Explore topic-wise InterviewSolutions in .

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.

251.

A gas will approach ideal behaviour atA. high temperature and high pressureB. low temperature and low pressureC. low temperature nad high pressureD. high temperature and low pressure

Answer» Correct Answer - D
A real gas behaves like an ideal gas when the molecualr attractions are minimum. This is possible when the molecules are as far apart as possible (low pressure) and are moving as fast as possible (high temperature).
252.

Calculate the pressure exerted by `22g` of `CO_(2)` in `0.5 dm^(3)` at `300 K` using (`a`) the ideal gas law and (`b`) the van der Waals equation. Given `a=300.0 kPa dm^(6) mol^(-2)` and `b=40.0 cm^(3) mol^(-1)`.

Answer» Moles of `CO_(2)=(w)/(Molar weight)=(22 g)/(44 g mol^(-1))=0.5 mol`
`V=0.5 dm^(3)`, `T=300 K`, `a=300.0 kPa dm^(6)mol^(-2)`
`b=40.0 cm^(3)mol^(-1)=0.04dm^(3)mol^(-1)`
(`a`) From ideal gas law `P=(nRT)/(V)`, we have
`P=((0.5)(8.314)xx300)/(0.5)=2494.2 Pa=2.49xx10^(3)kPa`
(`b`) From the van der Waals equation, we get
`P=(nRT)/(V-nb)-(n^(2)a)/(V^(2))`
`:. P=((0.5)xx(8.314)xx300)/((0.5)-(0.5)(0.04))-((0.5)^(2)xx(300))/((0.5)^(2))`
`=2598.12 kPa-300kPa=2298.12 kPa`
253.

The van der Waals constant `b` of `Ar` is `3.22xx10^(-5) m^(3) mol^(-1)`. Calculate the molecular diameter of `Ar`.

Answer» Use `b=4xx`volume occupied by the molecules in `1 mol` of a gas
`b=4xxN_(0)xx((4)/(3)pi r^(3))`
`3.22xx10^(-5)=4xx6.023xx10^(23)xx(4)/(3)xx(22)/(7)xxr^(3)`
`r=[(3.22xx10^(-5)xx3xx7)/(4xx6.023xx10^(23)xx4xx22)]^(1//3)`
`=0.1472xx10^(-9)m=0.1472 nm`
`d=2r=0.2944 nm`
254.

Calculate the pressure exerted by one mole of `CO_(2)` gas at `273 K` van der Waals constant `a=3.592 dm^(6) atm mol^(-2)`. Assume that the volume occupied by `CO_(2)` molecules is negligible.

Answer» The van der Waals equation for one mole of a gas is
`(P+(a)/(V^(2)))(V-b)=RT`
It is given that the volume occupied by `CO_(2)` molecules is negligible. Hence, the equation of state becomes
`(P+(a)/(V^(2)))(V)=RT`
or `p=(RT)/(V)-(a)/(V^(2))`
Assuming `V_(m)=22.414 dm^(3) mol^(-1)`, we get
`p=((8.314 kPa dm^(3) K^(-1) mol^(-1))(273 K))/((22.414 dm^(3) mol^(-1))-(3.592 dm^(6)atm mol^(-2))/((22.414 dm^(3) mol^(-1))^(2))`
`=101.246 kPa-7.15xx10^(-3)atm`
`=101.246 kPa-(7.15xx10^(-3)atm)((101.325 kPa)/(1 atm))`
`=101.264 kPa-0.724 kPa`
`=100.601 kPa`
255.

Two van der Waals gases have the same value of `b` but different values of `a`. Which of these will occupy greater volume under identical conditions. If the gases have the same value of `a` but different values of `b`, which of them will be more compressible?

Answer» When two gases have the same value of `b` but different values of `a`, the gas having a larger value of `a` will occupy lesser volume. This is because the gas with a larger value of `a` will have larger forces of attraction and hence lesser distance between its molecules.
When two gases have the same volume of `a` but different values of `b`, the smaller the value of `b`, the larger the compressibility because the gas with the smaller value of `b` will occupy lesser volume and hence will be more compressible.
256.

