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)
Equal volumes of two monoatomic gases, A and B at same temperature and pressure are mixed. The ratio of specific heats (Cp/Cv) of the mixture will be
A
0.83
B
1.50
C
3.3
D
1.67

Explanation

Cp for monoatomic gas mixture of same volume = $${5 \over 2}R$$

$$ \therefore $$ CV = $${3 \over 2}R$$

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

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
3

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 = ?

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

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

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

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