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Test: Gaseous State - 3 - Chemistry MCQ


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30 Questions MCQ Test Physical Chemistry - Test: Gaseous State - 3

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Test: Gaseous State - 3 - Question 1

If a gas is expanded at constant temperature:

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Test: Gaseous State - 3 - Question 2

The given graph represent the variations of Z (compressibility factor  versus p, for three real gases A, B and C. Identify the only incorrect statement:

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Test: Gaseous State - 3 - Question 3

A gas described by van der Waals’ equation:

Detailed Solution for Test: Gaseous State - 3 - Question 3

For a real gas

(P + an2/V2)(V – nb) = nRT

  1. Volume large => a=0 and V-nb = V. Hence it reduces to PV = nRT

  2. P large but V can’t be neglected. So, P(V-nb) = nRT

  3. a and b are temperature independent.

  4. For real gas a is not 0 but ideal gas has a = 0. Hence ideal gas exerts more pressure on the container walls.

From the above discussion, A, B and D are correct.

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Test: Gaseous State - 3 - Question 4

According to kinetic theory of gases:

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Test: Gaseous State - 3 - Question 5

The root mean square speeds at STP for the gases H2, N2, O2 and HBr are in the order

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Test: Gaseous State - 3 - Question 6

Which of the following are same for all ideal gas at same STP?

Detailed Solution for Test: Gaseous State - 3 - Question 6

A number of molecules in 1 mole for all ideal gas at same STP is 6.023×1023.
The Loschmidt's number is the number of particles (atoms or molecules) of an ideal gas in a given volume.

It is usually quoted at standard temperature and pressure and value is 2.68×1025 per cubic meter at 00C and 1 atm.
Similarly, average kinetic energy per molecule is  KE = 2/3​KT
And for one-mole gas, kinetic energy is KE = 2/3​RT

*Multiple options can be correct
Test: Gaseous State - 3 - Question 7

Rate of effusion of a gas depends upon:

Detailed Solution for Test: Gaseous State - 3 - Question 7

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Test: Gaseous State - 3 - Question 8

Which of the following statements are correct?

Detailed Solution for Test: Gaseous State - 3 - Question 8

a) Vapor refers to a gas phase at a temperature where the same substance can also exist in the liquid or solid statebelow the critical temperature of the substance. (For example, water has a critical temperature of 374 °C (647 K), which is the highest temperature at which liquid water can exist.)

 

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Test: Gaseous State - 3 - Question 9

Which of the following plots represents ideal gas laws:

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Test: Gaseous State - 3 - Question 10

Select the correct statements:

Detailed Solution for Test: Gaseous State - 3 - Question 10

a) Pressure of saturated vapours is independent of their volume at constant temperature.

b) Saturation vapor pressure (or the maximum amount of water vapor that can be in the air) increases exponentially as temperature increases. The sharp increase in saturation vapor pressure with increasing temperature continues beyond 40°C (104°F).

c) When a given volume of air is made more moist by adding water molecules, heavier molecules are replaced with lighter molecules. Therefore, moist air is lighter than dry air if both are at the same temperature and pressure.

d) "During a change in state the heat energy is used to change the bonding between the molecules. In the case of melting, added energy is used to break the bonds between the molecules. In the case of freezing, energy is subtracted as the molecules bond to one another. These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules.

"If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water. Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before, so their average kinetic energy remains the same, and, thus, their Kelvin temperature remains the same."

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Test: Gaseous State - 3 - Question 11

4.215 g of a metallic carbonate was heated in a hard glass tube, the CO2 evolved was found to measure 1336 mL at 27°C and 700 nm of Hg pressure. What is the equivalent weight of the metal:


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Test: Gaseous State - 3 - Question 12

3.7 g of a gas at 25°C occupied the same volume as 0.184 g of hydrogen at 17°C and at the same pressure. What is the molecular weight of the glass:


Test: Gaseous State - 3 - Question 13

A hydrocarbon contains 10.5 g of carbon per gram of hydrogen. 1 L of the vapour of the hydrocarbon at 127°C and 1 atm pressure weights 2.8 g. Find the molecular formula of the hydrocarbon.

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Test: Gaseous State - 3 - Question 14

The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 min when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight 79 in the molar ratio of 1 : 1 at a total pressure of 4000 mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 74 min.


Detailed Solution for Test: Gaseous State - 3 - Question 14

The change of pressure of oxygen in 47 min is 500 mm Hg. The change of pressure of oxygen after 74 min is

the 1:1 molar ratio mixture of oxygen and another gas, each of them has

an equal pressure of 2000 mm because the total pressure is given to be 

4000 mm Hg.
The pressure of oxygen left after 74 min is

Test: Gaseous State - 3 - Question 15

Calculate the average kinetic energy, in Joule per molecule in 8.0 g of methane at 27°C:

Detailed Solution for Test: Gaseous State - 3 - Question 15

Kinetic energy per molecule of a gas:

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Test: Gaseous State - 3 - Question 16

At room temperature, ammonium gas at 1 atm pressure and hydrogen chloride gas at p atm pressure are allowed to effuse through identical pin holes from opposite ends of a glass tube of one metre length and of uniform cross-sect ion. Ammonium chloride is first formed at a distance of 60 cm from the end through which HCl gas is sent in. What is the value of p:


Detailed Solution for Test: Gaseous State - 3 - Question 16

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Test: Gaseous State - 3 - Question 17

Calculate the pressure exerted by one mole of CO2 gas at 273 K, if the van der Waals’ constant ‘a’ = 3.592 dm6 atm mol–2. Assume that the volume occupied by CO2 molecules is negligible.


