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1

### AIPMT 2011 Prelims

A gaseous mixture was prepared by taking equal mole of CO and N2. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N2) in the mixture is
A
0.5 atm
B
0.8 atm
C
0.9 atm
D
1 atm

## Explanation

Number of moles $$\Rightarrow$$ nCO = nN2

Volume of container is same. So, VCO = VN2 .

Also, temperature is same for both the gases thus, TCO = TN2

According to ideal gas equation, PV = nRT

Now, V, n, R and T for both gases are same. So, PCO = PN2

Now, total pressure is 1 atm and according to Dalton’s sum of partial pressure,

PCO + PN2 = 1 atm

$$\Rightarrow$$ 2PN2 = 1 atm {$$\because$$ PCO = PN2 }

$$\Rightarrow$$ PN2 = 0.5 atm
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

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