A gas can be liquefied at temperature T and pressure p provided.

• T = T_c and p < p_c

• T < T_c and p > p_c

• T > T_c and p > p_c

• T > T_c and p < p_c

The dispersed phase and dispersion medium of fog respectively are

• solid, liquid

• liquid, liquid

• liquid, gas

• gas, liquid

For the reaction,
2SO₂ (g) + O₂ (g) → 2SO₃ (g) at 300 K, the value of $∆$G° is -690.9 R. The equilibrium constant value for the reaction at that temperature is (R is gas constant)

• 10 atm^-1

• 10 atm

• 10

• 1

204.

Four gases P, Q. R and S have almost same values of 'b' but their 'a' values (a, b are van der Waals' constants) are in the order Q < R < S < P. At a particular temperature, among the four gases, the most easily liquefiable one is

• P

• Q

• R

• S

The units of surface tension and viscosity of liquids respectively are

• kg m^-1s^-1, Nm^-1

• kg s^-2, kg m^-1s^-1

• Nm^-1, kg m^-1s^-2

• kg s^-1, kg m^-2s^-1

The compound that will have a permanent dipole moment among the following is

• I

• II

• III

• IV

The compressibility factor (Z) of one mole of a van der Waals' gas of negligible 'a' value is

• 1

• $\frac{\mathrm{bp}}{\mathrm{RT}}$

• $1+\frac{\mathrm{bp}}{\mathrm{RT}}$

• $1-\frac{\mathrm{bp}}{\mathrm{RT}}$

A van der Waals' gas may behave ideally when

• the volume is very low

• the temperature is very high

• the pressure is very low

• the temperature, pressure and volume all are very high

Mixing of two different ideal gases under isothermal reversible condition will lead to

• increase of Gibbs free energy of the system

• no change of entropy of the system

• increase of entropy of the system

• increase of enthalpy of the system

Two gases X (molecular weight M_X) and Y (molecular weight My; M_y > M_x) are at the same temperature T in two different containers. Their root mean square velocities are C_x and C_y respectively. If the average kinetic energies per molecule of two gases X and Y are E_x and E_y respectively then which of the following relation(s) is(are) true?

• E_x >E_y

• C_x > C_y

• E_x = E_y =(3/2)RT

• E_x = E_y =(3/2)k_BT