behavior of gases...gases which obey all gas laws under all conditions of pressure and temperature...
TRANSCRIPT
Behavior of Gases
Can we liquefy the Earth’s atmosphere? And do you know the
behavior of gases under very low temperature? Are there any laws that
govern the behavior of gases? Here we shall learn about the various
interesting properties of gases. Read out more in this section.
Behavior of Gases
There are 5 main states of matter: solid, liquid, gas, plasma and the
Bose-Einstein condensate. Out of these gases are a special state. Their
properties are easy to study. We see that gases follow certain laws
known as the gas laws. These laws tell us about the behavior of gases.
By that, we mean the values and relations of temperature, pressure and
volume etc. Let’s see what these laws are.
Browse more Topics under Kinetic Theory
● Specific Heat Capacity and Mean Free Path
● Law of Equipartition of Energy
● Kinetic Theory of an Ideal Gas
Gas Laws
In the periodic table of elements, we have the group of inert gases or
permanent gases which are very unreactive. Their properties are very
close to an ideal gas and hence their behavior resembles that of an
ideal gas. On the basis of certain experiments using inert gases, the
following laws governing the behavior of gases were established:
Boyle’s Law
Suppose you have some Helium in a gas container at a low pressure
and temperature. At a constant temperature, if you increase the
volume of the container, the pressure of the gas will decrease. This is
given by the Boyle’s law.
This law states that at a constant temperature, the volume (V) of a
given mass of gas is inversely proportional to its pressure (p). At
constant temperature, Boyle’s law can be stated as
V
∝
1/P or VP = constant ….(1)
The constant is a proportionality constant. Its values depend on the
mass, temperature and nature of the gas. If P1 and V1 are the initial
values of pressure and volume of any gas and P2 and V2 are another
set of values, then we can say that
P₁V₁ = constant …(2) and P2V2 = constant …(3)
Since the mass, temperature and nature of a gas are same throughout,
we say equation (2) and (3) represent the same quantity. Thus we
have:
P₁V₁ = P2V2
Charles’ Law
A similar relation is found between Volume and Temperature of an
ideal gas. We call it the Charle’s Law. This law states that at constant
pressure, the volume (V) of a given mass of gas is directly
proportional to its absolute temperature (T).
If V is the volume and T is the temperature of a gas at some constant pressure, then V
∝
T or V/T = constant. Following the same method as above, we can
write:
V₁/T₁ = V₂/T₂
Gay Lussacs’ or Regnault’s Law
This law states that at constant volume (V), the pressure (P) of a given mass of a gas is directly proportional to its absolute temperature (T). We can write: P
∝
T or P/T = constant. Also, we can write:
P₁/T₁ = P₂/T₂
Kinetic Theory of Gases
Avogadro’s Law
This law stat
es that equal volumes of different gases, under similar conditions of
temperature and pressure, contain equal number molecules. This
means that if you have two or more different gases, as long as they
have similar conditions of temperature and pressure, equal
concentrations of these gases will occupy equal portions of volume.
For example, at STP (Standard Temperature and Pressure) or NTP,
where T = 273K and p = 1 atm, 22.4L of each gas has NA = 6.023 x
10^23 molecules. In other words, one mole of any gas under STP
conditions occupies 22.4L volume.
Standard Gas Equation
Gases which obey all gas laws under all conditions of pressure and
temperature are called perfect gases or the ideal gases. Inert gases kept
under high temperature and very low pressure behave like ideal gases.
Equation of state for a perfect gas can be written as
PV=nRT
where, p = pressure, V = volume, T = absolute temperature, R =
universal gas constant = 8.31 J mol-1 K-1, n = number of moles of a
gas
Real Gases
None of the gases that exist in nature, follow the gas laws for all
values of temperature and pressure. So we see that the behavior of
gases that exist or the “real gases” differs from the behavior of the
ideal gases. These gases deviate from ideal gas laws because:
● Real gas molecules attract one another.
● Real gas molecules occupy infinite volume.
Hence the equations for such gases need modifications as discussed
below.
Real Gas Equation or Van der Waal’s Gas Equation
The equation of state for a real gas can be written as:
(P + a/V²) (V-b) = RT
where, a and b are Van der Waal’s constants.
Solved Examples For You
Q: Assertion: If the pressure of an ideal gas is doubled and volume is
halved, then its internal energy will remain unchanged.
