1

### AIPMT 2012 Mains

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)  O2, CO2, H2 and He
(II)  CH4. O2 and H2

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 < H2 < CO2 < O2   (II) CH4 > H2 > O2
B
(I) O2 < He < H2 < CO2   (II) H2 > O2 > CH4
C
(I) H2 < He < O2 < CO2   (II) CH4 > O2 > H2
D
(I) H2 < O2 < He < CO2   (II) O2 > CH4 > H2

## Explanation

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) H2 < He < O2 < CO2   (II) CH4 > O2 > H2
2

### AIPMT 2012 Mains

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

## Explanation

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 V1 = V2

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

Here t2 = 3t, M1 = 4 u, M2 = ?

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

$\Rightarrow 9 = {{{M_2}} \over 4} \Rightarrow {M_2} = 36\,u$
3

### AIPMT 2012 Prelims

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

## Explanation

According to Graham's law of diffusion,

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

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

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

${V_A} = {V_B},\,{T_A} = 150\,\sec$, TB = 200 sec, MB = 36, MA = ?

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

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

$\Rightarrow$ MA = ${{36} \over {4 \times 4}} \times 3 \times 3 = 20.25$
4

### AIPMT 2011 Prelims

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

## Explanation

Given V1 = 55 mL, V2 = ?

P1 = 715 – 15 = 700 mm, P2 = 760 mm

T1 = 300 K, T2 = 273 K

General gas equation,

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

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

$\Rightarrow$ V2 = 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 %