Calculate the molecular diameter of helium from its van der Waals constant `b`. `(b=24cm^(3)mol^(-1))`

Answer» Since `b=4xx` Volume occupied by the molecules in `1 mol` of a gas
or `b=4 N_(A)((4)/(3)pi r^(3))`
`:. R=((3b)/(16 N_(A)pi))^(1//3)`
`={(3xx24)/(16(6.023xx10^(23))(3.14))}^(1//3)`
`=1.335xx10^(-8)cm=133.5 Pm`
`:. D=2r=2xx133.5 Pm=267 Pm`
257.

The coefficient of viscosity `eta` of a liquid can be defined from the equationA. `F = eta u (d A)/(d x)`B. `F = eta A (du)/(dx)`C. `F = eta A (dx)/(du)`D. `F = eta u (dx)/(dA)`

Answer» Correct Answer - B
`F = eta A (du)/(dx)`
where `F` is the tangential force between any two adjacent layers of a liquid of area of contact `A , dx` is the distance between the layers, and `du` is the difference in their velocities. Viscosity coefficient is the tangetial force when velocity gradient is unity and the area of constant is unit area.
258.

A flask containing `12 g` of a gas relative molecular mass `120 at` a pressure of `100 atm` was evacuated by means of a pump until the pressure was `0.01 atm`. Which of the following in the best estimate of the number of molecules left in the flask `(N_(0)=6xx10^(23)mol^(-1))`?A. `6xx10^(9)`B. `6xx10^(18)`C. `6xx10^(17)`D. `6xx10^(13)`

Answer» `P_(1)V_(1)=n_(1)RT_(1)`
`P_(2)V_(2)=n_(2)RT_(2)`
`(n_(2))/(n_(1))=(P_(2))/(P_(1))`
`n_(2)=0.1xx(0.01)/(100)=10^(-5) mol= 6xx10^(18)` molecules
259.

The coefficient of viscosity `(eta)` of a fluid moving steadily between two surface is given by the formula `(f) = etaA dV//dx` where f is the frictional force on the flluid, A is the area in the fluid, and `dN//dx` is velocity gradient inside the fluid at that area. The SI unit of viscosity is given as :A. `kg m^(-1) s^(-1)`B. `Nm^(-2) s`C. NilD. Newtons

Answer» Correct Answer - B
is the correct answer.
260.

When `2 g` of a gas `A` is introduced into an evacuated flask kept at `25^(@)C`, the pressure is found to be `1 atm`. If `3 g` of another gas `B` is then heated in the same flask, the total pressure becomes `1.5 atm`. Assuming ideal gas behaviour, calculate the ratio of the molecular weights `M_(A)` and `M_(B)`.

Answer» Let the molecular masses of A and B be `M_(A)` and `M_(B)` respectively.
Pressure exerted by the gas `B = (1.5 - 1.0) = 0.5` atm. Volume and temperature are same in both the gases.
For gas A: `P = 1 atm, w = 2 g, M = M_(A)`
We know that, `PV = (w)/(M) RT`
`1 xx V = (2)/(M_(A)). RT " or " M_(A) = (2RT)/(V)`....(i)
For gas B: `P = 0.5 atm, w = 3g, M = M_(B)`
`0.5 xx V = (3)/(M_(B)).RT " or " M_(B) = (3RT)/(0.5 xx V)`....(ii)
Dividing Eq. (i) by Eq. (ii).
`(M_(A))/(M_(B)) = (2RT)/(V) xx (0.5 xx V)/(3RT)`
`= (2 xx 0.5)/(3) = (1)/(3)`
Thus, `M_(A) : M_(B) = 1 : 3`
261.

What is `SI` unit of viscosity coefficient `(eta)` ?A. pascalB. `N s m^(-2)`C. `km^(-2)s`D. ` N m^(-2)`

Answer» Correct Answer - B
262.

What is SI units of viscosity coefficient `(eta)` ?A. PascalB. Ns `m^(-2)`C. `km^(-2)`sD. N `m^(-2)`

Answer» Correct Answer - B
The SI unit of viscosity coefficient `(eta)` is N `m^(2)` s or N s `m^(-2)` page 5/63).
263.

