NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

AIPMT 2012 Mains

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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 %

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12