Gaseous State MCQ- 1 (Advanced)


20 Questions MCQ Test Chemistry for JEE Advanced | Gaseous State MCQ- 1 (Advanced)


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This mock test of Gaseous State MCQ- 1 (Advanced) for JEE helps you for every JEE entrance exam. This contains 20 Multiple Choice Questions for JEE Gaseous State MCQ- 1 (Advanced) (mcq) to study with solutions a complete question bank. The solved questions answers in this Gaseous State MCQ- 1 (Advanced) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Gaseous State MCQ- 1 (Advanced) exercise for a better result in the exam. You can find other Gaseous State MCQ- 1 (Advanced) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
*Multiple options can be correct
QUESTION: 1

Following graph is constructed for the fixed amount of the gas.

   

Incorrect statements are :

Solution:

*Multiple options can be correct
QUESTION: 2

Which of following statements are correct

Solution: (A)As we know Velocity is a vector quantity. If we keep the container at rest then this implies that the Center of mass is at rest and this means that the Velocity of each molecule cancel out other ones! Hence, (A) is absolutely correct option. (B)Each and every molecule have it's own speeds because of intermolecular forces and yes they are different for each molecule. Never-Ever Take Velocity and speed as the same quantity! Hence, (B) is wrong. (C)Let the initial moles be 1 and Pressure and Volume are constant. Then, n1 T1 =n2 T2 1*300=n*400 n=3/4 And Hence, the air which escaped out is 1-3/4= 1/4 Hence, (C) is constant (D)If z is less than 1 O.M.V less than Ideal molar Volume O.M.V less than nRT Observed Molar Volume less than 22.4 Hence, (D) is correct.
*Multiple options can be correct
QUESTION: 3

If the pressure of the gas contained in a closed vessel is increased by 20% when heated by 273°C then it's initial temperature must have been.

Solution:

Let initial temperature = T °C and initial pressure = p
Since volume is constant.
Therefore p∝T
P/(T+273)= 1.2P/(T+273+273)
1.2T+273×1.2=T+273+273
⇒0.2T=273×0.8
T=273×4 = 1092 °C . Its equivalent in K is 1092+273 = 1365 K

QUESTION: 4

 A mixture of H2 and O2 in 2:1 volume is allowed to diffuse through a porous partition what is the composition of gas coming out initially

Solution:
*Multiple options can be correct
QUESTION: 5

There are three closed containers in which equal amount of the gas are filled.If the containers are placed at the same temperatures then find the correct options:

Solution:
*Multiple options can be correct
QUESTION: 6

If the rms velocity of nitrogen and oxygen molecule are same at two different temperature and same pressure then.

Solution:

QUESTION: 7

X ml of H2 gas effuses through a hole in a container in 5 secs. The time taken for effusion of same volume of gas specified below under identical condition is

Solution:
*Multiple options can be correct
QUESTION: 8

With reference to the above graph, choose the correct alternatives

Solution:

At const temperature V α1/p

QUESTION: 9

The system shown in the figure is in equilibrium, where A and B are isomeric liquids and form an ideal solution at T.K. standard vapour pressure of A and B are PA° and PB° respectively at TK. We collect the vapour of A and B in two container of volume V, first container is maintained at 2TK and second container is maintained at 3T
K.
2
At the 
temperature greater than TK, both A and B exist in only gaseous form. Total vapour pressure of the system at TK is given as PT = PA°XA+ PB°XB
Where XA and XB are the mole fraction of A and B in liquid a mixture.
In container (1) We assume that collected gases behave ideally at 2TK and there can take place isomerisation reaction in which A converted into B by first order kinetics Reaction is given as 

In Container (II) At the given temperature 3T/2
, A and B are ideal in nature and non mixing in nature. A small pin hole is made container. We can determine the initial rate of effusion of both gases in vacuum by the expression

Where P = pressure difference between system and surrounding K = positive constant M = Molecular weight of the gas.

Q.

If partial vapour pressure of A is twice that of partial vapour pressure of B and total V.P. is equal to 2T at T = 50 K and V = B.21 lit, then number of mole of A and B in vapour phase is :

Solution:

PT = PA + PB = 3PB

QUESTION: 10

The system shown in the figure is in equilibrium, where A and B are isomeric liquids and form an ideal solution at T.K. standard vapour pressure of A and B are PA° and PB° respectively at TK. We collect the vapour of A and B in two container of volume V, first container is maintained at 2TK and second container is maintained at 3T
K.
2
At the 
temperature greater than TK, both A and B exist in only gaseous form. Total vapour pressure of the system at TK is given as PT = PA°XA+ PB°XB
Where XA and XB are the mole fraction of A and B in liquid a mixture.
In container (1) We assume that collected gases behave ideally at 2TK and there can take place isomerisation reaction in which A converted into B by first order kinetics Reaction is given as 

In Container (II) At the given temperature 3T/2
, A and B are ideal in nature and non mixing in nature. A small pin hole is made container. We can determine the initial rate of effusion of both gases in vacuum by the expression

Where P = pressure difference between system and surrounding K = positive constant M = Molecular weight of the gas.