Reason: The internal energy of an ideal gas is a function of
temperature only.
A. Both the Reason and Assertion are correct and Reason is the
correct explanation of Assertion.
B. Both are correct but Reason is not the correct explanation of
Assertion.
C. The assertion is correct but Reason is wrong.
D. The reason is correct but Assertion is wrong.
Solution: A). Let P’ = 2P and V’ = V/2. Then from the equation of the
state, P’V’ = nRT, we have from substitution PV = nRT. This shows
that temperature doesn’t change and as the Internal energy of an ideal
gas is a function of Temperature only, the internal energy of the gas
will remain same.
Specific Heat Capacity & Mean Free Path
Let us say that we want to heat equal weights of an iron rod and water.
After say 5 minutes, which of the two will be at a greater temperature?
The answer is the iron rod. But why is it so? The answer lies in the
Specific Heat Capacity of the substances. Different substances accept
heat differently. Let us learn more!Specific Heat Capacity
Specific Heat Capacity is the amount of energy required by a single
unit of a substance to change its temperature by one unit. When you
supply energy to a solid, liquid or gas, its temperature changes. This
change of temperature will be different for different substances like
water, iron, oxygen gas, etc.
This energy is known as the Specific Heat Capacity of the substance
and is denoted by ‘C’. Molar Specific Heat Capacity of a substance is
C and is calculated for one mole of the substance. Mathematically we
can write:
C = ΔQ/m
Further, when you supply energy to a substance, it may undergo a
change in volume and/ or pressure, especially in gaseous substances.
Hence, to determine the Specific Heat Capacity of gases, it is
important to pre-determine the pressure and volume under which you
want to calculate C since it can have infinite values (depending on the
values of pressure and volume). The Molar Specific Heat Capacity at
constant volume is denoted by Cv and that at constant pressure is
denoted by Cp.
Browse more Topics under Kinetic Theory
● Behaviour of Gases
● Law of Equipartition of Energy
● Behaviour of Gases
● Kinetic Theory of an Ideal Gas
Specific Heat Capacity of Gases
According to the first law of thermodynamics ΔQ = ΔU + ΔW
{change in heat of a system = change in internal energy + amount of
work done}. Change in heat of a system (ΔQ) can also be calculated
by multiplying Mass (m), Specific Heat Capacity (C) and change in
Temperature (ΔT):
ΔQ = mCΔT Or,
mCΔT = ΔU + ΔW ————(1)
Monatomic Gases (Monoatomic gases)
In monatomic gas, molecules have three translational degrees of
freedom. At temperature ‘T’ the average energy of a monatomic
molecule is (3/2)KBT. Now, let’s look at one mole of such a gas at
constant volume and calculate the internal energy (U):
U = (3/2) KBT x NA {where NA is Avogadro constant}
The total internal energy will by the internal energy of a single
molecule multiplied by the number of molecules in one mole of the
gas; which is Avogadro constant NA. Now, Boltzmann’s constant (KB)
is the Gas constant (R) divided by NA. Hence: U = (3/2)(R/NA)T × NA
U = (3/2)RT ———————(2)
In equation (1), since the energy is supplied at constant volume:
mCvΔT = ΔU + ΔW. For one mole of a gas, m = 1. Also, for
calculating Cv, ΔT = 1. Since the volume is constant, ΔW = 0.
Therefore, 1×Cv×1 = ΔU + 0
Cv = ΔU = (3/2)RT —————-[refer (2)]
So, the molar specific heat capacity to change the temperature by 1
unit would be Cv = (3/2)R. For an ideal gas, Cp – Cv = R (Gas
Constant). Therefore: Cp = R + Cv = R + (3/2)R = (5/2)R. The ratio of
Cp:Cv (γ) is hence 5:3.
Diatomic Gases
In case of diatomic gases, there are two possibilities:
● Molecule is a Rigid Rotator: In this scenario, the molecule will
have five degrees of freedom (3 translational and two
rotational). As defined by the Law of Equipartition of Energy,
the internal energy (U) can be calculated as:
U = (5/2)KBT * NA = (5/2)RT. Following the calculation used for
monatomic gases:
Cv = (5/2)R
Cp = (7/2)R
γ = 7:5
● Molecule is NOT a Rigid Rotator: In this scenario, the
molecule will have an additional vibrational degree of freedom.
