**Article Thermodynamic processes : Isothermal Adiabatic Isochoric Isobaric**

There are four thermodynamic processes, namely Isothermal, isochoric, isobaric and adiabatic processes.

**Isothermal Process (constant temperature)**

In an isothermal process, system temperature is kept constant. Theoretically, the analyzed system is an ideal gas. Ideal gas temperature is directly proportional to ideal internal gas energy (U = 3/2 n R T). T does not change, so U also does not change. Thus, if applied to the isothermal process, the first law of thermodynamic equation becomes:

ΔU = Q – W

Internal energy does not change, so ΔU = 0.

0 = Q – W

Q = W

Based on this equation, in the isothermal process (constant temperature), heat (Q) added to the system is used by the system to perform work (W).

Change in pressure and volume of the system in the isothermal process is illustrated by the graph below:

First system volume = V_{1} (small volume) and system pressure = P_{1} (big pressure). For the system temperature to be constant then after heat is added to the system, the system expands and does work on the environment. After system does work on the environment, system volume changes to V_{2} (system volume increases) and system pressure changes to P_{2} (system pressure decrease). The shape of the graph is curved because system pressure does not change regularly during the process. Amount of work done system = shaded area.

**Adiabatic Process**

In the adiabatic process, no heat is added to the system or leaves the system (Q = 0). Adiabatic processes can occur in well-isolated closed systems. For a well-insulated closed system, there is usually no heat that flows into the system or leaves the system. Adiabatic processes can also occur in closed systems that are not isolated.

In this case, the process must be done very quickly so that heat does not flow to the system or leave the system. If applied to an adiabatic process, first law Thermodynamic equation will change:

ΔU = Q – W

No heat is in or out of the system, so Q = 0.

ΔU = 0 – W

ΔU = -W

If the system is pressed quickly (work is done on the system), then work is negative. Because W is negative, then U is positive (energy in the system increases). Conversely, if the system expands quickly (system does work), then W is positive. Since W is positive, U is negative (energy in the system is reduced). The energy in the system (ideal gas) is proportional to the temperature (U = 3/2 n R T), therefore if the energy in the system increases then system temperature also increases. Conversely, if the energy in the system is reduced then system temperature decreases.

Change in system pressure and volume in the adiabatic process is illustrated by the graph below:

The adiabatic curve in this graph (curves 1-2) is steeper than the isothermal curve (curves 1-3). This steepness difference shows that for the same volume increase, the system pressure is reduced more in the adiabatic process than the isothermal process. System pressure decreases more in the adiabatic process because when adiabatic expansion occurs, the system temperature also decreases. The temperature is proportional to the pressure, therefore if the system temperature drops, the system pressure also decreases. In contrast to the isothermal process, the system temperature is always constant. Thus in the isothermal process, the temperature does not influence the pressure drop.

One example of an adiabatically process occurs in an internal combustion engine, such as a diesel engine and a gasoline engine. In diesel engines, air is inserted into the cylinder, and the air inside the cylinder is pressed quickly using a piston (work is done on air). The adiabatic compression process (system volume reduction) is illustrated by curve 2-1. Because of rapid adiabatic pressure, the temperature rises rapidly. At the same time, diesel is sprayed into the cylinder through the injector, and the mixture is instantly triggered. In a gasoline engine, a mixture of air and gasoline is inserted into the cylinder and then pressed quickly using a piston. Because it is rapidly pushed adiabatically, the temperature rises quickly. At the same time, spark plugs fire so that the process of burning occurs.

**Isochoric Process (constant volume)**

In the Isochoric process, system volume is kept constant. Since the system volume is always constant, the system can not work on the environment. The environment also can not do work on the system. If applied to the isochoric process, the first law of thermodynamic equation changes into:

ΔU = Q – W

The system does not work on the environment, so W = 0

ΔU = Q – 0

ΔU = Q

From this result, we can conclude that in the isochoric process (constant volume), heat (Q) added to the system is used to increase the energy in the system. Changes in system pressure and volume in the isochoric process are illustrated by the graph below:

At first system pressure = p_{1 }(small pressure). The additional heat of the system causes the energy in the system to increase. As the energy in system increases, system temperature (ideal gas) increases (U = 3/2 n R T). The temperature is directly proportional to pressure. Therefore if system temperature increases, then system pressure increases (p_{2}). Due to the constant system volume, there is no work done (no shaded area).

