States of Matter: Gases and Liquids — Page 6 | NEET Chemistry

Q51

Kinetic theory of gases proves

A only Boyle's law

B only Charle's law

C only Avogadro's law

D All of these

Correct Answer

Option D

Solution

Kinetic theory of gases proves all the given gas laws.

Q52

According to the kinetic theory of gases, in an ideal gas, between two successive collisions of a gas molecule travels

A in a wavy path

B in a straight line path

C with an accelerated velocity

D in a circular path

Correct Answer

Option B

Solution

According to kinetic theory the gas molecules are in a state of constant rapid motion in all possible directions colloiding in a random manner with one another and with the walls of the container and between two successive collisions molecules travel in a straight line path but show haphazard motion due to collisions.

Q53

At the sea level, the dry air mass percentage composition is given as nitrogen gas: 70.0 , oxygen gas: 27.0 and argon gas: 3.0 . If total pressure is 1.15 atm , then calculate the ratio of following respectively: (i) partial pressure of nitrogen gas to partial pressure of oxygen gas (ii) partial pressure of oxygen gas to partial pressure of argon gas (Given: Molar mass of N, O and Ar are 14, 16 and

40 \mathrm{~g} \mathrm{~mol}^{-1}

respectively.)

A

5.46,17.8

B

2.96,11.2

C

4.26,19.3

D

2.59,11.85

Correct Answer

Option B

Solution

Partial Pressure Ratio of Nitrogen to Oxygen : The partial pressure of a gas is calculated using its mole fraction.

The mole fraction is the ratio of the number of moles of the gas to the total number of moles of all gases.

For nitrogen ( $\text{N}_2$ ), the mole fraction $\dfrac{n_{\text{N}_2}}{n_{\text{N}_2} + n_{\text{O}_2} + n_{\text{Ar}}}$ can be simplified as $\dfrac{70/28}{70/28 + 27/32 + 3/40}$ .

For oxygen ( $\text{O}_2$ ), the mole fraction is $\dfrac{27/32}{70/28 + 27/32 + 3/40}$ .

Therefore, the ratio of the partial pressure of nitrogen to oxygen is :

\frac{P_{\text{N}_2}}{P_{\text{O}_2}} = \frac{\text{mole fraction of } \text{N}_2}{\text{mole fraction of } \text{O}_2} = \frac{70/28}{27/32} = 2.96

Partial Pressure Ratio of Oxygen to Argon : For argon ( $\text{Ar}$ ), the mole fraction is $\dfrac{3/40}{70/28 + 27/32 + 3/40}$ .

The ratio of the partial pressure of oxygen to argon is :

\frac{P_{\text{O}_2}}{P_{\text{Ar}}} = \frac{\text{mole fraction of } \text{O}_2}{\text{mole fraction of } \text{Ar}} = \frac{27/32}{3/40} = 11.25

Q54

Initially, the root mean square (rms) velocity of N2 molecules at certain temperature is

u

. If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the new rms velocity will be :

A u/2

B 2u

C 4u

D 14u

Correct Answer

Option B

Solution

Root mean square (rms) velocity of N2 molecule (u) =

\sqrt {{{3RT} \over M}}

. For N2 molecule, molecular mass, M = 28. Therefore,

u = \sqrt {{{3RT} \over {28}}}

.....

(1) When the temperature becomes doubled, that is, 2T.

Let the new root mean square (rms) velocity of N-atom be u'.

For N-atom, the molecular mass, M = 14.

Therefore,

u' = \sqrt {{{3RT} \over {14}}}

....... (2) On dividing Eq. (1) by Eq. (2), we get

{u \over {u'}} = \sqrt {{1 \over 4}}

= {1 \over 2} \Rightarrow u' = 2u

Q55

'a’ and `b’ are van der Waals’ constants for gases. Chlorine is more easily liquefied than ethane because

A a and b for Cl2 < a and b for C2H6

B a and b for Cl2 > a and b for C2H6

C a for Cl2 > a for C2H6 and b Cl2 < b for C2H6

D a for Cl2 < a for C2H6 and b Cl2 > b for C2H6

Correct Answer

Option C

Solution

The value of

a

is a measure of the magnitude of the attractive forces between the molecules of the gas. Greater the value of

'a',

larger is the attractive inter-molecular force between the gas molecules. The value of

b

related to the effective size of the gas molecules. It is also termed as excluded volume. The gases with higher value of

a

and lower value of

b

are more liquefiable, hence for

C{{\rm l}_2}

''a'''

should be greater than for

{C_2}{H_6}

but for it

b

should be less than for

{C_2}{H_6}.

