Isochoric Process

Isochoric Process in thermodynamics is referred to as a process in which the volume remains constant and the work done in this case is zero. This process is also called the Isometric Process, or Constant-Volume Process. The volume of gas in this process always remains constant and the work done in this thermodynamic process is zero.

In this article, we will learn about, the Isochoric Process Definition, its Formula, Examples, and others in detail.

What is an Isochoric Process

Isochoric process is also known as a constant-volume process. It is a thermodynamic process where the total volume of the system remains constant and so the work done in this process is zero.

An example of an isochoric process is when air is heated or called in a closed container its volume always remains constant.

Isochoric Process Formula

Isochoric process is defined as the thermodynamic proess in which the volume of the system remained constant so the the change in volume is zero and hence the work done by the gas is zero. Mathematically Isochoric Process is expressed as

For Isochoric Process,

Change in Volume(△V) = 0

Now, work done by the gas is,

W = P.△V = P.0 = 0

Thus, the work done by the gas in the Isochoric process is zero.

W = 0

Isochoric Process Characteristics

• It takes place at constant volume, also known as an isometric process.
• In an isochoric process, the work done is zero, as the volume of the system remains constant.
• This means that the change in volume ( ΔV ) is equal to zero, and as a result, the work done ( W ) equals zero.
• On a pressure-volume (P-V) diagram, an isochoric process appears as a straight vertical line.

Examples of Isochoric Process

Some important examples of isochoric processes include:

Otto Cycle: This thermodynamic cycle is applicable in car engines and represents heat transfer during the ignition process. In an Otto cycle the first and the last process is isochoric in nature, where the heat and pressure change without any change in the volume of the gas.

Heating of Water in a Tightly Sealed Flask: If we heat the water inside a closed flask then the volume of the water remains constant even though its temperature increases and its pressure also increases. This is the example of Isochoric process.

Pressure Cooker: When sealed closed, the volume inside cannot change, so when heat is added, pressure and temperature increase rapidly. However, pressure cookers expand slightly, and some gas is released from a valve on top.

Work Done and Heat Transfer in Isochoric Process

Since Volume of the system remains constant in an Isochoric process. This means that the system does no work, as work is defined as force times distance, and there is no change in distance (since the volume is constant). However, heat can still be transferred into or out of the system, which will cause a change in internal energy. The change in internal energy is equal to the quantity of heat transferred.

W = 0 (In case of Isochoric Process)

Now, the equation for the heat transferred in an isochoric process is:

Q = ΔU + W

W = 0

Q = ΔU

Mathematical Representation of Isochoric Process

The isochoric process can be represented mathematically by the following equation:

ΔU = Q

Equation is based on the first law of thermodynamics, which states that the total energy of an isolated system is constant. In an isochoric process, the total energy of the system is equal to the internal energy of the system, so the change in internal energy is equal to the heat added to the system.

The following equation can also represent the isochoric process:

C_vΔT = Q

where,

• C_v is Molar Heat Capacity at a Constant Volume
• ΔT is Change in Temperature of System

First Law of Thermodynamics in Isochoric Process

According to the first law of thermodynamics the total energy of an isolated system remains constant. This means that the energy within the system can only be transferred or transformed, not created or destroyed. In an isochoric process, the only energy transfer that occurs is heat transfer. Therefore, the change in the system's internal energy is equal to the amount of heat transferred into or out of the system.

The equation for the first law of thermodynamics for an isochoric process is:

ΔU = Q

Since the work done by the system is zero, the equation can also be written as:

ΔU = Q + W

Plugging in W = 0, we get the equation for the first law of thermodynamics for an isochoric process:

ΔU = Q

This equation means that the change in internal energy of the system is equal to the amount of heat transferred into or out of the system. If the system is heated, the internal energy increases and Q is positive. If the system is cooled, the internal energy decreases, and Q is negative.

Also Check, Second Law of Thermodynamics

Isochoric Process in Ideal Gases

For an ideal gas, the relationship between pressure (P), temperature (T), and volume (V) is given by the ideal gas law:

PV = nRT

Since V is constant for an isochoric process, we can rearrange the ideal gas law to get:

P/T = V = constant

This means that the pressure and temperature of an ideal gas are directly proportional to each other in an isochoric process. In other words, if the temperature of an ideal gas increases, the pressure will also increase, and vice versa

PV Diagrams of Isochoric Process

A vertical line represents an isochoric process on a PV diagram. This is because the volume of the system does not change, so the pressure can only change vertically. This is shown in the image added below as the PV diagram of Isochoric process,

👁 Isochoric-Process- PV Diagram

Isochoric processes are often used to model the heating or cooling of a gas in a sealed container. For example, when a gas is heated in a closed container, the pressure of the gas will increase. This is because the gas molecules move faster and hit the walls of the container more often.