In the isochoric process, the system cannot do work on the environment. Likewise, the environment cannot do work on the system. This is because, in an isochoric process, system volume is always constant. There are certain types of work that do not involve volume changes. So even though the volume of the system does not change, work can still be done on the system. For example, there is a fan + battery in a closed container. The fan can rotate using energy by the battery. In this case, fans, batteries, and air inside the container are considered as systems.

When the fan rotates, the fan does work on the air present in the container. At the same time, the kinetic energy of the fan turns into energy in the air. Electrical energy in the battery, of course, reduced because it has changed shape into energy in the air. This example shows that in the isochoric process (volume is always constant), work can still be done on the system (work that does not involve volume changes).

**Isobaric Process (constant pressure)**

In an isobaric process, system pressure is kept constant. Since pressure is constant, internal energy change (ΔU), heat (Q) and work (W) in the isobaric process are not zero. Thus, first law thermodynamics formula has not changed.

ΔU = Q − W

Changes in pressure and gas volume in the isobaric process are illustrated by the graph below:

First, system volume = V_{1} (small volume). Because pressure is kept constant so after heat is added to the system, the system expands and does work on the environment. After working on environment, system volume changes to V_{2} (system volume increases). Amount of work (W) done system = shaded area.

Question 1 :

Curves 1-2 in two diagrams below show gas expansion (gas volume increase) that occurs adiabatically and isothermally. In which process, work performed by gas is smaller?

Work performed by gases in the adiabatic process is less than work by gas in the isothermal process. Shaded area = work done by the gas during expansion process (increase in gas volume). The shaded area in the adiabatic process is less than the shaded area of the isothermal process.

Question 2 :

Thermodynamic processes are shown in the diagram below. Curves a-b and d-c = isochoric processes (constant volume). Curve b-c and a-d = isobaric process (constant pressure). In a-b process, heat (Q) 600 Joules is added to the system. In the b-c process, heat (Q) 800 Joule is added to the system.

a) Energy changes in a-b process

b) Energy changes in a-b-c process

c) Total heat added to the a-d-c process

P_{1 }= 2 x 10^{5} Pa = 2 x 10^{5} N / m^{2}

P_{2 }= 4 x 10^{5} Pa = 4 x 10^{5} N / m^{2}

V_{1} = 2 liter = 2 dm^{3 }= 2 x 10^{-3 }m^{3}

V_{2 }= 4 liters = 2 dm^{3} = 4 x 10^{-3} m^{3}

Solution

a) Internal energy changes in the a-b process

In a-b process, 600 J heat is added to the system. Process a-b = isochoric process (constant volume). In the isochoric process, the addition of heat in the system only increases energy in the system. Thus, energy changes in the system after receiving heat:

ΔU = Q

ΔU = 600 J

b) Energy changes in a-b-c process

Process a-b = isochoric process (constant volume). In a-b process, 600 J heat is added to the system. Due to constant volume, there is no work done by the system.

Process b-c = isobaric process (constant pressure). In b-c process, heat (Q) 800 Joules is added to system. In isobaric process, system can do work. The amount of work done by system in b-c process (isobaric process) is:

W = P (V_{2} – V_{1}) — constant pressure

W = P_{2} (V_{2 }– V_{1})

W = 4 x 10^{5} N/m^{2} (4 x 10^{-3} m^{3} – 2 x 10^{-3} m^{3})

W = 4 x 10^{5} N/m^{2} (2 x 10^{-3} m^{3})

W = 8 x 10^{2 }Joule

W = 800 Joule

Total heat added to the system in the a-b-c process is:

Q total = Q ab + Q bc

Q total = 600 J + 800 J

Q total = 1400 Joules

The total work performed by the system in the a-b-c process is:

W total = W ab + W bc

W total = 0 + W bc

W total = 0 + 800 Joule

W total = 800 Joule

Internal energy changes in process a-b-c = 600 J

c) Total heat added to a-d-c process

Total heat added to system can be calculated by equation below:

ΔU = Q – W

Q = ΔU + W

Total heat added to process a-d-c = internal energy changes in a-d-c + work in a-d-c process.

Heat and work are involved in energy transfer between systems and environment, while internal energy changes are the result of energy transfer between system and environment. Therefore a change in energy does not depend on the process of energy transfer. On the other hand, heat and work depend on the process. In the isochoric process (constant system volume), energy transfer is only in the form of heat, while work is not. In the isobaric process (constant pressure), energy transfer involves heat and work. Although not dependent on the process, energy changes in depending on initial state and state of the end system. If the initial and final states are the same, changes in energy are always the same, although the process is different.

