States of Matter : Gases & Liquids Test 10

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States of Matter : Gases & Liquids Test 10

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Question 1
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Direction (Q. Nos. 1-11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. In van der Waals’ equation of state for non-ideal gas, that includes correction for intermolecular forces is
[ITT JEE 1998, 2009]

A
(V - b)

B
RT

C

D
(RT)^-1

Solution

For correction in intermolecular forces, we have ∆P = an²/V²
So the actual pressure becomes, P+an²/V²
Question 2
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The value of van der Waals’ constant a for various gases are given

Q. The order of liquefaction of these gases is
[IITJEE 1989]

A
N₂ > O₂ > NH₃ > CH₄

B
NH₃ > CH₄ > N₂ > O₂

C
NH₃ < CH₄ < N₂ < O₂

D
N₂ < O₂ < NH₃ < CH₄

Solution

More the value of a, more be easy to liquify the gas.
So, the order of ease of liquefaction is, NH₃ > CH₄ > N₂ > O₂
Question 3
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A gas will approach ideal behavious at
[IIT JEE 1999]

A
low temprature and low pressure

B
low temprature and figh pressure

C
high temprature and low pressure

D
high temprature and high pressure

Solution

At high temperature and low pressure, the gas volume is infinitely large and both intermolecular force as well as molecular volume can be ignored. Under this condition postulates of kinetic theory applies appropriately and gas approaches ideal behavior.
Question 4
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The compressibility of a gas is less than unity for a gas at STP. Therefore,
[IIT JEE 2000]

A
V_m > 22.4 L

B
V_m < 22.4 L

C
V_m = 22.4 L

D
V_m = 44.8 L
Question 5
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Positive deviation from ideal behaviour takes place because of
[IIT JEE 2003]

A
finite size of atoms and

B
molecular interaction between atoms and (pV / nRT) < 1

C
molecular interaction between atoms and (pV / nRT) > 1

D
finite size of atoms and

Solution

Compressibility factor of an ideal gas :
z = pV/nRT = 1
If a gas is deviated from its ideal behaviour, therefore, z >1 or 1 > z.
Hence, for positive deviation of a gas from ideal behaviour z > 1.
Question 6
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Van der Waals ’ equation for one mole of CO₂ gas at low pressure will be
[JEE Advanced 2014]

A

B

C

D

Solution

At low pressure, volume correction for 1 mole of gas is negligible,
b=0 and van der Waals equation becomes
[P+a/V²]V=RT
So, the correct option is A
Question 7
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If V is the volume of one molecule of a gas under given conditions, then van der Waals ’ constant b (also called excluded volume or effective volume) is

A
4V

B

C

D
4VN₀

Solution

`b` is equal to 4 times the volume of molecules in one mole of a gas N₀ molecules
Volumes of one molecule = v
Volume of N₀ molecule =vN₀
Hence, b=4vN₀
Question 8
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Fraction of the effective volume occupied by molecules in 1 mole of a gas at STP (diameter of the molecule is 2 x 10^-10m) is

A
0.045%

B
0.011%

C
0.090%

D
0.022%
Question 9
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The compressibility factors for real gases at low pressure , high pressure and that of gases of low molar masses are Z₁ Z₂ and Z₃. These are
[AIEEE 2012 , JEE Main 2014]

A
a

B
b

C
c

D
d

Solution

The compressibility factor for real gases at low pressure, high pressure and that of gases of low molar masses are z₁, z₂and z₃.The values of z₁, z₂, z₃ can be derived as follows.
At low pressures, the gas equation can be written as,
(P + a/v2m) (Vm) = RT
or Z₁ = Vm / RT = 1 – a/VmRT
At higher pressure,
PV - Pb = RT
So, equating with real gas equation,
PV / RT = 1 + Pb/RT
Again, Z₂ = PV / RT
So, Z₂ = 1 + Pb / RT
Question 10
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a and b are van der Waals’ constants for gases. Chlorine is more easily liquefied than ethane because
[AIEEE 2011]

A
a

B
b

C
c

D
d

Solution

Van der waals constant ′a′ is a mixture of inter molecular forces and ′b′ is a measure of molecular size. A greater value of ′a′ and lesser value of ′b′ supports qualification of a given gas. Therefore, chlorine is more easily liquefied than ethane because
′a′ for Cl₂ > ′a′ for C₂H₆ but ′b′ for Cl₂ > ′b′ for C₂H₆.
Question 11
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For one mole of a van der Waals’ gas when b = 0 and T = 300 K, the pV versus 1/V plot is shown in figure. The value of the van der Waals' constant a (atm L² mol^-2) is

A
1.0

B
1.5

C
3.0

D
4.5

Solution

The correct answer is option B
Vanderwaal's equation for 1 mole of gas is
(P + a/V2 )( V − b ) = RT
But, b = 0......(given)
(P + a/V2) (V) = RT
∴PV= (−a × 1/V) + RT
We know the basic general equation of straight line, y= mx + c
Slope = tan( π − θ ) = −tan θ = −a
So, tanθ = a =
Question 12
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Direction (Q. Nos. 12) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.
Q. Match the parameters given in Column I with the units given in Column II and select the correct answer from the codes given below.

A
a

B
b

C
c

D
d

Solution

‘R’ = J mol^-1K^-1
‘a’ = atm L² mol^-2 = bar (dm³)² mol^-2
‘b’ = L mol^-1 = dm³ mol^-1
Question 13
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Direction (Q. Nos. 13 and 14) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.
Q.
Statement I : At constant temperature, pV versus p isotherm for real gases is not a straight line.
Statement II : At high pressure, all gases have Z > 1, but at intermediate pressure, most gases have Z < 1.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement i

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

Assertion is true. For real gas, pV vs p plot is not a straight line.
The reason for this is that for real gas, we have to consider the intermolecular forces which are ignored in ideal gas.
Reason is also correct.Z>1at higher pressure for all gas but at low pressure, some gases have Z<1. However this is not the correct explanation of assertion.
Question 14
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Statement I : In van der Waals’ equation, pressure correction is due to force of attraction between molecules.
Statement II : Rate of change of momentum is equal to force.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement i

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

Vander waals constant 'a' is a measure of attractive force
The van der waals constant 'a' represents the magnitude of the attractive forces present between gas molecules. Higher is the value of 'a', more easily the gas can be liquefied.
However it has no relation with statement II. Statement II is a general statement.