UNIT 18 – Thermodynamic Processes and The First Law

Beta Keramati

First and Second Laws of Thermodynamics

Last Update: 06/16/2022

the first law of thermodynamics

Thermodynamics is the branch of physics that deals with the transfer of heat energy and the relationship between heat energy and other forms of energy. Thermodynamics is useful in understanding the characteristics and behavior of macroscopic systems that contain a very large number of particles. An example of a thermodynamic system is a sample of gas in a container. In a thermodynamic system, the sum total of the kinetic and potential energies of all the particles in the system is called Internal Energy, U. Notice that the kinetic energy of the system itself or the location of the system as a whole and its potential energy does not affect the internal energy. For example, the internal energy of a gas in a container is not affected by moving the container from the floor to a table or by placing the container on a moving train.

The First Law of Thermodynamics is a statement of conservation of energy. In UNIT 10 we saw that work is a process in which energy is added or removed from a system. In UNIT 17 we learned that heat transfer is also a method of transferring energy to and from a system that occurs when there is a difference in temperature. Work and heat transfer are the only mechanisms through which the internal energy of a system can change. The First Law of Thermodynamics states that the change in the internal energy of a system, ΔU, is equal to the difference between the energy that goes in and the energy that comes out. Since the transfer of energy takes place in the form of heat and/or work, The First Law of Thermodynamics is expressed as:

$\Delta U=Q-W$ 1^st Law Of Thermodynamics – Conservation Of Energy

In this equation, Q is heat transferred into the system (energy going in), and W is work done by the system (energy coming out). By carefully using the correct signs for Q and W, this equation can be used for all situations regardless of whether heat is transferred into or out of the system or whether the system does work or work is done on the system by the surroundings.

We can rearrange this equation and write it as

$\boxed{Q=\Delta U+W}$ 1^st Law of Thermodynamics- Conservation of Energy

In this text, we will use $Q=\Delta U+W$ simply because it causes less confusion when it comes to determining the signs for Q and W. The sign conventions for Q and W are as follows.

When heat flows into the system, Q >0.

When heat flows out of the system, Q<0.

When the system does work on its surroundings, W>0.

When the surroundings do work on the system, W<0.

Example 18.1

Suppose there is a heat transfer of 40.00 J to a system, while the system does 10.00 J of work. Later, there is a heat transfer of 25.00 J out of the system while 4.00 J of work is done on the system. What is the net change in the internal energy of the system?

Solution

We could approach this problem in two different ways. One way is to find the total heat transfer, and the total work, and then use the 1st law of thermodynamics (conservation of energy) to determine the change in the internal energy. The second way is to determine the change in the internal energy for each process and then add the two changes together.

First Method

$Q=+40.00J-25.00J$

$Q=+15.00J$

$W=+10.00J-4.00J$

$W=+6.00J$

$Q=\Delta U+W$
$+15.00J=\Delta U+6.00J$

$\Delta U= +9.00J$

Second Method

$Q_1=(\Delta U)_1+W_1$

$+40.00J=(\Delta U)_1+10.00J$

$(\Delta U)_1=+30.00J$

$Q_2=(\Delta U)_2+W_2$

$-25.00J=(\Delta U)_2-4.00J$

$(\Delta U)_2=-21.00J$

$\Delta U=(\Delta U)_1+(\Delta U)_1$

$\Delta U=30.00J-21.00J$

$\Delta U=+9.00J$

Thermodynamics deals with the relationship between heat transfer and other forms of energy. Let’s consider a sample of an ideal gas as our thermodynamic system. By definition, the internal energy is the sum of all the energies, potential plus kinetic, of all the molecules in the gas. Since forces among molecules are negligible in an ideal gas, there is no potential energy. Therefore the only type of molecular energy to consider in an ideal gas is kinetic energy. This means the internal energy of a sample of an ideal gas is the sum of the kinetic energy of all the molecules in the gas. Now, we are going to find an expression for the internal energy of a sample of a monatomic ideal gas, and a sample of a diatomic ideal gas.

Internal Energy of a Sample of Monatomic Ideal Gas

In UNIT 16 we saw that absolute temperature is directly proportional to average translational kinetic energy per molecule.