Q56

Equal masses of methane and oxygen are mixed in an empty container at 25oC. The fraction of the total pressure exerted by oxygen is

A 2/3

B 1/2

C 1/3

\times

273/298

D 1/3

Correct Answer

Option D

Solution

Let the mass of methane and oxygen

=m

gm.

Mole fraction of

{O_2}

= {{Moles\,\,of\,\,{O_2}} \over {Moles\,\,of\,\,{O_2}\, + \,Moles\,\,of\,\,C{H_4}}}

= {{m/32} \over {m/32 + m/16}}

= {{m/32} \over {3m/32}} = {1 \over 3}

Partial pressure of

{O_2}

$=$ Total pressure $\times$ mole fraction of

{O_2},

{P_{{O_2}}} = P \times {1 \over 3} = {1 \over 3}P

Q57

0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K. Given R is the gas constant in JK

-

1mol

-

1, x is :

A

{{2R} \over {4 + R}}

B

{{2R} \over {4 - R}}

C

{{4 + R} \over {2R}}

D

{{4 - R} \over {2R}}

Correct Answer

Option D

Solution

We know, PV = nRT Given, P = 200 Pa V = 10 m3 T = 1000 K n = 0.5 + x $\therefore$ 200 $\times$ 10 = (0.5 + x) R $\times$ 1000 $\Rightarrow$ 0.5 + x =

{2 \over R}

$\Rightarrow$ x =

{2 \over R} - {1 \over 2}

$\Rightarrow$ x =

{{4 - R} \over {2R}}

Q58

At very high pressures, the compressibility factor of one mole of a gas is given by :

A

{{pb} \over {RT}}

B 1 +

{{pb} \over {RT}}

C 1

-

{{pb} \over {RT}}

D 1

-

{b \over {\left( {VRT} \right)}}

Correct Answer

Option B

Solution

According to van der Waals' equation, for one mole of a gas

\left( {p + {a \over {{V^2}}}} \right)(V - b) = RT

...... (1) At high pressure,

{a \over {{V^2}}}

can be neglected So,

p + {a \over {{V^2}}} \approx p

...... (2) From Eqs. (1) and (2), we get

p(V - b) = RT

pV - pb = RT

pV = RT + pb

Dividing both the sides by RT, we get

{{pV} \over {RT}} = 1 + {{pb} \over {RT}}

Z = 1 + {{pb} \over {RT}}

Q59

If Z is a compressibility factor, van der Waals equation at low pressure can be written as:

A Z = 1 +

RT \over Pb

B Z = 1 -

a \over VRT

C Z = 1 -

Pb \over RT

D Z = 1 +

Pb \over RT

Correct Answer

Option B

Solution

Compressibility factor

\left( Z \right) = {{PV} \over {RT}}

(For one mole of real gas) van der Waals equation

\left( {P + {a \over {{V^2}}}} \right)\left( {V - b} \right) = RT

At low pressure, volume is very large and hence correction term

b

can be neglected in comparison to very large volume of

V.

i.e.

\,\,\,V - b \approx V

\left( {P + {a \over {{V^2}}}} \right)V = RT;

PV + {a \over V} = RT

PV = RT - {a \over V};

{{PV} \over {RT}} = 1 - {a \over {VRT}}

Hence,

\,\,\,z = 1 - {a \over {VRT}}

Q60

An evacuated glass vessel weighs 40.0 g when empty, 135.0 g when filled with a liquid of density 0.95 g mL

-

1 and 40.5 g when filled with an ideal gas at 0.82 atm at 250 K. The molar mass of the gas in g mol

-

1 is : (Given : R = 0.082 L atm K

-

1 mol

-

1)

A 35

B 50

C 75

D 125

Correct Answer

Option D

Solution

Weight of empty glass vessel = 40 gm Weight of glass vessel filled with liquid = 135 gm $\therefore$ Weight of liquid = 135 $-$ 40 = 95 gm Given density of liquid = 0.95 gm ml $-$ 1 $\therefore$ Volume of liquid

={{95} \over {0.95}} = 100

ml Weight of glass filled with ideal gas = 40.5 gm $\therefore$ Weight of gas = 40.5 $-$ 40 = 0.5 g Let the Molar mass = M $\therefore$ Moles of gas

= {{0.5} \over M}

$\therefore$ Now applying ideal gas equation, pV = nRT

\Rightarrow 0.82 \times {{100} \over {1000}} = {{0.5} \over M} \times 0.082 \times 250

$\Rightarrow$ M = 0.5 $\times$ 250 = 125 g/mol