Detailed Solution for Test: Gaseous State - 3 - Question 17

According to the Van der Waal’s equation we have for one mole of a gas:

[P + a/V2] (V - b) = RT ….(1)

We have

R = 0.082 , T = 273K, V = 22.4 l for 1 mole of an ideal gas at 1 atm pressure.

[P + (3.592 L2 atm/mol2 ) (1.0 mol)2 /(22.4 L)2 ] [22.4 L – (1.0 mol)(0.04267 L/mol)] = (1.0 mol)(0.08206 L atm/mol K)(0+273 K)

P = 0.082 *273/22.4 – 3.592/(22.4)4

P=0.9922 atm

 

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Test: Gaseous State - 3 - Question 18

When 2 g of a gas A is introduced into an evacuated flask kept at 25°C, the pressure is found to be one atmosphere. If 3 g of another gas B is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of the molecule weights MA : MB.


Detailed Solution for Test: Gaseous State - 3 - Question 18

Test: Gaseous State - 3 - Question 19

Oxygen is present in one litre flask at a pressure of 7.6 × 10–10 mm Hg. Calculate the number of oxygen molecules in the flask at 0°C.

Detailed Solution for Test: Gaseous State - 3 - Question 19

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Test: Gaseous State - 3 - Question 20

Calculate the root mean square velocity in cm/s of ozone kept in a closed vessel at 20°C and 82 cm mercury pressure.


Detailed Solution for Test: Gaseous State - 3 - Question 20

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Test: Gaseous State - 3 - Question 21

A spherical balloon of 21 cm diameter is to be filled up with hydrogen at NTP from a cylinder containing the gas at 20 atmospheres at 27°C. If the cylinder can hold 2.82 L of water, calculate the number of balloons that can be filled up.


Detailed Solution for Test: Gaseous State - 3 - Question 21

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Test: Gaseous State - 3 - Question 22

A 4 : 1 molar mixture of He and CH4 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.


Detailed Solution for Test: Gaseous State - 3 - Question 22

Total moles = 4+1 =5, Total pressure = 20 bars
So, partial pressure of He = 4/5 x 20 = 16 bar
Or, partial pressure of CH4 = 20-16 = 4 bar
Now, molar mass of He = 4, molar mass of CH4 = 16

So, ratio of rate of effusion of He (rHe) and rate of effusion of methane (rCH4) will be:

Hence, as the rate of effusion of He is 8 times higher than CH4 so, initial mixture must contain the molar ratio of He: CH4​ to be 8:1.

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Test: Gaseous State - 3 - Question 23

At 27°C, hydrogen is leaked through a tiny hole into a vessel for 20 min. Another unknown gas at the same temperature and pressure as that of hydrogen is leaked through same bole for 20 min. After the effusion of the gases the mixture exerts a pressure of 6 atm. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 L. What is the molecular weight of the unknown gas?


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Test: Gaseous State - 3 - Question 24

A mixture of ethene (C2H6) and ethene (C2H4) occupies 40 L at 1.00 atm and at 400 K. The mixture reacts completely with 130 g of O2 to produces CO2 and H2O. Assuming ideal gas behaviour, Calculate the mole fractions of C2H4 and C2H6 in the mixture.


Detailed Solution for Test: Gaseous State - 3 - Question 24

Test: Gaseous State - 3 - Question 25

In the gas phase, the ratio of excluded volume to molecule volume for a spherical molecule is……………

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Test: Gaseous State - 3 - Question 26

In an ideal monoatomic gas, the speed of sound is given by . If the speed of sound in argon at 25°C us 1245 km h–1, the root mean square velocity in ms–1 is…………….


Detailed Solution for Test: Gaseous State - 3 - Question 26

From kinetic theory of ideal gas,

vs= sqrt[ 5RT/3M]

But,M=mxNo, where m is mass of Argon atom and No is Avogadro number.

Also, R= kNo, where k is Boltzmann constant. Therefore ,

vs=sqrt[ 5 No kT/3mNo]=sqrt[5kT/3m]……(1)

But, (1/2)m<v2 > =(3/2)kT OR

kT=m<v2>/3. Using this in equation (1),

vs= sqrt[5m<v2>/9m]= sqrt[(5/9)<v2] OR

Sqrt(<v2>)=vs x sqrt(9/5)=1245x(5/18)xsqrt(9/5) OR

v rms= 463.9 m/s.

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Test: Gaseous State - 3 - Question 27

A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 × 10–7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 × 10–10 m).


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Test: Gaseous State - 3 - Question 28

The compression factor (compressibility factor) for one mole of a van der Waal’s at 0oC and 100 atmospheric pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals constant a.     


Detailed Solution for Test: Gaseous State - 3 - Question 28

We know that, Compressibility factor, Z = PV/RT

0.5 = 100 *V/0.082 *273

∴ V = 0.1117L

NOTE : Further when volume of a gas molecule is negligible, van der Waal’s equation becomes

(P + a/V2) (V - 0) = RT

Or PV = RT – a/V or a = RTV – PV2

Substituting the values

A = (0.082 * 0.1119 *273) – (100 * 0.1119 * 0.1119)

= 1.253 atm L2 mol-2

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Test: Gaseous State - 3 - Question 29

At 400 K, the root mean square (rms) speed of a gas X (molecule weight = 40) is equal is equal to the most probable speed of gas Y at 60 K. The mo lecular weight of the gas Y is:


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Test: Gaseous State - 3 - Question 30

Using van der Waals’ equation, calculate the constant ‘a’ when two moles of a gas confined in a four litre flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 L mol–1:


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