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1.	A given mass of gas occupies `919.0` mL of dry state at N.T.P. The mass when collect over water at `15^(@)C` and `0.987 "bar"` pressure occupies one litre volume. Calculate the vapour pressure of water at `15^(@)C`.
Answer» Correct Answer - `0.005 "bar"` Step I. Vapour pressure of dry gas under experimently conditions. `{:("N.T.P. conditions",,"Experimental conditions"),(V_(1)=919mL,,V_(2)=1000mL),(P_(1)=1.013"bar",,P_(2)=?),(T_(1)=273K,,T_(2)=15+273=288K):}` `(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)) = ((1.013 "bar") xx (919 mL) xx (288 K))/((273 K) xx (1000 mL)) = 0.982 "bar"`. Step II. Pressure of water vapours (Aqueous tension) Pressure of water vapours = Pressures of moist gas - Pressure of dry gas `= 0.987 - 0.982 = 0.005 "bar"`

2.	A balloon filled with an ideal is taken from the surface of the sea deep to a depth of 100 m. What will be its volume in terms of its original volume ?
Answer» Correct Answer - A::C Pressure at the surface =76 cm of Hg`=76xx13.6 cm` of `H_(2)O=10.3" m"` of `H_(2)O` `:.` Pressure at 100 m depth= 100+10.3=110.3 m Applying `underset((At surface))P_(1)V_(1)=underset((At 100 m depth))P_(2)V_(2)` `10.3xxV=110.3xxV_(2)" or " V_(2)=0.093" V"=9.3%` of V

3.	A balloon filled with helium rises to a certain height at which it gets fully inflated to a volume of `10^(5)` litres. If at this altitude, temperature and atmospheric pressure are 268 K and `2xx10^(-3)` atm respectively, what weight of helium will be required to fully inflate the balloon ?
Answer» For the fully inflated balloon, we have `V=10^(5)" L", T=288" K", P=2xx10^(-3)" atm"` The aim is to find the mass of helium present in the balloon under these conditions. `PV=nRT=(w)/(M)RT` or `" " w=(MPV)/(RT)=(4 g mol^(-1)xx(2xx10^(-3)" atm")(10^(5)"L"))/((0.0821" L "atm K^(-1)mol^(-1))(268 K))=36.36 g`

4.	Hydrogen gas occupies a volume of `18` litres at `27^(@)C` and under a pressure of `0.92` bar. The number of moles present in the gas is :A. `0.56` molB. `0.67` molC. `0.35` molD. `0.87` mol
Answer» Correct Answer - B According to ideal gas equation `PV = nRT` or `n = (PV)/(RT)` `V = 18 L, P = 0.92` bar `T = 27+273 = 300 K`, `R = 0.083 L "bar" K^(-1) mol^(-1)` `n = ((0.92 "bar") xx (18 L))/((0.083 L "bar" K^(-1)) xx (300 K))` `= 0.067` mol.

5.	A spherical balloon of 21 cm diameter is to be filled with hydrogen at NTP from a cylinder containing the gas at20 atmosphere at `27^(@)C`.If the cylinder can hold 2.82 litres of water, calculate the number of balloons that can be filled up.
Answer» Volume of the balloon `=(4)/(3)pi r^(3)=(4)/(3)xx(22)/(7)xx((21)/(2))^(3)=4851 cm^(3)` Volume of the cylinder =2.82 litres=2820 `cm^(3)` Pressure =20 atm. Temperature =300 K Converting this to the volume at NTP, `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) :. (20xx2820)/(300)=(1xxV_(2))/(273)" or "V_(2)=51324" cm"^(3)`. When the pressure in the cylinder is reduced to one atmosphere,no more `H_(2)` will be released and hence 2820 `cm^(3)` of `H_(2)` will be left in it. Hence, volume of `H_(2)`used in filling the balloons `=51324-2820 cm^(3)=48504 cm^(3)`. Number of balloons filled =48504/4851=10

6.	At `27^(@)C` and under one atosphere pressure, a gas occupies a volume of V L. In case the temperature is increased to `177^(@)C` and pressure to `1.5` bar, the correponding volume will be :A. `V mL`B. `2V mL`C. `V//2 mL`D. `V//3 mL`
Answer» Correct Answer - A `V_(1) = V L , V_(2) = ?` `P_(1) = 1 "bar" , P_(2) = 1.5 "bar"` `T_(1) + 27 + 273 , T_(2) = 177 + 273` `= 300 K` , `=450K` According the ideal 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 "bar")xx (VL) xx (450 K))/((300K) xx (1.5 "bar")) = VL`