$\frac{1}{2}mv_{rms}^2 = \frac{3}{2}k_B T$

In a monatomic gas, the molecules can be treated as point particles and can only have translational kinetic energy. Therefore, we can easily find the internal energy of the system by simply multiplying the average translational kinetic energy per molecule by the number of molecules in the sample of gas.

$\boxed{U_{monatomic} = \frac{3}{2}Nk_B T}$ Internal Energy Of A Monatomic Gas

In this equation

$N$ is the number of molecules in the sample of gas

$k_B$ is the Boltzmann Constant. $k_B= 1.38\times 10^{-23}J/K$

$T$ is the absolute temperature of the gas in Kelvin

We can also express this equation in terms of the number of moles by multiplying and dividing the right-hand side by Avogadro’s Number, N_A, and replacing N/N_A with n, number of moles.

$U_{monatomic} = \frac{3}{2}\frac {N}{N_A} N_A k_B T$

$U_{monatomic} = \frac{3}{2}n( N_A k_B )T$

And since $R=N_A k_B$ , we get

$\boxed{U_{monatomic} = \frac{3}{2}nRT}$ Internal Energy Of A Monatomic Gas

In this equation

$n$ is the number of moles of gas

$R$ is the Universal Gas Constant. $R=8.31J/mol.K$

$T$ is the absolute temperature of the gas in Kelvin

Internal Energy of a Sample of Diatomic Ideal Gas

Since a diatomic molecule can rotate as well as move along a straight line, Its total kinetic energy is the sum of its translational kinetic energy and rotational kinetic energy. Including rotational kinetic energy yields the following equation for the internal energy of a diatomic gas

$\boxed{U_{diatomic} = \frac{5}{2}Nk_B T}$ Internal Energy Of A Diatomic Gas

In this equation

$N$ is the number of molecules in the sample of gas

$k_B$ is the Boltzmann Constant. $k_B= 1.38\times 10^{-23}J/K$

$T$ is the absolute temperature of the gas in Kelvin

We can also express this equation in terms of the number of moles.

$\boxed{U_{diatomic} = \frac{5}{2}nRT}$ Internal Energy Of A Diatomic Gas

In this equation

$n$ is the number of moles of gas

$R$ is the Universal Gas Constant. $R=8.31J/mol.K$

$T$ is the absolute temperature of the gas in Kelvin

Notice that for an ideal gas the internal energy is directly proportional to the absolute temperature of the gas.

Thermodynamic processes

A system is a section of the universe that is separated from its surroundings by real or hypothetical boundaries. A thermodynamic system is macroscopic and contains a large number of particles. For example, we can consider a gas confined to a cylinder to be a thermodynamic system. The quantities that describe the state of a thermodynamic system are known as state variables. In this book, we will only discuss thermodynamics as it pertains to ideal gases. For an ideal gas, state variables include (but are not limited to) pressure, volume, temperature, and the number of molecules or moles of the gas. The Ideal Gas Law, PV=nRT, is an equation that relates state variables for an ideal gas.

The state of a system can change as a result of its interaction with the environment. The interaction between a system and its environment involves the transfer of energy and this transfer of energy is governed by the First Law of Thermodynamics, Q=ΔU+W. Notice that internal energy, U, is the total energy of the particles in a system. At any given state the system has a certain amount of internal energy. Therefore, internal energy, U, is a state variable. On the other hand, heat transfer, Q, and work, W, represent energy flowing in and out of the system while the system is transitioning from one state to another state. Therefore Q and W are not state variables.

The manner in which the state of a system changes from an initial state to a final state is called a thermodynamic process. In this section, we discuss a few basic thermodynamic processes for an ideal gas.

isobaric process

An isobaric process is a process during which the pressure of the gas stays constant. For example, when a gas is confined to a cylinder with a piston that can move without friction, any heat that flows into the gas or out of the gas causes expansion or compression that occurs at constant pressure and is, therefore, isobaric. Let’s find an expression for work done in such an isobaric process.

Consider a sample of an ideal gas confined to a cylinder with a frictionless, moveable piston on top, as shown in Figure 18.1.

If heat is transferred into the system, the gas expands and pushes the piston upward. The work done in this process is the work done by force F applied by the gas to the piston to move it a distance d, as shown in Figure 18.1.