The internal energy can thus be calculated as:
U = [(5/2)KBT + KBT] * NA = [(5/2)(R/NA)T + (R/NA)T] * NA =
(7/2)RT
Following the calculation used for monatomic gases: Cv = (7/2)R
Cp = (9/2)R and hence γ = 9:7
Polyatomic Gases
The degrees of freedom of polyatomic gases are:
● 3 translational
● 3 rotational
● f vibrational
Deploying the Law of Equipartition of Energy for calculation of
internal energy, we get:
U = [(3/2)KBT + (3/2)KBT + fKBT] * NA = [(3/2)(R/NA)T +
(3/2)(R/NA)T + f(R/NA)T] * NA
U = (3 + f)RT
The molar specific heat capacities:
Cv = (3+f)R
Cp = (5+f)R
Specific Heat Capacity of Solids
Using the Law of Equipartition of energy, the specific heat capacity of
solids can be determined. Let us consider a mole of solid having NA
atoms. Each atom is oscillating along its mean position. Hence, the
average energy in three dimensions of the atom would be:
3 * 2 * (1/2)KBT = 3KBT
For one mole of solid, the energy would be:
U = 3KBT * NA = 3(R/NA)T * NA = 3RT —————–(3)
If the pressure is kept constant, then according to the laws of
thermodynamics
ΔQ = ΔU + PΔV
In case of solids, the change in volume is ~0 if the energy supplied is
not extremely high. Hence,
ΔQ = ΔU + P * 0 = ΔU
So, the molar specific heat capacity to change the temperature by 1
unit would be:
C = 3R ——————-[refer (3)]
Specific Heat Capacity of Water
For the purpose of calculation of specific heat capacity, water is
treated as a solid. A water molecule has three atoms (2 hydrogens and
one oxygen). Hence, its internal energy would be:
U = (3 * 3KBT)*NA = 9KBT*NA = 9(R/NA)T * NA = 9RT
And, following a similar calculation like solids: C = 9R
Mean Free Path
The Kinetic theory of gases assumes that molecules are continuously
colliding with each other and they move with constant speeds and in
straight lines between two collisions. Hence, a molecule follows a
chain of zigzag paths. Each of these paths is known as a free path
(since it lies between two collisions).
The average distance travelled by the molecule between two collisions
is known as the Mean Free Path. The number of collisions increases if
the gas is denser or the molecules are large in size.
If the molecules of a gas are spheres having a diameter – ‘d’; one
molecule is moving with an average speed – ‘v’ and the number of
molecules per unit volume is “n”, then the mean free path (l) can be
calculated by using the formula: l = 1/πnd2
The equation is arrived at under the assumption that all other
molecules are at rest, which is not actually the case. If we consider all
molecules to be moving in all directions with different speeds, then
the mean free path formula would be: l = 1/√2 πnd2
Question For You:
Q: One mole of an ideal monoatomic gas is mixed with 1 mole of an
ideal diatomic gas. The molar specific heat of the mixture at constant
volume is (in cal):
A) 22 cal B) 4 cal C) 8 cal D) 12 cal
Solution: B) As we know Cv = (3/2) R for a monoatomic gas and Cv =
(5/2) R for a diatomic case. Thus for the mixture, average of both is =
[(3/2) R + (5/2) R] /2 = 2R = 4 cal.
Law of Equipartition of Energy
A single atom is free to move in space along the X, Y and Z axis.
However, each of these movements requires energy. This is derived
from the energy held by the atom. The Law of Equipartition of Energy
defines the allocation of energy to each motion of the atom
(translational, rotational and vibrational). Before we understand this
law, let’s understand a concept called ‘Degrees of Freedom’.
Degrees of Freedom
Imagine a single atom. In a three dimensional space, it can move
freely along the X, Y and Z axis. Motion from one point to another is
also known as translation. Hence, this movement along the three axes
is called translational movement. If you have to specify the location of
this atom, then you need three coordinates (x, y, and z).
We can also say that a single atom has 3 Degrees of Freedom. Most
monoatomic molecules (i.e. molecules having a single atom like
Argon) have 3 translational degrees of freedom, provided their
movement is unrestricted.
(Source: Wikipedia)
Let’s now imagine a diatomic molecule (a molecule having two atoms
like O2 or N2). Apart from the three translational degrees of freedom,
these molecules can also rotate around the centre of mass. Two such
rotations are possible along the axis normal to the axis that joins the
two atoms.
