by `C^(**)` average speed by `barC` and root square speed by `C`, then for a large number of molecules, the ratios of these speeds areA. `C^(**) : barC : C = 1 : 1.225 : 1.128`B. `C^(**) : barC : C = 1.225 : 1.128 : 1`C. `C^(**) : barC : C = 1.128 : 1 : 1.225`D. `C^(**) : barC : C = 1 : 1.128 : 1.225` |
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Answer» Correct Answer - D `C^(*):barC:C=sqrt((2RT)/(M)) = sqrt((8RT)/(piM)) = sqrt((3RT)/(M))` `= sqrt(2) : sqrt((8)/(3.142)) : sqrt(3)` `1:1.128:1.225` |
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| 984. |
For gaseous state, if most probable speed is denoted by `C^(**)` average speed by `barC` and root square speed by `C`, then for a large number of molecules, the ratios of these speeds areA. `overset(.)(C) : bar(C) : C= 1 : 1.128 : 1.225`B. `overset(.)(C) : bar(C) : C = 1 : 1.225 : 1.128`C. `overset(.)(C) : bar(C) : C = 1.225 : 1.128 : 1`D. `overset(.)(C) : bar(C) : C = 1.128 : 1.225 : 1` |
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Answer» Correct Answer - A `overset(.)(C) : bar(C) : C = sqrt((2RT)/(m)) : sqrt((8RT)/(pi m)) : sqrt((3RT)/(m))` `= sqrt2 : sqrt((8)/(pi)) : sqrt3` `= 1.414 : 1.596 : 1.732` `= 1 : 1.128 : 1.225` |
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| 985. |
For gaseous state, if most probable speed is denoted by `C^(**)`, average speed by `vecC ` and mean square speed by C, then for a large number of molecules the ratios of these speeds are :A. `C^(**):overset(-)C:C=1:1.225:1.128`B. `C^(**):overset(-)C:C=1.225:1.128:1`C. `C^(**):overset(-)C:C=1.128:1.125:1`D. `C^(**):overset(-)C:C=1:1.128:1.225` |
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Answer» Correct Answer - D `C^(**)=sqrt((2RT)/(M)), bar(C )=sqrt((8RT)/(piM)),C=sqrt((3RT)/(M))` `:. C^(**):bar(C ):C=sqrt((8)/(pi)):sqrt(3)=1:sqrt((4)/(pi)):sqrt((3)/(2))` `=1:1.128:1.225` |
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| 986. |
Calculate the temperature at which the root mean square speed, average speed and most probable speed of oxygen gas are all equal to `1500 m s^(-1)` |
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Answer» Correct Answer - A::B::C::D `c=sqrt((3RT)/(M))" or "sqrt((3RT)/(M))=c^(2)" or "T=(c^(2)xxM)/(3R)=((1500" m s"^(-1))^(2)(32xx10^(-3)kg mol^(-1)))/(3xx8.314" kg m"^(2)s^(-1))=2886.7 K` `bar c=sqrt((8RT)/(piM))" or " T=(bar c^(2)xxpixxM)/(8R)=((1500)^(2)(3.14)(32xx10^(-3)))/(8xx8.314)=3399 K` `c^(**)=sqrt((2RT)/(M))" or "T=(c^(**2)M)/(2R)=((1500)^(2)(32xx10^(-3)))/(2xx8.314)=4330 K`. |
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| 987. |
At what temperature is the rms speed of `H_(2)` molecules the same as that of oxygen molecules at `1327^(@)C` ?A. `173 K`B. `100 K`C. 400 KD. 523 K |
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Answer» Correct Answer - B `(sqrt((3RT_(1))/(M_(1))))_(H_(2)) = (sqrt((3RT_(2))/(M_(2))))_(O_(2))` `sqrt((T_(1))/(2)) = sqrt((1600)/(32))` `T_(1) = 100 K` |
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| 988. |
The root mean square speed of molecules of nitrogen gas is v at a certain temperature. When the temperature is doubled, the molecules dissociate into individual atoms. Calculate the factor by which root mean square speed of the individual atoms increases. |
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Answer» `u_(rms)=sqrt((3RT)/(M))` `(u_(N))/(u_(N_(2)))=sqrt((T_(N))/(T_(N_(2)))xxM_(N_(2))/(M_(N_(2))))" or "(u_(N))/(v)=sqrt((2T)/(T)xx(28)/(14))=sqrt(4)=2` or `u_(N)=2v` Thus, speed of individual atoms becomes two times. |
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| 989. |
A gas bulb of 1 litre capacity contains `2.0xx10^(21)` molecules of nitrogen exerting a pressure of `7.57xx10^(3)" N "m^(-2)`. Calculate the root mean square (r.m.s) speed and the temperature of the gas molecules. |
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Answer» Volume of gas =1 litre `=10^(-3)m^(-3)` Pressure `=7.57xx10^(3)" N "m^(-2)` No. of moles of the gas `=(2.0xx10^(21))//(6.02xx10^(23))=0.00332` Applying ` " "` PV=nRT `(7.57xx10^(3))xx10^(-3)=0.00332xx8.314xxT" or " T=274.25" K"` Root Mean Square speed `=sqrt((2RT)/(M))=sqrt((3xx8.314xx10^(7)xx274.25)/(28))=49426.5" cm "s^(-1)` |
