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In this Physics tutorial, you will learn:

- What is/are the factor(s) affecting the internal energy of an ideal gas?
- What is molar specific heat at constant volume?
- How to calculate the molar specific heat at constant volume for an ideal monoatomic gas?
- What about the molar specific heat at constant volume if the gas is diatomic or polyatomic?
- The same for the molar specific heat at constant pressure.
- How do these two molar specific heats relate to each other?
- What are degrees of freedom and how they are related to the energy absorbed by ideal gas?
- How to find the number of degrees of freedom in various ideal gases?

Do you think all gases have the same ability to absorb or release heat energy? Why?

Do you know any other kind of motion besides the translational one described in the previous tutorial? Could you mention one of them?

What happens with the internal energy of a gas if it absorbs / releases heat from / to the surroundings?

We will explain in this tutorial how to answer correctly to the above questions not by mere intuition, rather, based on a scientific approach.

For simplicity, let's consider a monoatomic gas whose atoms move only in a translational way. Also, let's assume the internal energy of this gas is only due to the sum of translational KE of all atoms, neglecting the presence of any chemical form of energy in the system.

As explained in the previous tutorial, the translational kinetic energy of a single atom is given by the formula

KE_{1} = *3**/**2* × k × T

If we have a gas of n moles, it contains n × NA atoms. Therefore, the internal energy of gas (which corresponds to the average kinetic energy of all its atoms), is

U = n × N_{a} × KE_{1}

= n × N_{a} × *3**/**2* × k × T

= n × N

Given that the Boltzmann constant k is given by

k = *R**/**N*_{a}

we obtain for the internal energy of gas U:

U = *3**/**2* × n × R × T

Since n and R are constants, we conclude that:

**The internal energy of an ideal gas is a function of temperature only. In other words, it depends only on the temperature of gas.**

By definition, **Molar specific heat** of an ideal gas is the **heat** we provide to the gas to raise the temperature of one mole through 1K or 1°C. It differs from the specific heat capacity c discussed earlier, as the molar specific heat is calculated for 1 mole instead of 1 kg of material. We represent it as C and its unit is J mol/K.

Let's consider a gas inside a fixed cylinder, as the one shown in the figure below.

If the amount of gas inside the container is n moles, its initial temperature is T, pressure P and volume V, we consider these parameters as those of the initial state i.

If we provide a very small amount of heat energy to the gas, its temperature increases by ΔT and pressure increases by ΔP as well. Thus, the gas is brought to a final state f, which has new values for the gas parameters.

Given that volume is constant, we obtain for the P - V graph of the two states:

The heat absorbed by the gas during this process therefore is

Q = n × C_{v} × ∆T

where C^{V} is the molar specific heat at constant volume.

From the First Law of Thermodynamics, we know that

Q = ∆U + W

Rearranging, we obtain

∆U = Q - W

Thus, substituting the above value of Q, we obtain

∆U = n × C_{v} × ∆T - W

Since the volume of gas is constant, there is no work done on or by the gas. Hence,

∆U = n × C_{v} × ∆T

Or

C_{v} = *∆U**/**n × ∆T*

Substituting

∆U = *3**/**2* × n × R × ∆T

in the above equation, we obtain

C_{v} = *3**/**2* × n × R × ∆T*/**n × ∆T*

Simplifying, we obtain for the **molar specific heat** of ideal gas

C_{v} = *3**/**2* R

Since R = 8.31 J/K, we obtain for the value of molar specific heat of a monoatomic ideal gas:

C_{v} = *3**/**2* × 8.31 *J**/**mol × K*

≈ 12.5*J**/**mol × K*

≈ 12.5

When dealing with two or more atomic gases, we observe that the motion of molecules is not that simple, as atoms in monoatomic gases, because molecules can also spin or vibrate, not only move in a translational way. Therefore, the value of C^{V} will be greater than that obtained above for monoatomic gases.