This adds two additional degrees of freedom (rotational) to the
molecule. In simpler words, to specify the location of the molecule,
you would need the X, Y and Z coordinates along with the rotational
coordinates of the individual atoms.
It is important to note here that these diatomic molecules are not rigid
rotators (where molecules do not vibrate) at all temperatures. Along
with the translational and rotational movements, diatomic molecules
also oscillate along the interatomic axis like a single dimensional
oscillator. This adds a vibrational degree of freedom to such
molecules.
Hence, to specify the location of a diatomic molecule, you would
finally need the X, Y and Z coordinates along with the rotational and
vibrational coordinates. So, in a nutshell, Degrees of Freedom is
nothing but the number of ways in which a molecule can move. This
forms the basis of the Law of Equipartition of Energy.
Browse more Topics under Kinetic Theory
● Behaviour of Gases
● Specific Heat Capacity and Mean Free Path
● Behaviour of Gases
● Kinetic Theory of an Ideal Gas
Law of Equipartition of Energy
The law states that: “In thermal equilibrium, the total energy of the
molecule is divided equally among all Degrees of Freedom of
motion”. Before delving into the calculations, let’s understand the law
better. If a molecule has 1000 units of energy and 5 degrees of
freedom (which includes translational, rotational and vibrational
movements), then the molecule allocates 200 units of energy to each
motion.
Now, let us look at some equations!
Kinetic Energy of a single molecule: KE = 1/2 mv2. A gas in thermal
equilibrium at temperature T, the average Energy is:
Eavg = 1/2 mvx2 + 1/2 mvy2 + 1/2 mvz2 = 1/2KT + 1/2 KT + 1/2 KT =
3/2 KT
where K = Boltzmann’s constant. In case of a monoatomic molecule,
since there is only translational motion, the energy allotted to each
motion is 1/2KT. This is calculated by dividing total energy by the
degrees of freedom:
3/2 KT ÷ 3 = 1/2 KT
In case of a diatomic molecule, translational, rotational and vibrational
movements are involved. Hence the Energy component of
translational motion= 1/2 mvx2 + 1/2 mvy2 + 1/2 mvz2. Energy
component of rotational motion= 1/2 I1w12 + 1/2 I2w22 {I1 & I2
moments of inertia. w1 & w2 are angular speeds}
And, the energy component of vibrational motion= 1/2 m (dy/dt)2+
1/2 ky2. Where k is the force constant of the oscillator and y is the
vibrational coordinate. It is important to note here that this has both
kinetic and potential modes.
According to the Law of Equipartition of Energy, in thermal
equilibrium, the total energy is distributed equally among all energy
modes. While the translational and rotational motion contributes ½ KT
to the total energy, vibrational motion contributes 2 x 1/2KT = KT
since it has both kinetic and potential energy modes.
Some Questions for You:
Q: ‘N′ moles of a diatomic gas in a cylinder are at a temperature ′T′.
Heat is supplied to the cylinder such that the temperature remains
constant but n moles of the diatomic gas get converted into monatomic
gas. What is the change in the total kinetic energy of the gas?
A) 5/2 nRT B) 1/2 nRT C) 0 D) 3/2 nRT
Solution: D) Initial K.E. = (3/2) nRT . Number of moles in the final
sample = 2n
Since the gas is changed to monoatomic gas, we have: K.E. of the
final sample = (3/2)×2nRT
Hence, the change in the K.E. = 3nRT – (3/2) nRT = 3/2 nRT.
Behaviour of Gases
In gases, molecules are far from each other as compared to solids and
liquids. Also, mutual interaction between molecules is negligible
except when they collide. Hence, understanding the behaviour and
properties of gases is much easier than solids or liquids.
