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Equal weights of methane and oxygen are mixed in an empty container at 25ºC. The fraction of the total pressure exerted by oxygen is (1981 - 1 Mark)
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 obeys the ideal gas laws over a wide range of pressure is (1981 - 1 Mark)
The temperature at which a real gas behaves like an ideal gas is called Boyle’s temperature or Boyle’s point.
The ratio of root mean square velocity to average velocity of a gas molecule at a particular temperature is (1981 - 1 Mark)
Urms : Uav = = 1.086 : 1
Helium atom is two times heavier than a hydrogen molecule.At 298 K, the average kinetic energy of a helium atom is
(1982 - 1 Mark)
Average kinetic energy depends only on temperature and does not depend upon the nature of the gas. (∵ K.E. = 3/2 KT)
Equal weights of methane and hydrogen are mixed in an empty container at 25ºC. The fraction of the total pressure exerted by hydrogen is : (1984 - 1 Mark)
Pressure exerted by hydrogen will be proportional to its mole fraction.
Mole fraction of H2 =
Rate of diffusion of a gas is : (1985 - 1 Mark)
The average velocity of an ideal gas molecule at 27ºC is 0.3 m/sec. The average velocity at 927ºC will be: (1986 - 1 Mark)
or Uav2 = 0.6 m/sec.
In van der Waals equation of state for a non-ideal gas, the term that accounts for intermolecular forces is (1988 - 1 Mark)
= RTT; Here represents
the intermolecular forces.
A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends the white ammonium chloride ring first formed will be (1988 - 1 Mark)
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.
The values of van der Waals constant ‘a’ for the gases O2, N2, NH3 and CH4 are 1.360, 1.390, 4.170 and 2.253 L2 atm mol–2 respectively. The gas which can most easily be liquified is : (1989 - 1 Mark)
‘a’ is directly related to forces of attraction. Hence greater the value of ‘a’, more easily the gas is liquified.
The density of neon will be highest at (1990 - 1 Mark)
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.
The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is (1990 - 1 Mark)
or Mx = 64
According to kinetic theory of gases, for a diatomic molecule (1991 - 1 Mark)
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.
At constant volume, for a fixed number of moles of a gas the pressure of the gas increases with rise in temperature due to (1992 - 1 Mark)
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.
Longest mean free path stands for : (1995S)
The mean free path,
or where a = molecular diameter
∴ Smaller the molecular diameter, longer the mean free path. Hence H2 is the answer.
Arrange the van der Waals constant for the gases : (1995S)
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 ratio between the root mean square speed of H2 at 50 K and that of O2 at 800 K is, (1996 - 1 Mark)
The expression of root mean square speed is
X mL of H2 gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is : (1996 - 1 Mark)
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
One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4 (g) decomposes to NO2(g).The resultant pressure is : (1996 - 1 Mark)
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 for an ideal gas is (1997 - 1 Mark)
The compressibility factor of a gas is defined as
For an ideal gas, pVm = RT. Hence Z = 1
A gas will approach ideal behaviour at (1999 - 2 Marks)
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.
The rms velocity of hydrogen is times the rms velocity of nitrogen. If T is the temperature of the gas, then (2000S)
∴ TN2= 2TH2 or TN2 >TH2
The compressibility of a gas is less than unity at STP.Therefore, (2000S)
(PV)Observed / (PV)Ideal < 1
⇒ Vobs < Videal, Vobs < 22.4 litre.
At 100°C and 1 atm, if the density of liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 litre of steam at that temperature is (2000S)
Mass of 1 L of vapour = volume × density
= 1000 × 0.0006 = 0.6 g
V of liquid water =
The root mean square velocity of an ideal gas at constant pressure varies with density (d) as (2001S)
URMS = Using ideal gas equation,
where d is the
density of the gas
∴ URMS = at constant pressure,
Which of the following volume (V ) - temperature (T ) plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure ? (2002S)
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 ;
When the temperature is increased, surface tension of water (2002S)
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.
Positive deviation from ideal behaviour takes place because of(2003S)
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.
The root mean square velocity of one mole of a monoatomic gas having molar mass M is ur.m.s.. The relation between the average kinetic energy (E) of the gas and ur.m.s. is (2004S)
Average KE = E =
The ratio of the rate of diffusion of helium and methane under identical condition of pressure and temperature will be(2005S)
TIPS/Formulae : Use Grahams’ law of diffusion
When one mole of monoatomic ideal gas at T K undergoes adiabatic change under a constant external pressure of 1 atm volume changes from 1 litre to 2 litre. The final temperature in Kelvin would be (2005S)
TVγ-1 = Constant (Q change is adiabatic) 1
For monoatomic gas γ =
Hence A is correct.
A mono-atomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equals to 1.What is the molar heat capacity of the gas
In general, the molar heat capacity for any process is given by
when PVn = constant
Here = i.e. PV–1 = constant
For monoatomic gas,
The term that corrects for the attractive forces present in a real gas in the van der Waals equation is
Correction factor for attractive force for n moles of real gas is given by the term mentioned in (b).
For one mole of a van der Waal’s gas when b = 0 and T = 300 K, the PV vs, 1/V plot is shown below. The value of the van der Waal’s constant a (atm. liter2 mol–2) is : (2012)
PV + a/V = RT; PV = RT – a(V)
y = RT – a(x)
So, slope = a =
The qualitative sketch es I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH3OH and CH3(CH2)11OSO3– Na+ at room temperature. The correct assignment of the sketches is (JEE Adv. 2016)
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