Summarizing everything discussed above, we conclude that:

The actual internal energy of an ideal gas at constant volume is

U = n × C_{v} × T

and the change in the internal energy of an ideal gas at constant volume is

∆U = n × C_{v} × ∆T

0.3 moles of a monoatomic ideal gas are at 27°C inside a fixed container.

- What is the value of actual internal energy of this gas?
- By how many joules will the internal energy of gas increase when the temperature becomes 177°C?
- What will be the value of final internal energy of gas?

Clues:

n = 0.3 moles

T_{1} = 27°C = 27 + 273 = 300 K

T_{2} = 177°C = 177 + 273 = 450 K

C^{v} = 12.5 J/molK

Hence, we have

a) The actual (initial) internal energy of gas is

U_{i} = n × C_{v} × T

= 0.3 mol × 12.5*J**/**mol × K* × 300 K

= 1125 J

= 0.3 mol × 12.5

= 1125 J

b) Then increase in the internal energy of gas is

∆U = n × C_{v} × ∆T

= 0.3 mol × 12.5*J**/**mol × K* × (450 - 300)K

= 562.5 J

= 0.3 mol × 12.5

= 562.5 J

c) The value of final internal energy of gas will be

U_{f} = U_{i} + ∆U

= 1125 J + 562.5 J

= 1687.5 J

= 1125 J + 562.5 J

= 1687.5 J

The graph below shows three different paths representing three different processes that bring the ideal gas from temperature T to another temperature T + ΔT (one is at constant volume, one at constant pressure and the other is at changing volume and pressure).

It is obvious that the change in temperature is the same in all paths shown above and in many others. This means the change in internal energy is independent from the path followed during the process. Thus, we reached in a very important conclusion:

**Any change in the internal energy of an ideal gas enclosed within a container depends only on the change in temperature; it is independent from the type of process that has brought that change.**

Now, let's assume the container which holds the ideal gas is not fixed, i.e. its piston can slide freely up and down.

If we supply some heat to the gas enough to increase its temperature from T to T + ΔT like in the previous paragraph, the piston will move up as the gas expands due to the increase in temperature but the pressure remains constant. In both cases, the inner pressure of gas balances the effect of atmospheric pressure plus the pressure exerted by the piston's weight).

Similarly as in the process with constant volume, we have for the heat absorbed by the ideal gas at constant pressure

Q = n × C_{p} × ∆T

where C^{p} is known as the **molar specific heat at constant pressure**.

The value of C^{p} is numerically greater than the corresponding value of C^{V} for the same change in temperature as in the process with constant volume, a part of heat energy supplied goes for doing work for lifting the piston.

Given that the temperature increases without any change in pressure, we obtain for the P - V graph:

Let's find the relationship between the two molar specific heats C^{V} and C_{p}.

From the First Law of Thermodynamics, we know that

∆U = Q - W

Given that at constant pressure the work done by the gas to lift the piston is

W = P × ∆V = n × R × ∆T

we obtain

∆U = Q - n × R × ∆T

= n × C_{p} × ∆T - n × R × ∆T

= n × C

Also, given that

∆U = n × C_{v} × ∆T

we obtain

n × C_{v} × ∆T = n × C_{p} × ∆T - n × R × ∆T

Dividing both sides by n × ΔT, we obtain

C_{v} = C_{p} - R

Or

C_{p} = C_{v} + R

For a monoatomic gas, we have

C_{p} = *3**/**2* R + R

=*5**/**2* R

≈ 20.8*J**/**mol × K*

=

≈ 20.8

Therefore, we can write for the heat absorbed by a gas at constant pressure to increase its temperature by ΔT:

∆Q = *5**/**2* n × R × ∆T

This value is greater than the corresponding value of heat energy during a process at constant volume for the same increase in temperature (Q = ** 3/2** n R T), as we predicted earlier.

What is the heat energy required to increase by 50 K the temperature of 2 moles of an ideal monoatomic gas at constant pressure?

