1
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled ?
A
2.0
B
2.8
C
4.0
D
1.4

Explanation

Average velocity (vAV) = $$\sqrt {{{8RT} \over {\pi M}}} $$

$$ \Rightarrow $$ vAV $$ \propto $$ $$\sqrt T $$

$$ \therefore $$ $${{{{\left( {{v_{AV}}} \right)}_2}} \over {{{\left( {{v_{AV}}} \right)}_1}}} = \sqrt {{{2T} \over T}} $$ = 1.4
2
MCQ (Single Correct Answer)

AIPMT 2010 Mains

The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129oC is (Atomic masses : C = 12.01, H = 1.01 and R = 8.314 J K$$-$$1 mol$$-$$1 )
A
215216 Pa
B
13409 Pa
C
41648 Pa
D
31684 Pa

Explanation

PV = nRT

$$ \Rightarrow $$ $$P = {{nRT} \over V}$$ = $$ {w \over m}{{RT} \over V}$$

= $${6 \over {16.05}} \times {{8.314 \times 402} \over {0.03}}$$

= 41648 Pa
3
MCQ (Single Correct Answer)

AIPMT 2009

The energy absorbed by each molecule (A2) of a substance is 4.4 $$ \times $$ 10$$-$$19 J and bond energy per molecule is 4.0 $$ \times $$ 10$$-$$19 J. The kinetic energy of the molecule per atom will be
A
2.2 $$ \times $$ 10$$-$$19 J
B
2.0 $$ \times $$ 10$$-$$19 J
C
4.0 $$ \times $$ 10$$-$$20 J
D
2.0 $$ \times $$ 10$$-$$20 J

Explanation

Energy absorbed by each molecule = Bond energy per molecule + Kinetic energy per molecule

$$ \Rightarrow $$ 4.4 × 10–19 J = 4.0 × 10–19 J + Kinetic energy per molecule

$$ \Rightarrow $$ 0.4 × 10–19 = Kinetic energy per molecule

Kinetic energy per atom =
Kinetic energy per molecule
2


=
0.4 × 10–19
2
= 0.2 × 10–19 J

= 2 × 10–20 J
4
MCQ (Single Correct Answer)

AIPMT 2008

Volume occupied by one molecule of water (density = 1 g cm$$-$$3) is
A
3.0 $$ \times $$ 10$$-$$23 cm3
B
5.5 $$ \times $$ 10$$-$$23 cm3
C
9.0 $$ \times $$ 10$$-$$23 cm3
D
6.023 $$ \times $$ 10$$-$$23 cm3

Explanation

1 mole of water contains 6.023 × 1023 molecules of water

6.023 × 1023 molecules of water weigh = 18 g

So, 1 molecule of water weighs = $${{18} \over {6.023 \times {{10}^{23}}}}$$

Now, volume of 1 molecule of water

=
Mass of 1 molecule of water
Density of water
.

= $${{18} \over {6.023 \times {{10}^{23}}}} \times {1 \over {1\,g\,c{m^{ - 3}}}}$$

= 3 $$ \times $$ 10-23 cm3

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