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

Q11

For gaseous state, if most probable speed is denoted by C*, average speed by

\mathop C\limits^{\_\_}

and mean square speed by C, then for a large number of molecules the ratios of these speeds are:

A C*:

\mathop C\limits^{\_\_}

: C = 1.128 : 1.225 : 1

B C*:

\mathop C\limits^{\_\_}

: C = 1.225 : 1.128 : 1

C C*:

\mathop C\limits^{\_\_}

: C = 1 : 1.225 : 1.128

D C*:

\mathop C\limits^{\_\_}

: C = 1 : 1.128 : 1.225

Correct Answer

Option D

Solution

Most probable speed

\left( {{C^ * }} \right) = \sqrt {{{2RT} \over M}}

Average Speed

\left( {\overline C } \right) = \sqrt {{{8RT} \over {\pi M}}}

Root mean square velocity

\left( c \right) = \sqrt {{{3RT} \over M}}

{C^ * }:\overline C :C = \sqrt {{{2RT} \over M}} :\sqrt {{{8RT} \over {\pi M}}} :\sqrt {{{3RT} \over M}}

= 1:\sqrt {{4 \over \pi }} :\sqrt {{3 \over 2}}

= 1:1.128:1.225

Q12

The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. The pressures of the gases for equal number of moles are -

A 2PA = 3PB

B 3PA = 2PB

C PA = 3PB

D PA = 2PB

Correct Answer

Option A

Solution

Z = PV/nRT $\Rightarrow$ P =

{{ZnRT} \over V}

at constant T and mol(n), P $\propto$

{Z \over V}

$\Rightarrow$

{{{P_A}} \over {{P_B}}} = {{{Z_A}} \over {{Z_A}}} \times {{{V_B}} \over {{V_A}}}

=

\left( {{3 \over 1}} \right) \times \left( {{1 \over 2}} \right)

=

{3 \over 2}

$\therefore$ 2PA = 3PB

Q13

As the temperature is raised from 20°C to 40°C, the average kinetic energy of neon atoms changes by a factor of which of the following?

A 1/2

B 2

C

313\over293

D

\sqrt{313\over293}

Correct Answer

Option C

Solution

{{K.E\,\,of\,\,neon\,\,at\,\,{{40}^ \circ }C} \over {K.E\,\,of\,\,neon\,\,at\,\,{{20}^ \circ }C}}

= {{{3 \over 2}K \times 313} \over {{3 \over 2}K \times 293}}

= {{313} \over {293}}

Q14

An open vessel at 27oC is heated until two fifth of the air (assumed as an ideal gas) it has escaped from the vessel assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is -

A 750 K

B 500oC

C 750oC

D 500 K

Correct Answer

Option D

Solution

We know, PV = nRT As the vessel is open, so the pressure and volume is constant. $\therefore$ n1 R T1 = n2 R T2 $\Rightarrow$ n1 T1 = n2 T1

{2 \over 5}

th of the air escape from the vessel. So the remaining air is

{3 \over 5}

of the total air. $\therefore$ n $\times$ 300 =

{3 \over 5}

n $\times$ T2 $\Rightarrow$ T2 = 500 K

Q15

If 10–4 dm3 of water is introduced into a 1.0 dm3 flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established ? (Given : Vapour pressure of H2O at 300 K is 3170 Pa ; R = 8.314 J K–1 mol–1)

A 5.56 x 10–3 mol

B 1.53 x 10–2 mol

C 4.46 x 10–2 mol

D 1.27 x 10–3 mol

Correct Answer

Option D

Solution

From the ideal gas equation :

PV = nRT\,\,

or

\,\,\,n = {{PV} \over {RT}} = {{3170 \times {{10}^{ - 3}}} \over {8.314 \times 300}}

= 1.27 \times {10^{ - 3}}

Q16

At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as

p = {{RT} \over {V - b}}

at T. Here, b is the van der Waals constant. Which gas will exhibit steepest increase in the plot of Z (compression factor) vs p?

A Xe

B Ne

C Ar

D Kr

Correct Answer

Option A

Solution

Given

p = {{RT} \over {V - b}}

$\Rightarrow$

p\left( {V - b} \right) = RT

$\Rightarrow$

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

$\Rightarrow$ Z =

{b \over {RT}}p + 1

$\therefore$ Slope of Z vs p curve (straight line) =

{b \over {RT}}

$\therefore$ Higher the value of b, more steep will be the curve. As we know, b = 4vNA =

4 \times {4 \over 3}\pi {r^3} \times {N_A}

$\therefore$ b $\propto$ size of gas molecules Among Ne, Ar, Xe and Kr, size of Xe is the most as down the group in the periodic table size increases.

Q17

At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N2 ) at 4 bar. The molar mass of gaseous molecule is :

A 28 g mol

-

1

B 56 g mol

-

1

C 112 g mol

-

1

D 224 g mol

-

1

Correct Answer

Option C

Solution

Density =

{{Mass} \over {Volume}}

PV = RT $\Rightarrow$ V =

{{RT} \over P}

So, Density(d) =

{{MP} \over {RT}}

Now, d1 = x, P1 = 4, M1 = 28, d2 = 2x, P2 = 2, M2 = ? $\therefore$

{{{d_1}} \over {{d_2}}} = {{{M_1}{P_1}} \over {R{T_1}}} \times {{R{T_2}} \over {{M_2}{P_2}}} = {{{M_1}{P_1}} \over {{M_2}{P_2}}}

[As T1 = T2 ] $\Rightarrow$ M2 =

{{{M_1}{P_1}{d_2}} \over {{d_1}{P_2}}}

=

{{2x \times 28 \times 4} \over {2 \times x}}

= 112 g mol-1

Q18

Consider the van der Waals constants, a and b, for the following gases. .tg .tg Gas Ar Ne Kr Xe a/ (atm dm6 mol–2) 1.3 0.2 5.1 4.1 b/ (10–2 dm3 mol–1) 3.2 1.7 1.0 5.0 Which gas is expected to have the highest critical temperature?

A Ne

B Kr

C Xe

D Ar

Correct Answer

Option B

Solution

We know, Critical temperature (Tc) =

{{8a} \over {27bR}}

$\therefore$ Tc $\propto$

{a \over b}

So, species with greatest value of

{a \over b}

has greatest value of critical temperature. Accordig to the given data Kr have highest

{a \over b}

ratio.

Q19

Which one of the following statements is NOT true about the effect of an increase in temperature on the distribution of molecular speeds in a gas?

A The most probable speed increases

B The fraction of the molecules with the most probable speed increases

C The distribution becomes broader

D The area under the distribution curve remains the same as under the lower temperature

Correct Answer

Option B

Solution

Distribution of molecular velocities at two different temperature is given shown below.

NOTE : At higher temperature more molecules have higher velocities and less molecules have lower velocities.

As evident from fig. thus it is clear that With the increase in temperature the most probable velocity increase but the fraction of such molecules decreases.

Q20

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.