Clues:

ΔT = 50 K

n = 2

ΔQ = ?

Using the equation for heat absorbed by an ideal gas at constant pressure, we can write:

∆Q = *5**/**2* n × R × ∆T

=*5**/**2* × 2 moles × 8.31 *J**/**mol × K* × 50 K

= 2077.5 J

=

= 2077.5 J

With degrees of freedom, we understand independent ways of a particle's motion. For example, a train moving on a railway has only one degree of freedom as it can move only according one direction - the direction determined by the tracks.

A translational motion can be a combination of motion in three linear directions, which we usually call X, Y and Z according the names of the three well-known perpendicular axes we have already seen in previous tutorials.

All atoms in monoatomic gases are assumed to move only in a translational way as even if they spin around themselves, this is not observable and relevant. Indeed, when you watch a football match, you don't care about the spinning effect of ball but only about its translational motion. This is why the internal energy of monoatomic gases is equal to

U = *3**/**2* × n × R × T

The number 3 is because there are three degrees of freedom in translational motion. The division by 2 is because every individual direction includes two sub-directions in itself. (left-right, back-forth, up-down).

In diatomic (two-atomic) molecules, there is a spinning effect produced which causes a double direction of rotation.

Therefore, in total we have 5 degrees of freedom for the internal energy, i.e.

U = *5**/**2* × n × R × T

If molecules are three or more atomic (polyatomic), there is another spinning direction added because two points form a line (one direction) while three point form a plane (two directions). Therefore, the total internal energy becomes

U = *5 + 1**/**2* × n × R × T

=*6**/**2* × n × R × T

= 3 × n × R × T

=

= 3 × n × R × T

The table below shows the relationship between molar specific heats and degrees of freedom in ideal gases.

Based on the above analysis, Maxwell formulated the theorem of the equipartition of energy, which says:

**Every molecule has a certain number of degrees of freedom, which represent independent ways of storing energy. A portion of energy associated with a single degree of freedom is given by the expression**

U = ** 1/2** kT (for a single molecule) or U =

Thus, we can obtain a general form of the relationship between the number of degrees of freedom f and the **molar specific heat at constant volume** C^{V} for an ideal gas in terms of the ideal gas constant R:

C_{v} = *f**/**2* × R = f × *R**/**2*

= f ×*8.31**/**2* *J**/**mol × K*

= 4.16 × f*J**/**mol × K*

= f ×

= 4.16 × f

There is also a vibrational motion caused by the oscillations of diatomic or polyatomic molecules, whose behaviour is described through quantum theory, but which goes beyond the scope of this tutorial. It required higher amounts of energy to take place then the other two types of motion explained here. Therefore, we neglect this kind of motion when energy stored in molecules is relatively small.

**The internal energy of an ideal gas is a function of temperature only. In other words, it depends only on the temperature of gas.**

Mathematically, we have

U = *3**/**2* × n × R × T

By definition, **Molar specific heat** of an ideal gas is the **heat** we provide to the gas to raise the temperature of one mole through 1K or 1°C. It differs from the specific heat capacity c discussed earlier, as the molar specific heat is calculated for 1 mole instead of 1 kg of material. We represent it as C and its unit is J mol/K.

**Molar specific heat** of a monoatomic ideal gas is

C_{v} = *3**/**2* R

Since R = 8.31 J/K, we obtain for the value of molar specific heat of a monoatomic ideal gas:

C_{v} = *3**/**2* × 8.31 *J**/**mol × K*

≈12.5*J**/**mol × K*

≈12.5

When dealing with two or more atomic gases, we observe that the motion of molecules is not that simple, as atoms in monoatomic gases, because molecules can also spin or vibrate, not only move in a translational way. Therefore, the value of C^{V} will be greater than that obtained above for monoatomic gases. Thus,

The actual internal energy of an ideal gas at constant volume is

U = n × C_{v} × T

and the change in the internal energy of an ideal gas at constant volume is

∆U = n × C_{v} × ∆T

The above formula implies that:

**Any change in the internal energy of an ideal gas enclosed within a container depends only on the change in temperature; it is independent from the type of process that has brought that change.**

Similarly as in the process with constant volume, we have for the heat absorbed by the ideal gas at constant pressure

Q = n × C_{p} × ∆T

where C^{p} is known as the **molar specific heat at constant pressure**.