$W=Fdcos\theta$

Since $\theta=0 \Rightarrow cos\theta=1$ , and $W=Fd$ . Since pressure is defined as force per unit area, $P=\frac{F}{A}$ , we can replace $F$ with $PA$ , and get

$W=PAd$

Furthermore, $Ad=\Delta V$ , where $\Delta V$ is the change in volume of the gas (see Figure 18.1). Therefore,

$\boxed{W_{isobaric}=P\Delta V}$ Work Done In An Isobaric Process

Notice that if the gas expands, $\Delta V>0$ and $W>0$ . If, on the other hand, someone pushes down on the piston to compress the gas, then $\Delta V<0$ and $W<0$ .

We often show a thermodynamic process with a plot of pressure vs volume, called a PV diagram. Each point on a PV diagram corresponds to a specific pressure, volume, and temperature and represents a specific state of the gas. The PV diagram for an isobaric expansion from state A to state B is shown in Figure 18.2

It’s easy to see that the area under this line is equal to the work done in this process. This idea can be generalized to any process. The area under the plot of pressure vs volume is always equal to work done in that process.

isochoric (isometric) process

An isochoric (isometric) process is a process during which the volume of the gas stays constant. Since an isochoric process on a PV diagram is a straight vertical line, the work done in an isochoric process is zero. Applying the First Law of Thermodynamics we get

$Q=\Delta U$ For An Isochoric Process

To create an isochoric process, the gas must be confined to a rigid container with a fixed volume.

Example 18.2

Consider the expansions of an ideal gas between the equilibrium states A and C as shown in the PV diagram below. If 515 J of heat is added to the gas as it follows the path ABC, how much heat is required for the transition along ADC?

Assume that P₁=2.10×10⁵Pa, P₂=1.05×10⁵Pa, V₁=2.25×10^-3m³ , V₂=4.50×10^-3m³.

Solution

In both expansions, the initial state is A and the final state is C. Since the change in internal energy, ΔU, depends only on the change in temperature, ΔU is the same in both expansions.

Let’s apply the first law of thermodynamics to these two processes.

For the expansion along the path ABC. Since AB is an isobaric process, W_AB=P₁ΔV_AB

$Q=\Delta U +W$

$Q=\Delta U +P_1 (V_2 - V_1)$

$515J=\Delta U +2.10\times 10^{5}Pa (4.50 \times 10^{-3}m^3 - 2..25 \times 10^{-3}m^3)$

$\Delta U=42.5J$

Now, let’s do the same thing for the expansion along ADC.

$Q=\Delta U +W$

$Q=\Delta U +P_2 (V_2 - V_1)$

$Q=42.5J +1.05\times 10^{5}Pa (4.50 \times 10^{-3}m^3 - 2..25 \times 10^{-3}m^3)$

$Q=279J$

Discussion

The work calculations in this problem are made simple since no work is done along AD and BC (isometric processes). Along AB and DC; the pressure is constant over the volume change(isobaric processes), so the work done is simply PΔV.

Isothermal process

An isothermal process is a process during which the temperature of the gas stays constant. We can use the ideal gas law to predict the shape of an isothermal process on a PV diagram.

$PV=nRT$

$P=nRT\frac{1}{V}$

In an isothermal process $nRT=constant$ . This means P is inversely proportional to V. The graph of an inverse function ( $f(x)=\frac{1}{x}$ ) is a hyperbola as shown in Figure 18.4-a.

Using calculus, it can be shown that the area under an isothermal curve from state 1 to state 2 is equal to

$\boxed{W_{isothremal}=nRTln(\frac{V_2}{V_1})}$ Work In An Isothermal Process

As mentioned earlier, the internal energy of an ideal gas is directly proportional to its absolute temperature. Therefore, in an isothermal process, there is no change in the internal energy, and $\Delta U=0$ . Applying the First Law of Thermodynamics we get

$Q=W$ For An Isothermal Process

One way to maintain the temperature of a gas constant as the gas expands or contracts is by making the process occur very slowly in a container with walls that are an excellent conductor of heat, and placed in a reservoir at a constant temperature. For example, consider the following setup where a cylinder of gas with a movable piston is immersed in a large water tank whose temperature is maintained constant. Since the piston is freely movable, the pressure inside P_inis balanced by the pressure outside P_out by some weights on the piston.