7.	At `27^(@)C` and one atmospheric pressure, a gas has volume V. What will be its volume at `177^(@)C` and a pressure of 1.5 atmosphere ?
Answer» `{:("Initial conditions","Final conditions"),(V_(1)=V,V_(2)=?),(P_(1)=atm,P_(2)=1.5" atm"),(T_(1)=273+27=300 K,T_(2)=273+177=450 K):}` Applying gas equation, we have `(1xxV)/(300)=(1.5xxV_(2))/(450) :. V_(2)=(1xxVxx450)/(300xx1.5)=V :. "Volume of the gas"=V`

8.	Find the temperature at which 3 moles of `SO_(2)` will occupy a volume of 10 litre at a pressure of 15 atm `a = 6.71 atm "litre"^(2) mol^(-2), b = 0.0564 " litre " mol^(-1)`
Answer» `(P + (an^(2))/(V^(2))) (V = nb) = nRT` `(15 + (6.71 xx 9)/(100)) (10 - 3 xx 0.0564) = 3 xx 0.082 xx T` `T = 624 K`

9.	What is density of `SO_(2)` gas at `27^(@)C` and 2 atmospheric pressure ? (At. Wts. S=32, O=16, R=0.0821 L atm `K^(-1)mol^(-1)`)
Answer» `d=(PM)/(RT)=(2 atmxx64 g mol^(-1))/(0.0821" L atm K"^(-1)mol^(-1)xx300 K)=5.1969" g L"^(-1)`

10.	The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times.As a result, the diffusion coefficient of this gas increases `x` times. The value of `x` is........
Answer» Correct Answer - 4 Diffusion coefficient `prop` Mean free path `xx` Average speed `prop lamda xx v_(av)` `prop (kT)/(sqrt2 pi sigma^(2)P) xx sqrt((8RT)/(pi m))` `prop (T^(3//2))/(P)` `(("Diffusion coefficient")_("final"))/(("Diffusion coefficient")_("initial")) = ((4)^(3//2))/(2) = 4`

11.	Uranium isotopes have been separated by taking advantage of the different rates of diffusion of the two formsof uranium hexafluoride, one containing U-238 isotope and the other containing U-235. What are the relative rates of diffusion of these two molecules under ideal conditions ?
Answer» Correct Answer - A `(r(overset(238)""UF_(6)))/(r(overset(235)""UF_(6)))=sqrt((M(overset(235)""UF_(6)))/(M(overset(238)""UF_(6))))=sqrt((235+19xx6)/(238+19xx6))=sqrt((349)/(352))=0.9957:1`

12.	Calculate the relative rates of diffusion for `.^(235)UF_(6)` and `.^(238)UF_(6)`.
Answer» Mol. Mass `.^(235)UF_(6) = 235 + 6 xx 19 = 349` Mol. Mass `.^(238)UF_(6) = 238 + 6 xx 10 = 352` `(r_(1))/(r_(2)) = sqrt((M_(2))/(M_(1))) = sqrt((352)/(349)) = 1.0043` `r_(1) : r_(2) :: 1.0043 : 1.0000`

13.	Compare the rates of diffusion of `""^(235)UF_(6)` and `""^(238)UF_(6)`.
Answer» Molecular mass of `""^(235)UF_(6)=235+6xx19=349` , Molecular mass of `""^(238)UF_(6)=238+6xx19=352` `:. (r_(1)(""^(235)UF_(6)))/(r_(2)(""^(238)UF_(6)))=sqrt((M_(2))/(M_(1)))=sqrt((352)/(349))=1.004` Thus, `r(""^(235)UF_(6)) : r(""^(238)UF_(6))=1.004 : 1`

14.	Air open vessel at `127^(@)C` is heated until `1//5^(th)` of air in it has been expelled. Assuming that the volume of vessel remains constant the temperature to which the vessel has been heated is
Answer» In this problem, the volume of the vessel is constant. As the vessel is open, its pressure will also remain constant. According to ideal gas equation, `PV = nRT` `PV = n_(1)RT_(1)` and `PV = n_(2)RT_(2)` or `n_(1)RT_(1) = n_(2)RT_(2)` or `(n_(1))/(n_(2)) = (T_(2))/(T_(1))` Let the initial no. of moles `(n_(1)) = 1` `:.` Final no. of moles `(n_(2)) = 1 - (3)/(5) = (2)/(5) = 0.4` Initial temperature `(T_(1)) = 27+273 = 300 K` Final temperature `(T_(2)) = ?` Substituting the values, `T_(2) = (2) = (n_(1) xx T_(1))/(n_(2)) = (1 xx 300 K)/(0.4) = 750 K` `:.` Final temperature `(T_(2)) = 750-273 = 477^(@)C`.