At `100^(@)C`, liquid water and water vapour (stea) are present together in equilibrium. Comment on the average kinetic energy of water molecules in the liquid water and steam.

Answer» As both liquid water and steam are at the same temperature, they will have same average kinetic energy of their molecules.
264.

As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant ?A. IncreaseB. decreasesC. remains sameD. becomes half

Answer» Correct Answer - A
At constant volume, pressure increases in temperature of the gas.
265.

What is the ratio of average kinetic energy of oxygen molecules to that of ozone molecules at `27^(@)C` ?

Answer» Average kinetic energy of any gas depends only on temperature and not upon the nature of the gas. Hence, both the gases will have same average kinetic energy at `27^(@)C`, i.e., the ratio will be 1:1.
266.

A steel tank containing air at 15 pressure at `15^(@)C` is provided with a safety valve that will yield at a pressure of 30 atm. To what minimum temperature must the air be heated to blow the safety valve ?

Answer» Correct Answer - C
`(P_(1))/(T_(1))=(P_(2))/(T_(2)), i.e.,(15" atm")/(288 K)=(30" atm")/(V_(2))" or "T_(2)=576 K=(576-273)^(@)C=303^(@)C`
267.

Gases posses characteristic critical temperature which depends upon the magnitude of intermolecular forces upon the magnitude of intermolecular forces between the particles. Following are the critical temperature of some gases. `{:("Gases",H_(2),He,O_(2),N_(2)),("Critical temperature",,,,),("in Kelvin",33.2,5.3,154.3,126):}` From the above data what would be the order of liquefaction of these gases ? Start writing the order from the gas liquefying firstA. `H_(2).He,O_(2),N_(2)`B. `He, O_(2),H_(2),N_(2)`C. `N_(2),O_(2),He,H_(2)`D. `O_(2),N_(2),H_(2),He`

Answer» Correct Answer - D
Higher the critical temperature, more easily is the gas liquefied. Hence, order of liquefaction starting with the gas liquefying first will be : `O_(2),N_(2),H_(2),He`.
268.

Assertion (A) : The temperature at which vapour pressure of a liquid is equal to the external pressure is called boiling temperature. Reason (R ) : At high altitude, atmospheric pressure is high.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true R is not the correct explanation of A.C. A is true but R is false.D. A is false but R is true.

Answer» Correct Answer - C
Correct R. At high altitude, atmospheric pressure is low.
269.

As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant ?A. increasesB. decreasesC. remains sameD. becomes half

Answer» Correct Answer - A
At constant volume, as the temperature is increased, pressure increases.
270.

The variation of vapour pressure of different liquids with temperature is shown in the figure. (i) Calculate graophically boiling points of liquids A and B. (ii) If we take liquid C in a closed vessel and heat it continuosly, at what tempeature will it boil ? (iii) At high altitude, atmosopheric pressure is low (say `60 mm Hg`). At what temperature liquid D boils?

Answer» Please remember that a liquid boils at a temperature when the vapour pressure on its surface becomes equal to atmospheric pressure. It depends upon the height or altitude of the place.
(i) The boiling point of liquid A is approximatly 300 K and that of liquid B, it is approximately 315 K.
(iii) If the atmosophric pressure is close to 60 mm, the liquid D will boil at about 313 K.
271.

The pressure of a 1:4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen ?A. `0.8xx10^(5)` atmB. `0.008 N m^(-2)`C. `8xx10^(4) N m^(-2)`D. 0.25 atm

Answer» Correct Answer - C
Partial pressure of `O_(2)`=Mole fraction of `O_(2)`xxTotal pressure of mixture
`=(4)/(5)xx1 "atm"=0.8xx10^(5)" N "m^(-2)=8xx10^(4)" N "m^(-2)`.
272.

Why does the boundary between liquid phase and gaseous phase disappear on heating a liquid upto critical temperature in a closed vessel ? In this situation what will be the state of the substance ?

Answer» when the liquid is heated upto critical temperature. The density of the liquid and gaseous phase of a substance become equal. As a result, the liquid passes into the gaseous state continuously and the boundry of separation dissappears. The substance is said to be in the fluid state which includes both the liquid and gaseous states.
273.