Slowly removing the weights results in an isothermal expansion of the gas. This is because removing the weights on the piston leads to an imbalance of forces on the piston, which causes the piston to move up. As the piston moves up, the temperature is lowered momentarily, which causes heat to flow from the heat bath to the system. This flow of heat causes the temperature to go back up till thermal equilibrium is, once again, restored. So, effectively, the temperature of the gas stays constant as the gas slowly expands.

Note that all phase changes are also isothermal processes as discussed in UNIT 17.

As mentioned earlier, isothermal processes create hyperbolic curves on a PV diagram. The higher the temperature, the higher up is the position of the curve on the PV diagram, as shown below.

Adiabatic process

An adiabatic process is a process in which there is no transfer of heat between the system and the environment. Since $Q=0$ then

$W=-\Delta U$ For An Adiabatic Process

Notice that a change in internal energy, U, is always accompanied by a change in temperature. When an ideal gas is compressed adiabatically<span id="MathJax-Element-1925-Frame" class="MathJax" style="font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;text-transform: none;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px" role="presentation" data-mathml=" $(Q = 0),$ “> work is done on it and its temperature increases; in an adiabatic expansion, the gas does work, and its temperature drops. Therefore, if in a process, there is no transfer of heat to or from the system, then the temperature of the system must change.

One way to produce an adiabatic process is to make the process happen very quickly and leave no time for heat transfer. Adiabatic compressions actually occur in the cylinders of a car, where the compressions of the gas-air mixture take place so quickly that there is no time for the mixture to exchange heat with its environment. Nevertheless, because work is done on the mixture during the compression, its temperature does rise significantly. In fact, the temperature increases can be so large that the mixture can explode without the addition of a spark. Such explosions, since they are not timed, make a car run poorly—it usually “knocks.” Because ignition temperature rises with the octane of gasoline, one way to overcome this problem is to use higher-octane gasoline.

Another way to produce an adiabatic process is to use perfect insulation around a system to prevent heat transfer. For example, in Figure 18.5, even if the grains of sand are removed one at a time from the top of the piston to allow the gas to expand very slowly, insulation around the cylinder prevents heat transfer, and the temperature of the gas drops as it expands.

The PV diagram for an adiabatic process is a curve. Starting from the same pressure and volume, an adiabatic expansion produces a curve that is below the isothermal expansion starting from the same state. This is shown in Figure 18.6

In an adiabatic process from state 1 to state 2, pressure and volume are related according to

$\boxed{P_1V_1^\gamma = P_2V_2^\gamma}$ For An Adiabatic Process

where $\gamma = \frac{C_P}{C_V}$ .

$C_P$ is the specific heat at constant pressure, and $C_V$ is the specific heat at constant volume. In the next section, we explore the difference between the two values of specific heat for an ideal gas.

Example 18.3

An ideal gas expands isothermally along AB and does 700 J of work (see below).

(a) How much heat does the gas exchange along AB?

(b) The gas then expands adiabatically along BC and does 400 J of work. When the gas returns to A along CA, it exhausts 100 J of heat to its surroundings. How much work is done on the gas along this path?

Solution for (a)

Let’s apply the first law of thermodynamics to the process AB. But since the process is isothermal, there is no change in the internal energy of the gas, and we get

$Q=W$

$Q=700J$

Solution for (b)

Since process BC is adiabatic, Q=0. Therefore,, using the first law of thermodynamics, and using the fact that the gas does 400J of work, we get

$0=\Delta U + 400J$

$(\Delta U)_{BC} =- 400J$

Now let’s apply the first law for the process CA, and use the fact that 100J of heat is released.

$-100J=(\Delta U)_{CA} +W_{CA}$

$(\Delta U)_{CA}=-100J -W_{CA$

For the whole cyclic process of going from A to B to C and back to A, the change in internal energy must be zero because the gas started and ended at the same state and therefore at the same temperature. This means that

$(\Delta U)_{AB} +(\Delta U)_{BC}+(\Delta U)_{CA} =0$

$0+(-400J)+(-100J -W_{CA}) =0$

$W_{CA} = -500J$

Specific Heat at Constant Pressure (C_P) vs Specific Heat at Constant Volume (C_V)

As discussed in UNIT 17, we defined the specific heat of solids and liquids as the amount of heat needed to change the temperature of 1kg of a substance by 1°C. Since the behavior of gases and ideal gas law depend on the number of moles of the gas and not its mass, for gases, we use molar specific heat capacity. Molar specific heat capacity is defined as the amount of heat needed to change the temperature of 1mol of gas by 1°C.