Initial and final states for the a-b-c process in graph above = initial and final state of the a-d-c process. Thus, inner energy changes in the process of a-d-c = 600 J.

Total (W) work performed on a-d-c = W process on a-d + W on d-c process.

a-d process is an isobaric process (constant pressure), while the d-c process is an isochoric process (constant volume). Due to constant volume, there is no work done on the d-c process. First, we calculate work done on a-d process.

W ad = P (V_{2 }‐ V_{1})

W ad = P_{1} (V_{2} ‐ V_{1})

W ad = 2 x 10^{5} N/m^{2} (4 x 10^{‐3} m^{3} ‐ 2 x 10^{‐3} m^{3})

W ad = 2 x 10^{5} N/m^{2} (2 x 10^{‐3} m^{3})

W ad = 4 x 10^{2} Joule

W ad = 400 Joule

W total = W in proses a‐d + W in proses d‐c

W total = 400 Joule + 0

W total = 400 Joule

Thus, the amount of heat added to the a-d-c process is:

Q = ΔU + W

Q = 600 J + 400 J

Q = 1000 J

Question 3 :

1 liter of water turns to 1671 liters of steam when boiled at 1 atm pressure. Calculate change of internal energy and work done by water when evaporates … (Heat of vaporization for water = L_{V} = 22.6 x 10^{5} J/Kg)

Solution

Water density = 1000 kg/m^{3}

Heat of vaporization for water (L_{V}) = 22.6 x 10^{5 }J / Kg

P = 1 atm = 1.013 x 10^{5 }Pa = 1.013 x 10^{5} N/m^{2}

V_{1 }= 1 liter = 1 dm^{3} = 1 x 10^{-3} m^{3} (Water volume)

V_{2} = 1671 liters = 1671 dm^{3} = 1671 x 10^{-3} m^{3} (Volume of steam)

a) Changes in internal energy

Changes in internal energy = Heat added to water minus Work done by water when evaporated. First count Heat (Q) added to water …

Q = m L_{V}

Water density = water mass / volume of water

Water mass (m) = water density x water volume

Water mass (m) = (1000 kg/m^{3}) (1 x 10^{-3} m^{3})

Water mass (m) = (1000 kg/m^{3}) (0.001 m^{3})

Water mass (m) = 1 Kg

Q = (1 kg) (22.6 x 10^{5 }J / kg)

Q = 22.6 x 10^{5} J

Calculate Work (W) done by water when evaporate. Boiling of water occurs at constant pressure (isobaric process).

W = p (V_{2} – V_{1})

W = 1.013 x 10^{5} N/m^{2 }(1671 x 10^{-3} m^{3} – 1 x 10^{-3} m^{3})

W = 1.013 x 10^{5} N/m^{2} (1670 x 10^{-3} m^{3})

W = 1691.71 x 10^{2} Joule

W = 1.7 x 10^{5} Joules

Changes in the internal energy of water:

ΔU = Q – W

ΔU = 22.6 x 10^{5} J – 1.7 x 10^{5 }J

ΔU = 20.9 x 10^{5 }J

ΔU = 21 x 10^{5} J

21 x 10^{5 }Joule heat added to water is used to increase internal energy (overcoming tensile forces between molecules that keep water fluid). In other words, 21 x 10^{5} J is used to convert water into steam. When water has become steam, the remaining 1.7 x 105 J is used to do work.

Question 4 :

1 mole of gas in a cylinder expands rapidly adiabatically. At first, the gas temperature = 1000 K. After expanding, the gas temperature decreases to 500 K. Determine the work done by the gas …

Solution

Gas expansion occurs adiabatically. In the adiabatic process, no heat enters or exits the system. Thus, work performed gas = energy changes in gas.

ΔU = Q – W or W = Q – Δ U → Q = 0

W = 0 – ΔU

W = – ΔU

We can calculate the energy changes in the gas using the energy equation in the ideal gas:

ΔU = 3/2 (1 mol) (8.315 J / mol.K) (500 K – 1000 K)

ΔU = 3/2 (1 mol) (8.315 J / mol.K) (- 500 K)

ΔU = -6236.25 J

Thus, amount of work performed by gas is:

W = – ΔU

W = – (- 6236.25 J)

W = 6236.25 J