Behaviour of Gases
Gases which are at low pressures or high temperatures (higher than
their liquefaction of solidification range) approximately satisfy a
simple relation between their pressure, volume, and temperature given
by –
PV = KT … (1)
where T is the temperature in Kelvin, K is a constant for the given
sample (which can vary with the volume of the gas), V is the volume,
and P is the pressure of the gas. Let’s look at this now with the
perspective of atoms and molecules. Since the gas constant K varies
with the volume of the gas, it is proportional to the number of
molecules in the sample. Hence, we have
K = Nk
where N is the number of molecules in the sample and k is
Boltzmann’s constant which is also denoted as kB. From equation (1),
we have
PV = KT
⇒ K =
PV
T
⇒ NkB =
PV
T
⇒ kB =
PV
NT
Hence, for two samples 1 and 2, we have
P
1
V
1
N
1
T
1
=
P
2
V
2
N
2
T
2
= constant = kB … (2)
Form the above equation, we can conclude that if P, V, and T are same
then N is also the same for all gases. This is Avagadro’s Hypothesis
which states: ‘The number of molecules per unit volume is the same
for all gases at a fixed temperature and pressure’.
Avogadro’s Number
The number of molecules in 22.4 liters of any gas is 6.02 x 1023. This
is Avogadro’s number and is denoted by NA. The mass of 22.4 liters
of any gas is equal to its molecular weight at STP (Standard
Temperature and Pressure; Temperature = 273K and Pressure = 1
atm). This amount of substance is called a mole.
By studying chemical reactions, Avogadro had guessed the equality in
numbers in equal volumes of gas at fixed temperature and pressure.
This hypothesis is justified by Kinetic theory. Therefore, the perfect
gas equation is written as,
PV = μ RT … (3)
where μ is the number of moles and R = NA. kB is a universal
constant. The temperature T is absolute temperature. By choosing the
Kelvin scale for absolute temperature, we get, R = 8.314 J mol–1K–1.
Hence,
μ =
M
M
0
=
N
N
A
… (4)
where M is the mass of the gas containing N molecules, M0 is the
molar mass, and NA is Avogadro’s number. From equations (3) and
(4), we have
PV = kB NT or P = kB nT
where n is the number of molecules per unit volume. kB is the
Boltzmann’s constant having value 1.38 × 10–23 J K–1 (in SI units).
Equation (3) can also be written as
P =
pRT
M
0
… (5)
where p is the mass density of the gas.
Ideal Gas
Any gas that satisfies equation (3) at all pressures and temperatures is
called an ideal gas. However, this is for theoretical purposes only
since no gas can be truly ideal. The figure below shows the behaviour
of a real gas at three different temperatures.
You can observe that the real gas departs from the ideal gas behaviour
at three different temperatures. Also, all curves approach the ideal gas
line for low pressures and high temperatures. It simply means that
without molecular interactions, all gases behave like an ideal gas.
Boyle’s Law
If we fix μ and T in equation (3), we get
PV = constant … (6)
So, by keeping the temperature constant, the pressure of a given mass
of gas is inversely proportional to the volume of the gas. This is
Boyle’s law. The following figure shows a comparison of the
experimental and theoretical P-V curves as predicted by Boyle’s law.
From the figure, it is evident that the agreement is good at high
temperatures and low pressures.
Charles’ Law
Now, if you fix P, from equation (1), we get V ∝ T. Hence, at a fixed
pressure, the volume of a gas is inversely proportional to its absolute
temperature (Charles’ Law). This is represented in the figure below:
Dalton’s Law of Partial Pressures
Now consider a mixture of ideal gases which don’t interact with each
other in a vessel of volume V at temperature T and pressure P. Let the
mixture contain μ1 moles of gas 1, μ2 moles of gas2, etc. In this
mixture, we find that
PV = (μ1 + μ2 +… ) RT … (7)
i.e. P = (μ1
RT
V
+ μ2
RT
V
+ …) … (8)
Or P = (P1 + P2 + …) … (9)
Where P1 = μ1
RT
V
is the pressure gas 1 would exert at the same conditions of volume
and temperature if no other gases were present. This is called the
partial pressure of the gas. Hence, the total pressure of a mixture of
ideal gases is simply the sum of the partial pressures. This is Dalton’s
law of partial pressures.
Solved Examples For You
Q1. The figure below shows a plot of
PV
T
versus P for 1.00×10–3 kg of oxygen gas at two different
temperatures.
1. What does the dotted plot signify?
2. Which is true: T1 > T2 or T1 < T2?
3. What is the value of 4. PV
5. T
6. where the curves meet on the y-axis?
7. If we obtained similar plots for 1.00×10–3 kg of hydrogen, would we get the same value of
8. PV
9. T
10. at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of
11. PV
12. T
13. (for the low-pressure high-temperature region of the plot)?
(Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J
mo1–1 K–1.)
