The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles. Their magnitude depends upon-

1. Charge of interacting particles.

2. Mass of interacting particles.

3. Polarizability of interacting particles.

4. Strength of permanent dipoles in the particles.

Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess 'partial charges'.

The partial charge will be-

1. More than unit electronic charge.

2. Equal to unit electronic charge.

3. Less than unit electronic charge.

4. Double the unit electronic charge.

As the temperature increases, average kinetic energy of molecules increases. With increase in temperature at constant volume, the pressure -  

1. Increases

2. Decreases

3. Remains same

4. Becomes half

Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases.

From the above data, the order of liquefaction of the given gases is-
(write the order from the gas liquefying first)
1. ${\mathrm{H}}_2,$ $\mathrm{He},$ ${\mathrm{O}}_2,$ ${\mathrm{N}}_2$
2. $\mathrm{He},$ ${\mathrm{O}}_2,$ ${\mathrm{H}}_2,$ ${\mathrm{N}}_2$
3. ${\mathrm{N}}_2,$ ${\mathrm{O}}_2,$ $\mathrm{He,}$ ${\mathrm{H}}_2$
4. ${\mathrm{O}}_2,$ ${\mathrm{N}}_2,$ ${\mathrm{H}}_2,$ $\mathrm{He}$

 In the given figure, the curve representing an ideal gas is-

1.  B only

2.  C and D only

3.  E and F only 

4.  A and B only

With increase in temperature the surface tension of a liquid will:

1. Remains same

2. Decreases

3. Increases

4. No regular pattern is followed

The state of a gas can be described by quoting the relationship between:

1.  Pressure, volume, temperature

2.  Temperature, amount, pressure

3.  Amount, volume, temperature

4.  Pressure, volume, temperature, amount

The density of a gas is found to be 5.46 g/dm³ at 27^°C at 2 bar pressure. The density of gas at STP would be -

$1. 2.0 \mathrm{g} {\mathrm{dm}}^{-3}$
$2. 3.0 \mathrm{g} {\mathrm{dm}}^{-3}$
$3. 4.5 \mathrm{g} {\mathrm{dm}}^{-3}$
$4. 1.5 \mathrm{g} {\mathrm{dm}}^{-3}$

34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. The molar mass of phosphorus would be -

$1.$ $1247.5$ $\mathrm{g}$ ${\mathrm{mol}}^{-1}$
$2.$ $2234.7$ $\mathrm{g}$ ${\mathrm{mol}}^{-1}$
$3.$ $3126.9$ $\mathrm{g}$ ${\mathrm{mol}}^{-1}$
$4.$ $1134.6$ $\mathrm{g}$ ${\mathrm{mol}}^{-1}$

A plot of volume versus temperature (T) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in the figure given below.

       

The correct order of pressure is -

         Temperature (K) 

1. ${\mathrm{p}}_1>{\mathrm{p}}_2>{\mathrm{p}}_3>{\mathrm{p}}_4$

2. ${\mathrm{p}}_1={\mathrm{p}}_2={\mathrm{p}}_3={\mathrm{p}}_4$

3. ${\mathrm{p}}_1<{\mathrm{p}}_2<{\mathrm{p}}_3<{\mathrm{p}}_4$

4. ${\mathrm{p}}_1<{\mathrm{p}}_2={\mathrm{p}}_3<{\mathrm{p}}_4$