1
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

What is the density of N2 gas 227oC and 5.00 atm. pressure? (R = 0.082 L atm K$$-$$1 mol$$-$$1)
A
1.40 g/mL
B
2.81 g/mL
C
3.41 g/mL
D
0.29 g/mL

Explanation

PV = nRT

$$ \Rightarrow $$ PV = $${W \over M}RT$$

$$ \Rightarrow $$ $$P = {W \over M} \times {{RT} \over V}$$

$$ \Rightarrow $$$$P = {{dRT} \over M}$$   [Density = $${{Mass} \over {Volume}}$$]

$$ \Rightarrow d = {{PM} \over {RT}} = {{5 \times 28} \over {0.0821 \times 500}} = 3.41\,g/ml$$
2
MCQ (Single Correct Answer)

AIPMT 2012 Mains

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

$$ \Rightarrow {{{C_P}} \over {{C_V}}} = {{{5 \over 2}R} \over {{3 \over 2}R}} = {5 \over 3} = 1.67$$
3
MCQ (Single Correct Answer)

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
4
MCQ (Single Correct Answer)

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 = ?

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

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

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