1

### AIPMT 2011 Mains

A bubble of air is underwater at temperature 15oC and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25oC and the pressure is 1.0 bar, what will happen to the volume of the bubble?
A
Volume will become greater by a factor of 1.6.
B
Volume will become greater by a factor of 1.1
C
Volume will become smaller by a factor of 0.70
D
Volume will become greater by a factor of 2.5

## Explanation

Given

P1 = 1.5 bar, T1 = 273 + 15 = 288 K, V1 = V

P2 = 1.0 bar, T1 = 273 + 25 = 298K, V2 = ?

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

$\Rightarrow$ ${{1.5 \times V} \over {288}} = {{1 \times {V_2}} \over {298}}$

$\Rightarrow$ V2 = 1.55V

$\therefore$ Volume of bubble will be almost 1.6 time to initial volume of bubble.
2

### 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
3

### 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
4

### 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