Consider two identical samples of an ideal gas, with the same pressure, temperature, volume, and the number of moles. One sample is confined to a rigid container with a fixed volume (Vessel A in Figure 18.7a), and the other one is in a cylinder equipped with a freely moving piston (Vessel B in Figure 18.7b)

Which sample needs to receive more heat for its temperature to go up by 1°C? Since the gas in vessel A cannot expand and do work, all the energy added to the system in the form of heat is going to increase the average kinetic energy of the molecules, and, as a result, increase the temperature. In vessel B, however, some of the energy added in the form of heat is leaving the system in form of work as the gas expands. This means that only some of the heat energy is contributing to the increase in the average kinetic energy of the molecules and causing the temperature to go up. Clearly, more heat needs to be added to vessel B to cause the same 1°C temperature change as in vessel A. Therefore, heat capacity at constant pressure is greater than heat capacity at constant volume.

$C_P>C_V$

Using the Ideal Gas Law it is possible to derive the values given for C_P and C_V in Table 18.1.

Using the values in Table 18.1, we see that

$\gamma_{monatomic}=1.67$ and $\gamma_{diatomic}=1.40$ For Ideal Gases

Example 18.4

2.00 mol of He gas that is initially at a temperature of 20.0ºC expands to three times its initial volume.

a) Calculate the work done, if the process was isothermal.

b) Calculate the work done, if the process was adiabatic.

Solution for (a)

Work done in an isothermal process is

$Work=nRTln(\frac{V_2}{V_1})$

$Work=(2.00mol)(8.315J/molK)(293K)ln(\frac{3V}{V})$

$Work=5.35 \times 10^3 J$

Solution for (b)

In an adiabatic process, there is no transfer of heat and Q=0. Using the first law of thermodynamics we get:

$0=\Delta U +W$

$W=-\Delta U$

Since He is a monatomic gas, the change in its internal energy is given by

$\Delta U= \frac{3}{2}nR\Delta T$

So we need to find the final temperature. If we write the ideal gas law for the initial and final state and take into account that the number of moles, n, doesn’t change, we get

$P_1 V_1 = nRT_1$ and $P_2 V_2 = nRT_2$ ⇒ $\frac{P_1 V_1}{T_1} =\frac{P_2 V_2}{T_2}$

Since V₂ = 3V₁ we get

$\frac{P_1}{T_1} =\frac{3P_2 }{T_2}$ OR $T_2=T_1 \frac{3P_2}{P_1}$

For an adiabatic process, inital and final pressures and volumes are related according to

$P_1V_1^\gamma = P_2V_2^\gamma$

This results in

$\frac{P_2}{P_1}=\frac{V_1^\gamma}{V_2^\gamma}$

For a monatomic gas, γ=1.67.

$\frac{P_2}{P_1}=\frac{V_1^{1.67}}{(3V_1)^{1.67}}$

$\frac{P_2}{P_1}=0.160$

Therefore T₂ is

$T_2=T_1(3 \times 0.160)$

$T_2=(293K)(3 \times 0.160)$

$T_2=141K$

Now we can calculate the change in the internal energy of the gas during the expansion.

$\Delta U= \frac{3}{2}nR\Delta T$

$\Delta U= \frac{3}{2}(2.00mol)(8.315J/molK)(141K-293K)$

$\Delta U=-3.79 \times 10^3 J$

Since the process is adiabatic and Q=0,

$W=-\Delta U$

$W=3.79 \times 10^3 J$

Discussion

As demonstrated in Figure 18.6 (reproduced below), the curve for an adiabatic expansion is below the curve for an isothermal expansion on a PV diagram. Since work is the area under the plot of Pressure vs Volume, we expect the work done in the adiabatic expansion to be less than the work done in the isothermal expansion. This agrees with the numerical results in this problem.