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| 990. |
Calculate the root mean square, average and most probable speeds of oxygen molecules at `27^(@)C`. |
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Answer» Correct Answer - A::B::C::D Proceed as in Solved Sample Problem 2(b), page 5/41 |
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| 991. |
Oxygen at 1 atmosphere and `0^(@)C` has a density of `1.4290 g L^(-1)`. Find the rms speed of oxygen molecule. |
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Answer» `P = 1 atm = 101.3 xx 10^(3) Pa` `d = 1.4290 g L^(-1) = 1.4290 kg m^(-3)` We know that, `c = sqrt((3P)/(d)) = sqrt((3 xx 101.3 xx 10^(3))/(1.4290)) = 461.15 ms^(-1)` |
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| 992. |
At what temperature will hydrogen molecules have the same root mean square speed as nitrogen molecules at `27^(@)C` ? |
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Answer» `sqrt((3RT)/(M_(H))) = sqrt((3R xx 300)/(M_(N)))` or `(T)/(M_(H)) = (300)/(M_(N)) " or " T = (300)/(28) xx 2 = 21.43 K` |
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| 993. |
Assertion (A) Liquids tend to have maximum number of molecules at their surface. Reason (R) Small liquid drops have spherical shape.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true and R is not the correct explanation of A.C. A is true but R is false.D. A is false and R is true. |
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Answer» Correct Answer - D Correct assertion. Liquids tend to have minimum number of molecules at their surface. |
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| 994. |
Liquids like ether and acetone are kept in coll places. |
| Answer» Both these liquids are volatile in nature which means that their boilinb points are quite low. They are expected to evaporate at a fast rate in case temperature is high . Therefore, the temperature has to be lowered or they are to be kept in cool place to check their evaporations. | |
| 995. |
In the compound AX, the radius of `A^+` ion is 95 pm and that of `X^-` ion is 181 pm. Predict the crystal structure of AX and write the coordination number of of each of the ions. |
| Answer» Radius ratio, `r_+ // r_-` =95/181=0.525 which lies in the range 0.414-0.732. Hence, the structure is octahedral and coordination number is 6. | |
| 996. |
Assertion: Noble gases can be liquefied. Reason: Attractive forces can exist between nonpolar molecules.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 Noble gases also have force of attraction and can be liquified. |
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| 997. |
Zinc blende has ________ arrangement of `S^(2-)` ions whereas wurtzite has _____ arrangement of `S^(2-)` ions. |
| Answer» Correct Answer - ccp/fcc,hcp | |
| 998. |
The ratio `gamma` for inert gases isA. `1.33`B. `1.66`C. `2.13`D. `1.99` |
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Answer» Correct Answer - B `r=(C_(P))/(C_(V))=(5)/(3)= 1.66` ( For monoatomic as `He, Ne, Ar)` |
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| 999. |
A cylinder contains accetylene gas at `27^(@)C` and `4.05M pa`. The pressure in the cylinder after half the mass of gas is used up and temperature has fallen to `12^(@)C` will be:A. `4.05 MP a`B. `2.025 MP a`C. `3.84 MP a`D. `1.92 MP a` |
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Answer» Correct Answer - D `PV=(w)/(M_(w))RT " " (P)/(WT)="constant"` `(P_(1))/(w_(1)T_(1))=(P_(2))/(w_(2)T_(2)) " " (4.05)/(wxx300)=(P_(2))/((w)/(2)xx285)` `P_(2)= 1.92MPa` |
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| 1000. |
A container of 1 L capacity contains a mixture of 4 g of `O_(2)` and 2 g `H_(2)` at `0 .^(@)C` . What will be the total pressure of the mixture ?A. 50 . 42 atmB. 25 . 21 atmC. 15 . 2 atmD. 12 . 5 atm |
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Answer» Correct Answer - B Applying `PV=(w)/(M)RT` `P_(O_(2))=(4)/(32)xx(0.0821xx273)/(1)=2.81` atm `P_(H_(2))=(2)/(2)xx(0.0821xx273)/(1)=22.4` atm `P_("total")=P_(O_(2))+P_(H_(2))=2.81+22.4=25.21` atm |
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