States of Matter: Gases and Liquids — Page 2 | NEET Chemistry

Q11

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 onehalf of the hydrogen to escape?

A 3/8

B 1/2

C 1/8

D 1/4

Correct Answer

Option C

Solution

Let the number of moles of each gas = x Fraction of hydrogen escaped =

{1 \over 2}x

{{{r_{{O_2}}}} \over {{r_{{H_2}}}}} = \sqrt {{{{M_{{H_2}}}} \over {{M_{{O_2}}}}}}

\Rightarrow {{{n_{{O_2}}}/t} \over {{x \over 2}/t}} = \sqrt {{2 \over {32}}} = \sqrt {{1 \over {16}}} = {1 \over 4}

\Rightarrow {n_{{O_2}}} = {1 \over 8}x

Then, fraction of oxygen escaped =

{1 \over 8}

Q12

A gas such as carbon monoxide would be most likely to obey the ideal gas law at

A low temperaures and high pressures

B high temperatures and hgh pressures

C low temperatures and low pressures

D high temperature and low pressures.

Correct Answer

Option D

Solution

Real gas shows ideal gas behaviour at high temperature and low pressure.

Q13

What is the density of N 2 gas 227 o C and 5.00 atm. pressure? (R = 0.082 L atm K

-

1 mol

-

1 )

A 1.40 g/mL

B 2.81 g/mL

C 3.41 g/mL

D 0.29 g/mL

Correct Answer

Option C

Solution

PV = nRT $\Rightarrow$ PV =

{W \over M}RT

$\Rightarrow$

P = {W \over M} \times {{RT} \over V}

$\Rightarrow$

P = {{dRT} \over M}

[Density =

{{Mass} \over {Volume}}

]

\Rightarrow d = {{PM} \over {RT}} = {{5 \times 28} \over {0.0821 \times 500}} = 3.41\,g/ml

Q14

Maximum deviation from ideal gas is expected from

A CH 4(g)

B NH 3(g)

C H 2(g)

D N 2(g)

Correct Answer

Option B

Solution

The compressibility factor is the term which is measured for a gas to study its deviation from the ideal behaviour.

Compressibility factor,

Z = {{PV} \over {nRT}}

Greater is the difference in the value of Z from 1, greater is the deviation of the gas from ideal behaviour.

Among the given molecules, NH 3 is easily liquefiable gas which deviates from ideal behaviour to the maximum extent.

Q15

For real gases van der Waals equation is written as

\left( {p + {{a{n^2}} \over {{V^2}}}} \right)

(V

-

nb) = n RT where

a

and

b

are van der Waals constants. Two sets of gases are (I) O 2 , CO 2 , H 2 and He (II) CH 4 . O 2 and H 2 The gases given in set-I in increasing order of b and gases given in set-II in decreasing order of

a

, are arranged below. Select the correct order from the following

A (I) He < H 2 < CO 2 < O 2 (II) CH 4 > H 2 > O 2

B (I) O 2 < He < H 2 < CO 2 (II) H 2 > O 2 > CH 4

C (I) H 2 < He < O 2 < CO 2 (II) CH 4 > O 2 > H 2

D (I) H 2 < O 2 < He < CO 2 (II) O 2 > CH 4 > H 2

Correct Answer

Option C

Solution

Van der Waal gas constant '

a

' represent intermolecular force of attraction of gaseous molecules and Van der Waal gas constant 'b' represent effective size of molecules .

Therefore order should be (I) H 2 < He < O 2 < CO 2 (II) CH 4 > O 2 > H 2

Q16

A certain gas takes three times as long to effuse out as helium. Its molecular mass will be

A 27 u

B 36 u

C 64 u

D 9 u

Correct Answer

Option B

Solution

According to Graham's law of diffusion

r \propto {1 \over {\sqrt d }} \propto {1 \over {\sqrt M }}

\Rightarrow {{{r_1}} \over {{r_2}}} = \sqrt {{{{M_2}} \over {{M_1}}}}

Rate of diffusion =

{{Volume\,of\,gas\,diffused\,(V)} \over {Times\,taken\,(t)}}

$\therefore$

{{{V_1}/{t_1}} \over {{V_2}/{t_2}}} = \sqrt {{{{M_2}} \over {{M_1}}}}

If same volume of two gases diffuse then V 1 = V 2 $\Rightarrow$

{{{t_1}} \over {{t_2}}} = \sqrt {{{{M_2}} \over {{M_1}}}}

Here t 2 = 3t, M 1 = 4 u, M 2 = ?