Solution ● The dotted line is parallel to P. Hence, it remains unchanged
even when P is changed. Therefore, it corresponds to the
behaviour of an ideal gas.
● The dotted line represents an ideal gas. From the figure, we can
see that T1 is closer to the dotted line than T2. Now, a real gas
approaches the behaviour of an ideal gas at high temperatures.
Hence, T1>T2.
● When the two curves meet, the value of ● PV
● T
● is μR. This is because the ideal gas equation for μ moles is: PV
= μRT.
The molecular mass of oxygen = 32 g. Therefore,
μ =
1×
10
−3
32×
10
−3
kg =
1
32
We also know that the value of the universal constant R is 8.31
Jmole-1K-1. Therefore,
PV
T
=
1
32
× 8.31 = 0.26 JK-1
● If we obtained similar plots for 1 x 10-3 kg of hydrogen, then we will not get the same value of
● PV
● T
● at the point where the curve meets the axis. The reason is that
the molecular mass of hydrogen is different from that of
oxygen.
Let’s calculate the mass of hydrogen needed to obtain the same value of
PV
T
. We have,
PV
T
= 0.26 JK-1
Molecular mass of hydrogen = M = 2.02 u.
We know that μ =
m
M
… where m is the mass of hydrogen. Also, we know that
PV
T
= μR at constant temperature. Therefore,
PV
T
=
m
M
R
m =
PV
T
×
M
R
=
0.26×2.02
8.31
= 6.32 x 10-2 gram or 6.32 x 10-5 kg.
Kinetic Theory of an Ideal Gas
Can you imagine a collection of spheres that collide with each other
but they do not interact which each other. Here the internal energy is
the kinetic energy. In this article, we shall understand what kinetic
theory of an ideal gas is.
Basics of Kinetic Theory
It says that the molecules of gas are in random motion and are
continuously colliding with each other and with the walls of the
container. All the collisions involved are elastic in nature due to which
the total kinetic energy and the total momentum both are conserved.
No energy is lost or gained from collisions.
Ideal Gas Equation
(Source: Pinterest)
The ideal gas equation is as follows
PV = nRT
the ideal gas law relates the pressure, temperature, volume, and
number of moles of ideal gas. Here R is a constant known as the
universal gas constant.
Browse more Topics under Kinetic Theory
● Behaviour of Gases
● Specific Heat Capacity and Mean Free Path
● Law of Equipartition of Energy
● Behaviour of Gases
Assumptions
1. The gas consists of a large number of molecules, which are in
random motion and obey Newton’s laws of motion.
2. The volume of the molecules is negligibly small compared to
the volume occupied by the gas.
3. No forces act on the molecules except during elastic collisions
of negligible duration.
At the ordinary temperature and pressure, the molecular size is very
very small as compared to the intermolecular distance. In case of gas,
the molecules are very far from each other. So when the molecules are
far apart and the size of the molecules is very small when compared to
the distance between them. Therefore the interactions between the
molecules are negligible.
In case there is no interaction between the molecules than there will be
no force acting on the molecule. This is because it is not interacting
with anything. Newton’s first law states that an object at rest will be
at rest and an object will be in motion unless an external force acts
upon it.
So in this case, if the molecule is not interacting with any other
molecule then there is nothing that can stop it. But sometimes when
these molecules come close they experience an intermolecular force.
So this basically something we call as a collision.
Solved Question For You
Question: The number of collisions of molecules of an ideal gas with
the walls of the container is increasing per unit time. Which of the
following quantities must also be increasing?
I.pressure
II. temperature
III. the number of moles of gas.
A. I only
B. I and II only
C. II only
D. II and III only
Solution: B. If there are more collisions between the molecules and
the walls of the container, there must be more pressure against the
wall. If there are more collisions than the molecules must have high
average kinetic energy. Since kinetic energy is proportional to
temperature, the temperature is also increasing.
Question: When the volume of a gas is decreased at constant
temperature the pressure increases because of the molecules
A. strike unit area of the walls of the container more often
B. strike unit area of the walls of the container with higher speed
C. move with more kinetic energy
D. strike unit area of the walls of the container with less speed
Solution: A. The kinetic theory of the molecules depends on the
temperature and since here the temperature remains constant, the
pressure cannot increase due to the other options mentioned. So option
A is correct as more pressure is generated here and hence pressure
increases.
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