JEE Advanced (Single Correct MCQs): States of Matter - JEE MCQ

TIPS/Formulae : Mole fraction of O₂ = 

Partial pressure of O₂ = Mole fraction of O₂

Mole fraction of O₂ =

The temperature at which a real gas behaves like an ideal gas is called Boyle’s temperature or Boyle’s point.

U_rms : U_av = = 1.086 : 1

Average kinetic energy depends only on temperature and does not depend upon the nature of the gas. (∵ K.E. = 3/2 KT)

Pressure exerted by hydrogen will be proportional to its mole fraction.
Mole fraction of H₂ =

or U_av2 = 0.6 m/sec.

= RTT; Here  represents
the intermolecular forces.

TIPS/Formulae :

Rate of diffusion  

∵ Molecular mass of HCl > molecular mass of NH₃

∴ HCl diffuses at slower rate and white ammonium chloride is first formed near HCl bottle.

‘a’ is directly related to forces of attraction. Hence greater the value of ‘a’, more easily the gas is liquified.

TIPS/Formulae :

It means density of gas is directly proportional to pressure and inversely proportional to temperature.
Density of neon will be maximum at highest pressure and lowest temperature.
∴ (b) is correct answer.

    or    M_x = 64

Pressure exerted by the gas,    ...(1)

Here, u = root mean square velocity m = mass of a molecule, n = No. of molecules of the gas Hence (a) & (b) are clearly wrong.
Again       [explained from (1)]

Here, M = Molecular wt. of the gas;

Hence (c) is wrong

Further, Average K.E. =  KT;  Hence (d) is true.

Due to increase in the temperature, the kinetic energy of the gas molecules increases resulting in an increase in average molecular speed. The molecules are bombarded to the walls of the container with a greater velocity resulting in an increase in pressure.

The mean free path,

or  where a = molecular diameter

∴ Smaller the molecular diameter, longer the mean free path. Hence H₂ is the answer.

NOTE : The value of ‘a’ indicates the magnitude of attractive forces between gas molecules.
Value of ‘a’ µ size of molecule.
∴ inert gas will have minimum value of ‘a’ followed by H₂O, C₆H₆ and C₆H₅CH₃

The expression of root mean square speed is

Hence,

= 1

Under identical conditions,   

As rate of diffusion is also inversely proportional to time, we will have,   

(a) Thus,  For He,

(b) For O₂, t₂ 

(c) For CO, t₂ 

(d) For CO₂,  t₂  

According to ideal gas equation, at two conditions At 300 K;        
P₀V = n₀RT₀            1 × V = 1.08 ×R × 300                   …(i)
At 600 K;      
P₁V = n₁RT₁ P₁× V = (0.86 +0.43) × R × 600      …(ii)
Divide (ii) by (i),

= 2.38 atm. 

The compressibility factor of a gas is defined as

For an ideal gas, pV_m = RT. Hence Z = 1

For an ideal-gas behaviour, the molecules of a gas should be far apart. The factors favouring this condition are high temperature and low pressure.

TIPS/Formulae :

∴ T_N2= 2T_H2 or T_N2 >T_H2

(PV)_Observed / (PV)_Ideal < 1

⇒ V_obs < V_ideal, V_obs < 22.4 litre.

Mass of 1 L of vapour = volume × density                    
= 1000 × 0.0006 = 0.6 g

V of liquid water = 

U_RMS =  Using ideal gas equation,

where d is the
density of the gas

∴  U_RMS = at constant pressure,

TIPS/Formulae : Find the volume by either
V = RT/P (PV = RT) or P₁V₁ = P₂V₂ and and match it with the values given in graph to find correct answer.
Volume of 1 mole of an ideal gas at 273 K and 1 atm is 22.4 L and that at 373 K and 1 atm pressure is calculated as ;

Upon increase of temperature the internal energy of water or any system increases resulting in decrease in intermolecular force and hence decrease in surface tension. Surface tension decreases with increase in mobility due to increase in temperature.

For positive deviation: PV = nRT + nPb

Thus, the factor nPb is responsible for increasing the PV value,  above ideal value. b is actually the effective volume of molecule. So, it is the finite size of molecules that leads to the origin of b and hence positive deviation at high pressure.

Average KE = E = 

TIPS/Formulae : Use Grahams’ law of diffusion