\therefore {{3{t_1}} \over {{t_1}}} = \sqrt {{{{M_2}} \over 4}} \Rightarrow 3 = \sqrt {{{{M_2}} \over 4}}

\Rightarrow 9 = {{{M_2}} \over 4} \Rightarrow {M_2} = 36\,u

Q17

Equal volumes of two monoatomic gases, A and B at same temperature and pressure are mixed. The ratio of specific heats (C p /C v ) of the mixture will be

A 0.83

B 1.50

C 3.3

D 1.67

Correct Answer

Option D

Solution

C p for monoatomic gas mixture of same volume =

{5 \over 2}R

$\therefore$ C V =

{3 \over 2}R

\Rightarrow {{{C_P}} \over {{C_V}}} = {{{5 \over 2}R} \over {{3 \over 2}R}} = {5 \over 3} = 1.67

Q18

50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is 36, the molecular mass of gass A will be

A 96

B 128.74

C 20.25

D 64.42

Correct Answer

Option C

Solution

According to Graham's law of diffusion,

{{{r_1}} \over {{r_2}}} = \sqrt {{{{d_2}} \over {{d_1}}}} = \sqrt {{{{M_2}} \over {{M_1}}}}

{r_A} = {{{V_A}} \over {{T_A}}},\,{r_B} = {{{V_B}} \over {{T_B}}}

{{{V_A}/{T_A}} \over {{V_B}/{T_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}}

{V_A} = {V_B},\,{T_A} = 150\,\sec

, T B = 200 sec, M B = 36, M A = ?

{{{T_B}} \over {{T_A}}} = \sqrt {{{{M_B}} \over {{M_A}}}} \Rightarrow {{200} \over {150}} = \sqrt {{{36} \over {{M_A}}}}

\Rightarrow {4 \over 3} = \sqrt {{{36} \over {{M_A}}}} \,\,or\,\,{{4 \times 4} \over {3 \times 3}} = {{36} \over {{M_A}}}

$\Rightarrow$ M A =

{{36} \over {4 \times 4}} \times 3 \times 3 = 20.25

Q19

A bubble of air is underwater at temperature 15 o C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25 o C and the pressure is 1.0 bar, what will happen to the volume of the bubble?

A Volume will become greater by a factor of 1.6.

B Volume will become greater by a factor of 1.1

C Volume will become smaller by a factor of 0.70

D Volume will become greater by a factor of 2.5

Correct Answer

Option A

Solution

Given P 1 = 1.5 bar, T 1 = 273 + 15 = 288 K, V 1 = V P 2 = 1.0 bar, T 1 = 273 + 25 = 298K, V 2 = ?

{{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}

$\Rightarrow$

{{1.5 \times V} \over {288}} = {{1 \times {V_2}} \over {298}}

$\Rightarrow$ V 2 = 1.55V $\therefore$ Volume of bubble will be almost 1.6 time to initial volume of bubble.

Q20

In Duma's method of estimation of nitrogen 0.35 g of an organic compound gave 55 mLof nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (aqueous tension at 300 K = 15 mm).

A 15.45

B 16.45

C 17.45

D 14.45

Correct Answer

Option B

Solution

Given V 1 = 55 mL, V 2 = ? P 1 = 715 – 15 = 700 mm, P 2 = 760 mm T 1 = 300 K, T 2 = 273 K General gas equation,

{{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}

$\Rightarrow$

{{700 \times 55} \over {300}} = {{760 \times {V_2}} \over {273}}

$\Rightarrow$ V 2 = 46.098 mL Now, 22400 mL of nitrogen = 1 mole $\therefore$ 46.098 mL of nitrogen =

{{1 \times 46.098} \over {22400}}

mol Weight of nitrogen =

{{1 \times 46.098} \over {22400}}

$\times$ 28 = 0.057 g Percent composition of nitrogen in 0.35 g of compound =

{{0.0576} \over {0.35}} \times 100

= 16.45 %