Pick out the wrong statement (s). (i) Vapour pressure of a liquid is the measure of the strength of intermolecular attractive forces . (ii) Surface tension of a liquid acts perpendicular to the surface of the liquid . (iii) Vapour pressure of all liquids is same at their freezing points . (iv) Liquids with stronger intermolecular attractive forces are more viscous than those with weaker intermolecular force .A. (ii) , (iii) and (iv)B. (ii) and (iii)C. (i) , (ii) and (iii)D. (iii) only

Answer» Correct Answer - D
Vapour pressure of all liquids are different at their freezing points.
274.

A 4.0 `dm^(3)` flask containing `N_(2)` at 4.0 bar was connected to a 6.0 `dm^(3)` flask containing helium at 6.0 bar, and the gases were allowed to mix isothermally, then the total pressure of the resulting mixture will beA. 10.0 barB. 5.2 barC. 1.6 barD. 5.0 bar

Answer» Correct Answer - B
At constant temperature,
`P_(1)V_(1)+P_(2)V_(2)=P_(3)(V_(1)+V_(2))`
`(4.0" bar")(4.0" dm"^(3))+(6.0" bar")(6.0" dm"^(3))`
`=P_(3)(4.0+6.0" dm"^(3))`
or `P_(3)=(16+36)/(10)=(52)/(10)=5.2" bar"`
275.

Give reasons for the following : (i) The size of weather balloon becomes larger and larger as it ascends into higher altitudes. (ii) Tyres of automobiles are inflated to lesser pressure in summer than in winter.

Answer» (i) As we go to higher altitudes, the atmospheric pressure decreases. Thus, the pressure outside the balloon decreases. To regain equilibrium with the external pressure, the gas inside expands to decrease its pressure. Hence, the size of the balloon increases.
(ii) In summer, due to higher temperature, the average kinetic energy of the air molecules inside the typre increases, i.e., molecules start moving faster. Hence, the pressure on the walls of teh tube increases. If pressure inside is not kept low at the time of inflation, at higher temperature, teh pressure may become so high that the tyre may burst.
276.

Why are tyres of automobiles inflated to lesser pressure in summmer than in water ?

Answer» The pressure of the air is directly proportional to the temperature. Since the temperature is higher in summer than in winter. the pressure of the air in the tube of the tyre is likely to tbe quite high as compared to winter. It is quite likely that the tube may burst under high pressure in summer. Therefore, its is advisable to inflate the tyres to lesser pressure in summer than in winter.
277.

The equalitative sketches I, II and III given below show the variation of surface tension with molar concentration of three diferent aqueous solutions of `KCl, CH_(3)OH` and `CH_(3)(CH_(2))_(11)OSO_(3)^(-)Na^(+)` at room temperature. The correct assignment of the sketches isA. `KCl, CH_(3)OH, CH_(3)(CH_(2))_(11)OSO_(3)^(-)Na^(+)`B. `CH_(3)(CH_(2))_(11)OSO_(3)^(-)Na^(+), CH_(3)OH, KCl`C. `KCl, CH_(3)(CH_(2))OSO_(3)^(-)Na^(+), CH_(3)OH`D. `CH_(3)OH, KCl, CH_(3)(CH_(2))_(1))OSO_(3)^(-)Na^(+)`

Answer» Correct Answer - D
278.

Calculate the pressure exerted by `8.5 g` of ammonia `(NH_(3))` contained in a `0.5 L` vessel at `300 K`. For ammonia, `a=4.0 atm L^(2)mol^(-2)`, `b=0.036 L mol^(-1)`.

Answer» Number of moles of ammonia
`n=(8.5)/(17)=0.5 mol`
According to van der Waals equation
`(P+(an^(2))/V^(2))(V-nb)=nRT`
`P=(nRT)/((V-nb))-(an^(2))/(V^(2))=(0.5xx0.082xx300)/((0.5-0.5xx0.036))-(4(0.5)^(2))/((0.5)^(2))`
`=21.51 atm`
279.

If `X` is the total number of collisions that a gas molecule registers with other molecules per unit time under particular conditions, then what is the collision frequency of the gas containing `N` molecules per unit volume is ?