Q.

Vapour is collected and passed into a container of volume 8.21 lit, maintainer at 50 K and after 5 min number of mole of B = 8/3. Then calculate pressure develop into the container after two half lives.

Solution:

A  →  B

QUESTION: 11

The system shown in the figure is in equilibrium, where A and B are isomeric liquids and form an ideal solution at T.K. standard vapour pressure of A and B are PA° and PB° respectively at TK. We collect the vapour of A and B in two container of volume V, first container is maintained at 2TK and second container is maintained at 3T
K.
2
At the 
temperature greater than TK, both A and B exist in only gaseous form. Total vapour pressure of the system at TK is given as PT = PA°XA+ PB°XB
Where XA and XB are the mole fraction of A and B in liquid a mixture.
In container (1) We assume that collected gases behave ideally at 2TK and there can take place isomerisation reaction in which A converted into B by first order kinetics Reaction is given as 

In Container (II) At the given temperature 3T/2
, A and B are ideal in nature and non mixing in nature. A small pin hole is made container. We can determine the initial rate of effusion of both gases in vacuum by the expression

Where P = pressure difference between system and surrounding K = positive constant M = Molecular weight of the gas.

Q.

If vapour is collected in container of volume 8.21 lit maintained at 75 K then ratio of initial rate of effusion of Gas A and B is given as:

Solution:

QUESTION: 12

Gas 'A' (Molar Mass of A= 128 g mol-1) is taken in a closed container at the initial total pressure of 1000 mm of Hg. Pressure of the gas decreases to 900 torr in 5 seconds due to the diffusion though square cross section. Another similar size container is taken in which gaseous mixture of A and B are taken. (Molar mass of the mix is 472/5)
at  the total pressure of 5000 torr (molar mass of B = 72 g moll. A rectangular crosssection is made in this container and gases are allowed to duffuse. Width of this rectangular cross-section is same as the side of the square cross section and length of the rectangular cross-section is 50% more than the width. Assume that the gases A and B are non reacting and rate of diffusion of the gases are only dependent upon the initial total pressure and it is independent of the change in the pressure due to diffusion. Assume all other conditions to be identical.

Q.

Gas diffused out initially from 2nd container has composition:

Solution:

QUESTION: 13

Gas 'A' (Molar Mass of A= 128 g mol-1) is taken in a closed container at the initial total pressure of 1000 mm of Hg. Pressure of the gas decreases to 900 torr in 5 seconds due to the diffusion though square cross section. Another similar size container is taken in which gaseous mixture of A and B are taken. (Molar mass of the mix is 472/5)
at  the total pressure of 5000 torr (molar mass of B = 72 g moll. A rectangular crosssection is made in this container and gases are allowed to duffuse. Width of this rectangular cross-section is same as the side of the square cross section and length of the rectangular cross-section is 50% more than the width. Assume that the gases A and B are non reacting and rate of diffusion of the gases are only dependent upon the initial total pressure and it is independent of the change in the pressure due to diffusion. Assume all other conditions to be identical.

Q.

Ratio of the number of moles of A and B left in the container after 10 seconds from the diffusion starts is:

Solution:

QUESTION: 14

Gas 'A' (Molar Mass of A= 128 g mol-1) is taken in a closed container at the initial total pressure of 1000 mm of Hg. Pressure of the gas decreases to 900 torr in 5 seconds due to the diffusion though square cross section. Another similar size container is taken in which gaseous mixture of A and B are taken. (Molar mass of the mix is 472/5)
at  the total pressure of 5000 torr (molar mass of B = 72 g moll. A rectangular crosssection is made in this container and gases are allowed to duffuse. Width of this rectangular cross-section is same as the side of the square cross section and length of the rectangular cross-section is 50% more than the width. Assume that the gases A and B are non reacting and rate of diffusion of the gases are only dependent upon the initial total pressure and it is independent of the change in the pressure due to diffusion. Assume all other conditions to be identical.

Q.

What is the time after which container will have same number of moles of A and B:

Solution:

QUESTION: 15

When a sample of ideal gas is changed from an initial state to a final state, various curves can be plotted for the process like P - V curve, V - T curve, P - T curve etc.
For example, P - V curve for a fixed amount of an ideal gas at constant temperature is a rectangular hyperbola, V - T curve for a fixed amount of an ideal gas at constant pressure is a straight line and P - T curve for a fixed amount of an ideal gas at constant volume is again a straight line. However, the shapes may vary if the constant parameters are also changed.
Now, answer the following questions:

Q.

