JEE Advanced
JEE Main
BITSAT
View complete IIT JEE section
AIPMT
AIIMS
View Complete Medical section
Medical Exam Calendar
NTSE
KVPY
Olympiads
CBSE
ICSE
UAE
Saudi Arabia
Qatar
Oman
Bahrain
Kuwait
Indonesia
Malaysia
Singapore
Uganda
View complete NRI section
Physics
Chemistry
Maths
Revision notes
View complete study material
Buy IIT JEE/AIPMT study material
Buy CBSE Grade 8/9/10 study material
Deviation From Ideal Gas Behavior A gas which obeys the gas laws and the gas equation PV = nRT strictly at all temperatures and pressures is said to be an ideal gas.The molecules of ideal gases are assumed to be volume less points with no attractive forces between one another. But no real gas strictly obeys the gas equation at all temperatures and pressures. Deviations from ideal behaviour are observed particularly at high pressures or low temperatures. The deviation from ideal behaviour is expressed by introducing a factor Z known as compressibility factor in the ideal gas equation. Z may be expressed as Z = PV / nRT
The deviation from ideal behaviour is expressed by introducing a factor Z known as compressibility factor in the ideal gas equation. Z may be expressed as Z = PV / nRT
In case of ideal gas, PV = nRT ∴ Z = 1
In case of real gas, PV ≠ nRt ∴ Z ≠ 1
Thus in case of real gases Z can be < 1 or > 1
When Z < 1, it is a negative deviation. It shows that the gas is more compressible than expected from ideal behaviour.
When Z > 1, it is a positive deviation. It shows that the gas is less compressible than expected from ideal behaviour.
The causes of deviations from ideal behaviour may be due to the following two assumptions of kinetic theory of gases.
The volume occupied by gas molecules is negligibly small as compared to the volume occupied by the gas.
The forces of attraction between gas molecules are negligible.
The first assumption is valid only at low pressures and high temperature, when the volume occupied by the gas molecules is negligible as compared to the total volume of the gas. But at low temperature or at high pressure, the molecules being in compressible the volumes of molecules are no more negligible as compared to the total volume of the gas.
The second assumption is not valid when the pressure is high and temperature is low. But at high pressure or low temperature when the total volume of gas is small, the forces of attraction become appreciable and cannot be ignored.
1 mole of SO_{2} occupies a volume of 350 ml at 300K and 50 atm pressure. Calculate the compressibility factor of the gas.
Solution:
P = 50 atm
V = 350 ml = 0.350 litre
n = 1 mole
T = 300L Z = PV / nRT
∴ Z = 50 × 0.350 / 1 × 0.082 × 300 = 0.711
Thus SO_{2 }is more compressible than expected from ideal behaviour.
This equation can be derived by considering a real gas and 'converting ' it to an ideal gas.
Hence, Ideal volume
V_{i} = V – nb …..........(i)
n = Number of moles of real gas
V = Volume of the gas
b = A constant whose value depends upon the nature of the gas
The Van der Waals constant b (the excluded volume) is actually 4 times the volume of a single molecule.
b = 4 × volume of a single molecule = 4 × 6.023 × 10^{23} ×(4/3)r^{3}, where r is the radius of a molecule.
Let us assume that the real gas exerts a pressure P. The molecules that exert the force on the container will get attracted by molecules of the immediate layer which are assumed not to be exerting pressure.
It can be seen that the pressure the real gas exerts would be less than the pressure an ideal gas would have exerted. The real gas experiences attractions by its molecules in the reverse direction. Therefore if a real gas exerts a pressure P, then an ideal gas would exert a pressure equal to P + p (p is the pressure lost by the gas molecules due to attractions).
This small pressure p would be directly proportional to the extent of attraction between the molecules which are hitting the container wall and the molecules which are attracting these.
Therefore p ∝ n/v (concentration of molecules which are hitting the container's wall)
P ∝ n/v (concentration of molecules which are attracting these molecules) p ∝ n^{2}/v^{2}
P = an^{2}/v^{2} where a is the constant of proportionality which depends on the nature of gas.
A higher value of 'a' reflects the increased attraction between gas molecules.
Hence ideal pressure
P_{i} = (P + an^{2} / V^{2}) …...........(ii)
Here,
a = A constant whose value depends upon the nature of the gas
Substituting the values of ideal volume and ideal pressure in ideal gas equation i.e. pV=nRT, the modified equation is obtained as
Vander Waals constant for attraction (a) and volume (b) are characteristic for a given gas. Some salient feature of a & b are:
For a given gas Vander Waal’s constant of attraction ‘a’ is always greater than Vander Waals constant of volume (b).
The gas having higher value of ‘a’ can be liquefied easily and therefore H_{2} & He are not liquefied easily.
The units of a = litre^{2} atm mole^{–2} & that of b = litre mole ^{–1}
The numerical values of a & b are in the order of 10^{–1} to 10^{–2} & 10^{–2} to 10^{–4} respectively.
Higher is the value of ‘a’ for a given gas, easier is the liquefaction.
‘V’ is large and ‘b’ is negligible in comparison with V.
The Vander Waals equation reduces to:
(P + a / V^{2}) V = RT;
PV + a/ V = RT
PV = RT – a/V or PV < RT
This accounts for the dip in PV vs P isotherm at low pressures.
a/V^{2} may be neglected in comparison with P.
The Vander Waals equation becomes
P (V – b) = RT
PV – Pb = RT
PV = RT + Pb or PV > RT
This accounts for the rising parts of the PV vs P isotherm at high pressures.
V becomes so large that both b and a/V^{2} become negligible and the Vander Waals equation reduces to PV = RT. This shows why gases approach ideal behaviour at very low pressures.