Answer» We know that collision number
`Z_(1)=sqrt(2)pir^(2)u_(av)N^(*)=X`
While collision frequency
`Z_(11)=(1)/(2)Z_(1)N`
Hence, collision frequency will be
`Z_(11)=(1)/(2)NX`
280.

Calculate the density of `NH_(3)` at `30^(@)C` and `5 atm` pressure.

Answer» Density of a gas `=(PM)/(RT)`
`=(5xx17)/(0.0821xx303)=3.417 g L^(-1)`
281.

If `3.7 g` of a gas at `25^(@)C` occupies the same volume as `0.814 g` of hydrogen at `17^(@)C` and at the same pressure, then what is the molecular weight of the gas?

Answer» `PV=nRT`
`P` and `V` are the same.
So, `n_(1)T_(1)=nT_(2)`
or `(w_(1))/(M_(1))T_(1)=(w_(2))/(M_(2))T_(2)`
or `(3.7)/(M_(1))xx298=(0.184)/(2)xx290`
or `M_(1)=41.33 g`
282.

The number of collisions made by a single molecule with other molecules per `cm^(3)` per second. Is `Z_(1)`. At constant temperature by how much will `Z_(1)` change if the pressure is doubled in the vessel.A. Increase `2` timesB. Decrease `2` timesC. Increase `0.5` timesD. Decrease `0.5` times

Answer» `Z_(1)=sqrt(2) pi sigma^(2)overline(c ) N^(*)` (`overline(c ) =` average velocity)
(`N^(*)=` Number of molecules per `cm^(3)`)
`Z_(1) prop P` (when `T` is constant)
`Z_(1) prop (1)/(sqrt(T))` (when `P` is constant)
`Z_(1) prop sqrt(T)` (when volume is constant)
283.

Two gaseous molecules `A` and `B` are traveling towards each other. Let the mean free path of the molecule be `sigma` and `Z` be the collision number with other molecules at pressure `1 atm`. Answer the following questions If the mean free path is `sigma` at `1 atm` pressure, then its value at `5 atm` pressure isA. `5sigma`B. `(2)/(5)sigma`C. `(sigma)/(5)`D. None

Answer» `sigma prop (1)/(P)`.
Therefore, `sigma` at `5 atm` is `(sigma)/(5)`.
284.

The number of bimolecular collisions per `cm^(3)` per second is `Z_(11)`. At constant temperature, by how much will `Z_(11)` change if the pressure is tripled in the vessel?A. Increase `3` timesB. Decrease `3` timesC. Increase `9` timesD. Decrease `9` times

Answer» (`i`) `Z_(11)=(1)/(2)(Z_(1)N^(*))=(1)/(sqrt(2))pi sigme^(2)overline(c ) N^(*2)`
(` overline(c )=` average velocity)
(`N^(*)=` Number of molecules per `cm^(3)`)
If collision involves two unlike molecules, then `Z_(12) per cm^(3)` per second is
`Z_(12)=pi sigma_(12)^(2)overline(c )n_(1)n_(2)`, `overline(c )=sqrt((8KT)/(pi mu))`
`n_(1)` and `n_(2)` are the numbrs of molecules `per cm^(3)` of the two types of molecules, `sigma_(12)` is the average diameter of the two molecules.
i.e., `sigma_(12)=((sigma_(1)+sigma_(2))/(2))`
If `mu` is the reduced mass,
`(1)/(mu)=(1)/(m_(1))+(1)/(m_(2))`
(`ii`) `Z_(11) prop P^(2)` (if `T` is constant)
`Z_(11) prop (1)/(T^(3//2))` (if `P` is constant)
`Z_(11) prop sqrt(T)` (if `V` is constant)
`Z_(11) prop sqrt(P)` (if `V` is constant)
285.

Critical temperature of a substance is defined as :A. the temperature above which a substance decomposesB. the substance can exist only as a gasC. melting point of the substanceD. boiling point of the substance.

Answer» Correct Answer - B
Above the critical temperature, a substance exists only as a gas and cannot be liquefied.
286.