Which of the following statements is correct regarding a fixed amount of ideal gas undergoing the following process:

Solution:

(A) Temperature decreases A to B, so RMS speed decreases

(B) dgas = PM/RT. Upon moving from A to B, K increases and temperature decreases. So, density of gas increases.

(C)This graph is possible if during the process: .

∵ PT = Constant ∴   = Constant  ∵        P2V = Constant or PV1/2 = Constant

(D) If PB = 4PA, then VA = 16VB (according P2V = Constant for process).

QUESTION: 16

When a sample of ideal gas is changed from an initial state to a final state, various curves can be plotted for the process like P - V curve, V - T curve, P - T curve etc.
For example, P - V curve for a fixed amount of an ideal gas at constant temperature is a rectangular hyperbola, V - T curve for a fixed amount of an ideal gas at constant pressure is a straight line and P - T curve for a fixed amount of an ideal gas at constant volume is again a straight line. However, the shapes may vary if the constant parameters are also changed.
Now, answer the following questions:

Q.

Two moles of an ideal gas is changed from its initial state (16 atm, 6L) to final state (4 atm, 15L) in such a way that this change can be represented by a straight line in P - V curve. The maximum temperature attained by the gas during the above change is: (Take R = 1/12L atm K1 mo1-1)

Solution:

Equation of straight line

QUESTION: 17

When a sample of ideal gas is changed from an initial state to a final state, various curves can be plotted for the process like P - V curve, V - T curve, P - T curve etc.
For example, P - V curve for a fixed amount of an ideal gas at constant temperature is a rectangular hyperbola, V - T curve for a fixed amount of an ideal gas at constant pressure is a straight line and P - T curve for a fixed amount of an ideal gas at constant volume is again a straight line. However, the shapes may vary if the constant parameters are also changed.
Now, answer the following questions:

Q.

Which of the following graphs is not possible for a fixed amount of ideal gas upon moving from initial state A to final state B:

Solution:

(A) This graphs is possible if temperature of gas is kept constant and pressure is increased from A to B.

(B) This graphs is possible if temperature of gas is increased continuously during the process A to B.

                        (C) This graphs is possible if during the process: 

QUESTION: 18

In the Figure, isotherms of CO2 at several temperatures near the critical point are shown. At the critical point (critical state), the distinction between the liquid and gaseous states disappear and the density of the gaseous substance is equal to that in the liquid state. For every gas this occurs at specific values of temperature and pressure, called critical temperature and critical pressure respectively. At temperatures and pressures above the critical point value, a gas is said be in a supercritical state. Critical constants are evaluated by solving the Vander W aals equation which is a cubic in volume. The values are Tc 

In the supercritical region the behaviour is studied by plotting Z vs P plots. The variation in this region is studied with respect to boyles temperature a

At temperature 
higher than CB the variation is linear (Plot I) and at temperature lower than CB the variation is non linear (Plot II) With reference to the passage answers the following questions:

Q.

For a real gas at a given temperature, which of the following facts is not correct?

Solution:
*Multiple options can be correct
QUESTION: 19

In the Figure, isotherms of CO2 at several temperatures near the critical point are shown. At the critical point (critical state), the distinction between the liquid and gaseous states disappear and the density of the gaseous substance is equal to that in the liquid state. For every gas this occurs at specific values of temperature and pressure, called critical temperature and critical pressure respectively. At temperatures and pressures above the critical point value, a gas is said be in a supercritical state. Critical constants are evaluated by solving the Vander W aals equation which is a cubic in volume. The values are Tc 

In the supercritical region the behaviour is studied by plotting Z vs P plots. The variation in this region is studied with respect to boyles temperature a

At temperature 
higher than CB the variation is linear (Plot I) and at temperature lower than CB the variation is non linear (Plot II) With reference to the passage answers the following questions:

Q.

In the figure representing variation of the compressibility fac tor z of a real gas with pressure:

Solution:
QUESTION: 20

In the Figure, isotherms of CO2 at several temperatures near the critical point are shown. At the critical point (critical state), the distinction between the liquid and gaseous states disappear and the density of the gaseous substance is equal to that in the liquid state. For every gas this occurs at specific values of temperature and pressure, called critical temperature and critical pressure respectively. At temperatures and pressures above the critical point value, a gas is said be in a supercritical state. Critical constants are evaluated by solving the Vander W aals equation which is a cubic in volume. The values are Tc 

In the supercritical region the behaviour is studied by plotting Z vs P plots. The variation in this region is studied with respect to boyles temperature a

At temperature 
higher than CB the variation is linear (Plot I) and at temperature lower than CB the variation is non linear (Plot II) With reference to the passage answers the following questions:

Q.

A gas at C2° is condensed to liquid following the path XYAB. The liquid appears at the point:

Solution:

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