States of Matter: Gases and Liquids Questions — JEE Chemistry

Q1

Value of gas constant R is

A 0.082 litre atm

B 0.987 cal mol-1 K-1

C 8.3 J mol-1 K-1

D 83 erg mol-1 K-1

Correct Answer

Option C

Solution

Value of gas constant

\left( R \right) = 0.0821L\,atm\,{K^{ - 1}}\,mo{l^{ - 1}}

= 8.314 \times {10^7}\,\,ergs\,{K^{ - 1}}\,mo{l^{ - 1}}

= 8.314J{K^{ - 1}}\,mo{l^{ - 1}}

= 1.987\,cal\,{K^{ - 1}}\,mo{l^{ - 1}}

Q2

Among the following, the incorrect statement is :

A At low pressure, real gases show ideal behaviour.

B At very low temperature, real gases show ideal behaviour.

C At very large volume, real gases show ideal behaviour.

D At Boyle’s temperature, real gases show ideal behaviour.

Correct Answer

Option B

Solution

A real gas do not show ideal behaviour at low temperature, high pressure and low volume.

So according to the question, at low temperature real gas show ideal behaviour.

This statement is wrong.

Q3

The interaction energy of London forces between two particles is proportional to rx, where r is the distance between the particles. The value of x is :

A 3

B

-

3

C

-

6

D 6

Correct Answer

Option C

Solution

For London dispersion forces.

E \propto {1 \over {{r^6}}}

Hence, x = $-$ 6

Q4

The unit of the van der Waals gas equation parameter 'a' in

\left( {P + {{a{n^2}} \over {{V^2}}}} \right)(V - nb) = nRT

is :

A kg m s

-

2

B dm3 mol

-

1

C kg m s

-

1

D atm dm6 mol

-

2

Correct Answer

Option D

Solution

{{a{n^2}} \over {{V^2}}} = atm \Rightarrow a = atm \times {{d{m^6}} \over {mo{l^2}}}

Q5

For an ideal gas, number of moles per litre in terms of its pressure P, gas constant R and temperature T is

A PT/R

B PRT

C P/RT

D RT/P

Correct Answer

Option C

Solution

PV = nRT\,\,

(number of moles

= n/V

) $\therefore$

\,\,\,n/V = P/RT.

Q6

In Van der Waals equation of state of the gas law, the constant ‘b’ is a measure of

A intermolecular repulsions

B intermolecular collisions per unit volume

C Volume occupied by the molecules

D intermolecular attraction

Correct Answer

Option C

Solution

In van der Waals equation

'b'

is for volume correction.

Q7

The compressibility factor for a real gas at high pressure is :

A 1 + RT/pb

B 1

C 1 + pb/RT

D 1–pb/RT

Correct Answer

Option C

Solution

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

at high pressure

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

can be neglected

PV - Pb = RT\,\,\,

and

\,\,\,PV = RT + Pb

{{PV} \over {RT}} = 1 + {{Pb} \over {RT}}

z = 1 + {{Pb} \over {RT}};Z > 1\,\,\,

at high pressure

Q8

Consider the following table : .tg .tg Gas a/(k Pa dm6 mol-1) b/(dm3 mol-1) A 642.32 0.05196 B 155.21 0.04136 C 431.91 0.05196 D 155.21 0.4382 a and b are vander Waals constants. The correct statement about the gases is :

A Gas C will occupy more volume than gas A; gas B will be more compressible than gas D

B Gas C will occupy lesser volume than gas A; gas B will be more compressible than gas D

C Gas C will occupy lesser volume than gas A; gas B will be lesser compressible than gas D

D Gas C will occupy more volume than gas A; gas B will be lesser compressible than gas D

Correct Answer

Option A

Solution

van der Walls equation of state is,

z = 1 + {{Pb} \over {RT}} - {a \over {{V_m}RT}}

b = intermolecule volume of gases.

So when the value of 'b' is higher then the higher value of ‘b’ will occupy higher volume and according to van der Walls equation of state 'z' will be higher hence gas will be less compressible.

a

= intermolecule force of attraction The value of 'a' is higher means intermolecule force of attraction is higher then the volume of gas will be lesser and according to van der Walls equation of state 'z' will be smaller hence gas will have higher compressibility.

Q9

Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is : (Atomic wt. of Cl = 35.5 u)

A 1.46

B 0.46

C 1.64

D 0.64

Correct Answer

Option B

Solution

We know, PV = nRT n = no. of moles =

{m \over M}

So, PV =

{m \over M}RT

$\Rightarrow$

\,\,\,\,

P =

{m \over V} \times {{RT} \over M}

$\Rightarrow$

\,\,\,\,

P = d $\times$

{{RT} \over M}

[ d = density =

{m \over V}

] at constant temperature and pressure d $\propto$ M Now let d1 and d2 are the density of ammonia and HCl.

\therefore\,\,\,\,

{{{d_1}} \over {{d_2}}} = {{{M_{N{H_3}}}} \over {{M_{HCl}}}}

$\Rightarrow$

\,\,\,\,

{{{d_1}} \over {{d_2}}}

=

{{17} \over {36.5}}

= 0.46

Q10

A mixture of one mole each of H2 , He and O2 each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of H2 is 2 atm, the total pressure of the gases in the cylinder is :

A 14 atm

B 38 atm

C 6 atm

D 22 atm

Correct Answer

Option C

Solution

According to Dalton’s law of partial pressure, pi = xi × PT pi = partial pressure of the ith component xi = mole fraction of the ith component pT = total pressure of mixture $\Rightarrow$ 2 atm =

\left( {{{{n_{{H_2}}}} \over {{n_{{H_2}}} + {n_{He}} + {n_{{O_2}}}}}} \right)

$\times$ pT $\Rightarrow$ pT =

\left( {{{1 + 1 + 1} \over 1}} \right)

$\times$ 2 = 6 atm