At the critical temperature , the substance exists as aA. liquidB. gasC. both (1) and (2)D. fluid

Answer» Correct Answer - D
When liquid is heated in a closed vessel , vapor pressure increases continuously. At first , a clear boundary is visible between the liquid and vapor phases because liquid is more dense than vapor. As the temperature increases , more vapor rises. At the same time , liquid becomes less dense. It expands because the molecules move apart. When the density of liquid and vapors becomes the same, the clear boundary between liquid an dvapors disappears. This temperature is called critical temperature. At the critical temperature , there is no fundamental distinction between a liquid and a gas - we simply have a fluid.
Note that intermolecular forces are independent of temperature, the kinetic energy of molecules increases with temperature.
287.

Calculate the values of `sigma`, `l` (mean free path),`Z_(1)` and `Z_(11)` for oxygen at `300 K` at a pressure of `1 atm`. Given `b=3.183xx10^(-2) dm^(3)mol^(-1)`.

Answer» Number of molecules per unit volume
`N^(*)=(P)/(kT)`, where `k=` Boltzmann constant
`:. N^(*)=(1 atm)/((1.38xx10^(-23)JK^(-1))(300K))`
`=0.241xx10^(22)m^(-3)`
`=2.41xx10^(22)m^(-3)`
The van der Waals constant is
`b=4N_(A)((4)/(3)pir^(3))`
`:. r=((3b)/(16 pi N_(A)))^(1//3)`
`=((3xx3.183xx10^(-2)dm^(3) mol^(-1))/(16xx3.14xx6.023xx10^(23)mol^(-1)))^(1//3)`
`=1.467xx10^(-9)dm`
Therefore,
`sigma` (collision diameter)`=2r`
`=2xx1.467xx10^(-9) dm`
`=2.934xx10^(-10)m`
Average speed
`u_(av)=swrt((8RT)/(pi M))`
`={(8(8.314JK^(-1)mol^(-1))(300 K))/((3.14)(0.032 kg mol^(-1)))}^(1//2)`
`=445.6 m s^(-1)`
Mean free path
`l=(1)/(sqrt(2)pi sigma^(2)N^(*))`
`=(1)/((1.414)(3.14)(2.934xx10^(-10)m)^(2)(241xx10^(23)m^(-3)))`
`=3.18xx10^(-7)m`
`Z_(1)=sqrt(2)pi sigma^(2)u_(av)N^(*)`
`=(1.414)(3.14)(2.934xx10^(-10)m)^(2)(445.6 m s^(-1))xx(241.0xx10^(23)m^(-3))`
`=1.397xx10^(9)s^(-1)`
`Z_(11)=(1)/(2)Z_(1)N^(*)`
`=(1)/(2)(1.397xx10^(9)s^(-1))(241.0xx10^(23)m^(-3))`
`=1.68xx10^(34)m^(-3)s^(-1)`
288.

The critical temperature of a substance isA. The temperature above which a substance can exist only as a gasB. Boiling point of the substanceC. All are wrong.D.

Answer» The critical temperature of a substance is the temperature above which a substance can exist only as a gas.
289.

The magnitude of surface tension of liquid depends on the attractive forces between the molecules. Arrange the following in increasing order of surface tension : water, alcolhol `(C_(2)H_(5)OH)` and hexane `[CH_(3)(CH_(2))_(4)CH_(3)]`.

Answer» Attractive forces are minimum in hexane (only London forces). Hydrogen bonding in `H_(2)O` is stronger than that in `C_(2)H_(5)OH`. Hence the order of surface tension is : hexane `lt` alcohol`lt`water.
290.

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape ?A. `3//8`B. `1//2`C. `1//8`D. `1//4`

Answer» Correct Answer - C
Let the moles of each gas `= x`
Fraction `H_(2)` escaped `= y_(2)x`.
`(r_(O_(2)))/(r_(H_(2)))=sqrt((M_(H_(2)))/(M_(O_(2))))=(nO_(2)//t)/(x/2H_(2)//t) = sqrt((2)/(32)) = 1/4`
`n_(o_(2)) = 1/4 xx x/2 = x/8`
`n_(o_(2)) = (1)/(4) xx x/2 = x/8`
`:.` Fraction of `O_(2)` escaped `= 1//8`
291.