The value of C^{p} is numerically greater than the corresponding value of C^{V} for the same change in temperature as in the process with constant volume, a part of heat energy supplied goes for doing work for lifting the piston.

From the First Law of Thermodynamics, we know that

∆U = Q - W

Substituting, we obtain

n × C_{v} × ∆T = n × C_{p} × ∆T-n × R × ∆T

After simplifying and rearranging, we obtain for the molar specific heat at constant pressure.

C_{p} = C_{v}+R

For a monoatomic gas, we have

C_{p} = *3**/**2* R + R

=*5**/**2* R

≈ 20.8*J**/**mol × K*

=

≈ 20.8

Therefore, we can write for the heat absorbed by a monoatomic gas at constant pressure to increase its temperature by ΔT:

∆Q = *5**/**2* n × R × ∆T

With degrees of freedom, we understand independent ways of a particle's motion.

All atoms in monoatomic gases are assumed to move only in a translational way as even if they spin around themselves, this is not observable and relevant.

The internal energy of monoatomic ideal gases is equal to

U = *3**/**2* × n × R × T

The number 3 is because there are three degrees of freedom in translational motion. The division by 2 is because every individual direction includes two sub-directions in itself. (left-right, back-forth, up-down).

In diatomic (two-atomic) molecules, there is a spinning effect produced which causes a double direction of rotation.

Therefore, in total we have 5 degrees of freedom for the internal energy, i.e.

U = *5**/**2* × n × R × T

If molecules are three or more atomic (polyatomic), there is another spinning direction added because two points form a line (one direction) while three point form a plane (two directions). Therefore, the total internal energy becomes

U = *5 + 1**/**2* × n × R × T

=*6**/**2* × n × R × T

= 3 × n × R × T

=

= 3 × n × R × T

Maxwell formulated the theorem of the equipartition of energy, which says:

**Every molecule has a certain number of degrees of freedom, which represent independent ways of storing energy. A portion of energy associated with a single degree of freedom is given by the expression**

U = ** 1/2** kT (for a single molecule) or U =

Thus, we can obtain a general form of the relationship between the number of degrees of freedom f and the **molar specific heat at constant volume** C^{V} for an ideal gas in terms of the ideal gas constant R:

C_{v} = *f**/**2* × R = f × *R**/**2*

= f × 8.3*1**/**2* *J**/**mol × K*

= 4.16 × f*J**/**mol × K*

= f × 8.3

= 4.16 × f

1) 0.4 moles of a monoatomic ideal gas inside a fixed contained absorb 800 J of heat energy from a heat source. If the initial temperature of gas was 77°C, what will be the final temperature in Kelvin degree at the end of process?

- 160 K
- 510 K
- 237 K
- 587 K

**Correct Answer: B**

2) Three moles of O2 at 27°C are heated at constant volume until the temperature of gas reaches 400 K. What is the amount of heat energy absorbed by the gas during this process? Assume the O2 behaviour as that of an ideal gas.

- 13948 J
- 3739.5 J
- 23247 J
- 6232.5 J

**Correct Answer: D**

3) 176 g of CO2 are heated by 40 K at constant pressure. What is the increase in the internal energy of CO2 if it is assumed as an ideal gas? Take the molar mass of CO2 as 44 g/mol.

- 5318.4 J
- 324010 J
- 3988.8 J
- 175507.2 J

**Correct Answer: A**

We hope you found this Physics tutorial "Molar Specific Heats and Degrees of Freedom" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Thermodynamics with our Physics tutorial on Gas Laws.

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