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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