Equal moles of hydrogen and oxygen gas are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half fo the hydrogen to escape ?A. `3//8`B. `1//2`C. `1//8`D. `1//4`

Answer» Correct Answer - C
Suppose the number of moles of each gas taken
=s moles
`H_(2)` escaped in time `t=(x)/(2)` mole
Suppose number of moles of `O_(2)` diffused in the same time=n times
Then `(r_(O_(2)))/(r_(H_(2)))=sqrt((M_(H_(2)))/(M_(O_(2))))`
i.e., `(n//t)/((x//2)//t)=sqrt((2)/(32))" or " (2n)/(x)=(1)/(4)" or " n=(1)/(8)x`
`:.` Fraction of `O_(2)` escaped`=(1)/(8)`.
292.

A type tube with a pin hole is first filled with oxygen to a pressure of 30 Ib/sq inch and allowed to leak out. Then it is filled with `N_(2)` gas to the same pressure and allowed to leak out again. In which case, the time taken will be more and why ?

Answer» The time taken will be more in case the type tube is filled with `O_(2)`. This is because `O_(2)` has higher molar mass than `N_(2)` and rate of effusion is inversely proportional to the square root of molar mass
293.

Calculate the temperature at which the root mean square velocity, the average velocity, and the most proable velocity of oxygen gas are all equal to `1500 m s^(-1)`.

Answer» (`a`) Root mean square velocity is
`sqrt((3RT_(1))/(M))=1.5xx10^(3)m s^(-1)`
`:. T_(1)=((1.5xx10^(3) m s^(-1))^(2)(32xx10^(3) kg mol^(-1)))/(3xx8.314 J K^(-1) mol^(-1))`
(Molar mass of oxygen `= 32 g mol^(-1)`)
`2886 K`
(`b`) Average velocity `=((8RT_(2))/(pi M))^(1//2)=1.5xx10^(3) m s^(-1)`
`:. T_(2)=((1.5xx10^(3) m s^(-1))^(2)(32xx10^(-3) kg mol^(-1))(3.1416))/(8xx8.314 J K^(-1) mol^(-1))`
`=3399 K`
(`c`) Most probable velocity `= sqrt((2RT_(3))/(M))=1.5xx10^(-3)ms^(-1)`
`:. T_(3)=((1.5xx10^(-3) m s^(-1))^(2)(32xx10^(-3) kg mol^(-1)))/(2xx8.314 J K^(-1) mol^(-1))`
`=4330 K`
294.

A vessel of volume `8.0 xx 10^(-3) m^(3)` contains an ideal gas at `300 K` and `210 kPa`. The gas is allowed to leak till the pressure falls to `135 kPa`. Calculate the amount of the gas (in moles) leaked assuming the temperature to remain constant.

Answer» There is no change in volume and temperature of the gas during leakage. However, the number of moles of the gas or its mass changes.
According to ideal gas equation, `PV = nRT` or `n = (PV)/(RT)`
Before leakage, `n_(1) = (P_(1)V)/(RT)`
After leakage, `n_(2) = (P_(2)V)/(RT)`
`:.` Change in no. of moles i.e., `(n_(1) + n_(2)) = (V(P_(1) - P_(2)))/(RT)`
According to available data,
`V = 8.0 xx 10^(-3)m^(3) , P_(1) = 210 xx 10^(3) Nm^(-2)`,
`P_(2) = 135 xx 10^(3) Nm^(-2) , R = 8.314 Nm K^(-1) mol^(-1)`,
`T = 300 K`
Substituting the values, `Deltan = ((8.0 xx 10^(-3)m^(3)) xx (210 - 135) xx (10^(3) Nm^(-2)))/((8.314 NmK^(-1)mol^(-1)) xx 300K) = 0.240 mol`.
`:.` No. of moles of gas leaked `= 0.240` mol
295.

Calculate the root mean square, average, and most proable speeds of `H_(2)` molecules. The density of the gas at `101.325 kPa` is `0.09 g dm^(-3) (0.09 kg m^(-3))`. Assume ideal behaviour.