These are two lightest gases known. Their molecules have very small masses. The attractive forces between such molecules will be extensively small. So a/V^{2} is negligible even at ordinary temperatures.
Thus PV > RT. Thus Vander Waals equation explains quantitatively the observed behaviour of real gases and so is an improvement over the ideal gas equation. Vander Waals equation accounts for the behaviour of real gases.
At low pressures, the gas equation can be written as,
(P + a/v^{2}_{m}) (V_{m}) = RT
or
Z = V_{m} / RT = 1 – a/V_{m}RT
Where Z is known as compressibility factor. Its value at low pressure is less than 1 and it decreases with increase of P. For a given value of V_{m}, Z has more value at higher temperature.
At high pressures, the gas equation can be written as
P (V_{m} – b) = RT
Z = PV_{m} / RT = 1 + Pb / RT
Here, the compressibility factor increases with increase of pressure at constant temperature and it decreases with increase of temperature at constant pressure. For the gases H_{2} and He, the above behaviour is observed even at low pressures, since for these gases, the value of ‘a’ is extremely small.
One litre of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors found are 1.072 & 1.375 respectively at initial and final states. Calculate the final volume.
P_{1}V_{1} = Z_{1}nRT_{1} and P_{2}V_{2} = Z_{2}nRT_{2}
P_{2}V_{2} / T_{2 }× T_{1} / P_{1}V_{1} = Z_{2} / Z_{1}
or V_{2} = Z_{2} / Z_{1} × T_{1} / T_{1} × P_{1}V_{1} / P_{2} = 1.375 / 1.072 × 273 / 473 × 300 × 1 / 600 = 370.1 ml
At a given temperature it is given by equation
RH = (partial pressure of water in air) / (vapour pressure of water )
Temperature at which real gas obeys the gas laws over a wide range of pressure is called Boyle’s Temperature. Gases which are easily liquefied have a high Boyle’s temperature [T_{b}(O_{2})] = 46 K] whereas the gases which are difficult to liquefy have a low Boyle’s temperature [T_{b}(He) = 26K].
Boyle’s temperature(T_{b}) = a / Rb = 1/2 T_{1}
where T_{i} is called Inversion Temperature and a, b are called van der Waals constant.
It (T_{c}) is the maximum temperature at which a gas can be liquefied i.e. the temperature above which a gas can’t exist as liquid.
T = 8a / 27Rb
It is the minimum pressure required to cause liquefaction at T_{c}
P_{c} = a/27b^{2}
It is the volume occupied by one mol of a gas at T_{c} and P_{c}
V_{c} = 3b
Specific heat c, of a substance is defined as the amount of heat required to raise the temperature of is defined as the amount of heat required to raise the temperature of 1 g of substance through 1^{0}C, the unit of specific heat is calorie g^{-1} K^{-1}. (1 cal is defined as the amount of heat required to raise the temperature of 1 g of water through 1^{0}C)
Molar heat capacity C, is defined as the amount of heat required to raise the temperature of 1 mole of a gas trough 1^{0}C. Thus,
Molar heat capacity = Sp. Heat molecular wt. Of the gas
For gases there are two values of molar heats, i.e., molar heat at constant pressure and molar heat at constant molar heat at constant volume respectively denoted by C_{p} and C_{v}. C_{p} is greater than C_{v} and C_{p}-R = 2 cal mol^{-1} K^{-1}
From the ratio of C_{p} and C_{v}, we get the idea of atomicity of gas.
For monatomic gas C_{p} = 5 cal and C_{v} =3 cal
∴ (γ = 5/3 = 1.67 (γ is poisson's ratio = C_{p} / C_{v})
For diatomic gas C_{p} = 7 cal and C_{v} = 5 cal
γ = 7/5 = 1.40
For polyatomic gas C_{p} = 8 cal and C_{v}= cal
γ = 8/6 = 1.33
also C_{p} = C_{p}m,
where, C_{p} and C_{v} are specific heat and m, is molecular weight.
Calculate Vander Waals constants for ethylene
T_{C} = 282.8 k; P_{C} = 50 atm
b = 1/8 RT_{c} / P_{c} = 1/8 × 0.082 × 282.8 / 50 = 0.057 litres/mole
a = 27 / 64 R^{2} × T^{2}_{C} / P_{c} = 27/64 × (0.082)^{2} × (282.8)^{2} × 1/50 = 4.47 lit^{2} atm mole^{–2}
Question 1: When Z < 1 , it showns
a. a negative deviation
b. a positive deviation.
c. that the gas is showing ideal behaviour
d. that the gas is less compressible than expected from ideal behaviour
Question 2: A real gas behaves ideally at
a. high temperature and low pressure
b. high temperature and high pressure
c. low temperature and high pressure
d. low temperature and low pressure
Question 3: Which of the following equations represents the correct form of Vandear waal’s equation at very low pressure?
a. PV = RT – a/V
b. pV = nRT
c. P (V_{m} – b) = RT
d. P (V – b) = RT
Question 4: The maximum temperature at which a gas can be liquefied is called
a.boiling point.
b. melting point
c. Boyle’s temperature
d. critical temperature
Q.1
Q.2
Q.3
Q.4
a
b
d
Click here to go through JEE Chemistry Syllabus
Look here for Reference books for IIT JEE
You can also refe to the Ideal Gas Law & Dalton’s Law of Partial Pressure
OR
Complete Your Registration (Step 2 of 2 )
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Graham’s Law of Diffusion & Gas...
Kinetic Molecular Theory of Gases The various gas...
Liquid State The liquid molecules are relatively...
Properties of Gases and Gas Laws General...
Solved Problems Question 1: A gas occupies one...
Ideal Gas Law & Dalton’s Law of Partial...
Van der Waals Forces Intermolecular forces are the...