Attribution

This chapter contains material taken from Openstax College Physics-Thermodynamics, and Openstax University Physics Volume 2-The First Law of Thermodynamics, and is used under a CC BY 4.0 license, . Download these books for free at Openstax

To see what was changed, refer to the List of Changes.

Problems

[openstax univ. phys. vol.2 – 3.23]It takes 500 J of work to compress 0.500 mol of an ideal gas to one-fifth its original volume while the temperature of the gas stays constant. Determine that temperature.
[openstax univ. phys. vol. 2 – 3.25] An ideal gas with a volume of 20.0L and pressure of 0.800atm expands isobarically. In this process, 500 J of work is done by the gas. Determine the final volume of the gas.
In a thermodynamic process, 0.750mol of an ideal gas expands from 0.500L to 4.00 L at a constant temperature of 300K.
1. What is the change in the internal energy of the gas?
2. How much heat is absorbed by the gas in this process?
[openstax univ. phys. vol. 2 – 3.48] An ideal monatomic gas at a pressure of 2.00×10⁵Pa and a temperature of 300K undergoes an isobaric expansion from 2.00×10³cm³ to 4.00×10³cm³.
1. What is the work done by the gas?
2. What is the temperature of the gas after the expansion?
3. How many moles of gas are there?
4. What is the change in the internal energy of the gas?
5. How much heat is added to the gas?
[openstax univ. phys. vol.2 – 3.54-modified] Two (2.00) moles of a diatomic ideal gas at 5.00×10⁶Pa and 5.00 L is expanded isothermally until the volume is doubled (step 1). Then it is cooled isochorically until the pressure is 1.00×10⁶Pa (step 2). The temperature drops in this process. The gas is now compressed isothermally until its volume is back to 5.00 L, but its pressure is now 2.00×10⁶Pa (step 3). Finally, the gas is heated isochorically to return to the initial state (step 4).
1. Draw the four processes in the PV plane.
2. Find the total work done from beginning to end. Is this work done on the gas or by the gas?
3. Find the heat exchanged between the gas and the environment in each process. For each process, indicate whether heat is absorbed or released by the gas.
4. Determine the total heat exchanged between the gas and the environment in this cycle.
5. Determine the total change in the internal energy of the gas in this cycle.
[openstax univ. phys. vol. 2 – 3.67(edited)] A monatomic ideal gas undergoes an adiabatic expansion in which its volume is doubled. How is the pressure of the gas changed?
[openstax univ. phys. vol. 2 – 3.78] Two (2.00) moles of a monatomic ideal gas such as helium are compressed adiabatically from a state (3.00 atm, 5.00 L) to a state with a pressure of 4.00 atm.
1. Find the volume of the final state.
2. Find the temperature of the initial and final states of the gas.
3. Find the work done by the gas in the process.
4. Find the change in the internal energy of the gas in the process.
[openstax univ. phys. vol. 2 – 3.87] Four (4.00) moles of a monatomic ideal gas in a cylinder at 27.0ºC expands at constant pressure of1.00 atm until its volume doubles.
1. What is the change in internal energy?
2. How much work was done by the gas in the process?
3. How much heat was transferred to the gas?
Consider the PV diagram of a series of thermodynamic processes shown below.
1. Find the work done from A to B. Is this work done on the gas or by the gas?
2. Find the work done from B to C. Is this work done on the gas or by the gas?
3. Find the work done in the cycle.
A ideal gas with pressure P₁ and volume V₁expands to a state with volume V₂ once isothermally, and a second time adiabatically.
1. Draw these processes on a PV diagram.
2. In which case there is more work done by the gas? Explain.
A ideal gas with pressure P₁ and volume V₁is compressed to a state with volume V₂ once isothermally, and a second time adiabatically.
1. Draw these processes on a PV diagram.
2. In which case there is more work done on the gas? Explain.

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the first law of thermodynamics

Internal Energy of a Sample of Monatomic Ideal Gas

Internal Energy of a Sample of Diatomic Ideal Gas

Thermodynamic processes

isobaric process

isochoric (isometric) process

Isothermal process

Adiabatic process

Specific Heat at Constant Pressure (CP) vs Specific Heat at Constant Volume (CV)

Attribution

Problems

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Specific Heat at Constant Pressure (C_P) vs Specific Heat at Constant Volume (C_V)