Answer» `u_(rms)=sqrt((3RT)/(M))=sqrt((3PV)/(M))=sqrt((3P)/(d))={(3(101.325xx10^(3)Pa))/((0.09kg m^(-3)))}^(1//2)`
`=1838 m s^(-1)`
`u_(av)=sqrt((8RT)/(piM))=sqrt((8PV)/(piM))=sqrt((8P)/(pid))={(8(101.325xx10^(3)Pa))/(3.14(0.09kg m^(-3)))}^(1//2)`
`=1694 m s^(-1)`
`u_(MP)=sqrt((2RT)/(M))=sqrt((2PV)/(M))=sqrt((2P)/(d))={(2(101.325xx10^(3)Pa))/((0.09kg m^(-3)))}^(1//2)`
`=1501 m s^(-1)`
296.

A `3:2` molar mixture of `N_(2)` and `CO` is present in a vessel at `500 "bar"` pressure. Due to hole in the vessel, the gas mixture leaks out. The composition of mixture effusing out initially isA. `n_(N_(2)):n_(CO): : 1:2`B. `n_(N_(2)):n_(CO): : 6:1`C. `n_(CO):n_(N_(2)): : 1:2`D. `n_(CO):n_(N_(2)): : 2:3`

Answer» Molar ratio of `N_(2)` and `CO` is `3:2`, i.e., `300 bar` and `200 bar`, respectively.
`(n_(N_(2)))/(n_(CO))=sqrt((m_(CO))/(mN_(2)))xx(P_(N_(2)))/(P_(CO))=(300)/(200)=(3)/(2)`
297.

A `4:1` mixture of helium and methane is contained in a vessel at 10 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The composition of the mixture effusing out initially isA. `8:1`B. `8:3`C. `4:1`D. `1:1`

Answer» Correct Answer - A
Pressure of helium=8 bar
Pressure of `CH_(4)=2` bar
`(r_(He))/(r_(CH_(4)))=(P_(1))/(P_(2))sqrt((M_(CH_(4)))/(M_(He)))=(8)/(2)sqrt((16)/(2))=(8)/(1)=8:1`
298.

An iron cylinder contains helium at a pressure of 250 kPa at 300K. The cylinder can withstand a pressure of `1xx10^(6)Pa`. The room in which cylinder is placed catches fire. Predict whether the cylinder will blow up before it melts or melts or not. (M.P. of the cylinder=1800 K)

Answer» `P_(1)=250 K Pa," "T_(1)=300 K`
`P_(2)=? " "T_(2)=1800 K`
Applying pressure-temperature law, `(P_(1))/(T_(2))=(P_(2))/(T_(2))" "," "(250)/(300)=(P_(2))/(1800)" or "P_(2)=1500 kPa`
As the cylinder can withstand a pressure of `10^(6)Pa=10^(3) kPa=1000 kPa`, hence it will blow up.
299.

A `4:1` molar mixture of `He` and `CH_(4)` is contained in a vessel at `20 bar` pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially?

Answer» Pressure `=20 bar`
Molar ratio `=4:1` (`He` and `CH_(4)`)
`P_(He)=(4)/(5)xx20=16 bar`
`P_(CH_(4))=(1)/(5)xx20=4 bar`
`(r_(1))/(r_(2))=(P_(1))/(P_(2))xxsqrt((M_(2))/(M_(1)))`
`(r_(He))/(r_(CH_(4)))=(P_(He))/(P_(CH_(4)))xxsqrt((M_(CH_(4)))/(M_(He)))=(16)/(4)xxsqrt((16)/(4))=8`
Therefore, the ratio of moles of heilum and methane coming out initially is `8:1`.
300.

A gas with molecular formula `C_(n)H_(2n + 2)` diffuses through a porous plug at a rate `1//6th` of the rate of diffusion of hydrogen gas under similar condtition. The formula of the gas is:A. `C_(2)H_(6)`B. `C_(10)H_(22)`C. `C_(5)H_(12)`D. `C_(6)H_(14)`

Answer» Correct Answer - C
`(r_(1))/(r_(2)) = sqrt((M_(2))/(M_(1))) , (r_(1))/(r_(1)//6) = sqrt((M_(2))/(2)) , M_(2) = 72`
`:.` The forumla of gas will be `C_(5)H_(12)`