Applications of the Isochoric Process

Here are some real-world examples of isochoric processes:

• Heating or Cooling a Confined gas: A common example of an isochoric process is heating or cooling a gas in a sealed container. For instance, consider a balloon filled with air and placed in a refrigerator. As the air cools, its internal energy decreases, and its pressure and temperature drop. However, the volume of the balloon remains constant due to its rigid container.

• Heating a Material in a Bomb Calorimeter: A bomb calorimeter is a device used to measure a substance's combustion heat. The substance is placed within a sealed chamber, and its combustion is initiated. The heat released during the combustion causes the temperature of the substance and the surrounding water to rise, but the volume remains constant.

• Gas Compression in a Piston-Cylinder Device with a Locked Piston: In a piston-cylinder device, the piston can move up and down, changing the volume of the enclosed gas. However, if the piston is locked in place, the volume remains constant, and any gas compression or expansion would be an isochoric process.

• Isometric Exercise: It involves contracting muscles against an immovable object, such as pushing against a wall. During these exercises, the muscles exert tension, but since there is no movement, no work is done, and the process is considered isochoric.

• Gas Storage in Tanks: Gas storage tanks, such as those used for storing propane or compressed air, typically maintain a constant volume. When gas is added or removed from the tank, the pressure fluctuates, but the volume remains constant, making it an isochoric process.

In addition to these examples, isochoric processes play a role in various thermodynamic cycles, such as the Otto and Diesel cycles, commonly used in internal combustion engines. Understanding isochoric processes is essential for analyzing the efficiency and performance of these engines.

Read More,

• Thermodynamic Processes
• Basic Concepts of Thermodynamics

Isochoric Process JEE Questions

Q1. What is true regarding the work performed by a system whose volume remains constant and undergoes an increase in pressure?

The system does no work. The answer to this question, is given using formula, W = −PΔV

Q2. Which is Correct?

1. In an Isobaric Process, ∆P = 0
2. In an Isochoric Process, ∆W = 0
3. In an Isothermal Process, ∆T = 0
4. In an Isothermal Process, ∆Q = 0

Option (4) In an Isothermal Process, ∆Q = 0 is Correct

Q3. In which thermodynamic process, volume remains same

1. Isobaric
2. Isothermal
3. Adiabatic
4. Isochoric

Option (4) Isochoric is Correct

Q4.In a reversible Isochoric change

1. ∆W = 0
2. ∆Q = 0
3. ∆T = 0
4. ∆U = 0

Option (1) ∆W = 0 is Correct

Q5.In an Isochoric process if T₁ = 27° C and T₂ = 127° C, then P₁/P₂ is equal to

1. 9/59
2. 2/3
3. 3/4
4. 7/57

Option (3) 3/4 is Correct

At Constant Volume P ~ T,

P₁/P₂ = T₁/T₂

P₁/P₂ = 300/400 = 3/4

Isochoric Process Numericals

Numericals on Isochoric Process are added below,

Example 1: 500 g of water is heated from 30°C to 60°C. Determine the water's change in internal energy, ignoring its small expansion? (specific heat of water 4184 J/kg.K)

Solution:

• Water's volume barely changes when it heats up from 30°C to 60°C.
• Work done by the system is zero in an Isochoric process.
• Heat supplied to it is used to increase only the internal energy.

∆U = Q = msv ∆T

Given,

• Mass of Water = 500 g = 0.5 kg
• Change in Temperature = (60 - 30) = 30 K
• s = 4184 J/kg.K

∆U = Q = 0.5×4184×30 = 62.76 kJ

Thus, change in internal energy is 62.76 kJ

Example 2: A gas in a 10 L volume cylinder is subjected to a pressure of 1 atm. The gas experiences a rise in temperature from 34 °C to 60 °C in an isochoric process. If molar specific heat C_v = 2.5 × R where R = 8.31 J/mol K. Calculate the change in Internal energy.

Solution:

Given,

• P = 1 atm = 101325 Pa
• V = 10 l = 0.01 m³
• ΔT = 60 - 34 = 26 °C = 26 K
• C_v = 2.5 × R = 2.5 × 8.31 = 20.775

Q_v = nC_v ΔT

Using Ideal Gas Equation

PV = nRT

n = PV/RT

n = ( 101325).(0.01)/(8.31).(26)

n= 0.39 moles

Therefore,

ΔU = Q_v = nC_v ΔT

ΔU = 0.39 x 2.5x 8.31x 26

ΔU = 210.65 J

Example 3: In a sealed container, 0.1 moles of monoatomic gas are kept at 27°C. The container is then heated to 500K. Calculate the change in internal energy of gas. (R = 8.315 J/mol.K)

Solution:

Given,

• n = 0.1 mol
• T_i = 27°C + 273 = 300 K
• T_f = 500 K
• R = 8.315 J/mol.K

For an isochoric process, ΔV = 0. Therefore work done is zeo.

From Frst law of Thermodynamics,

ΔU = ΔQ

For Monoatomic Gas,

ΔU = 3/2nRΔT

ΔU = 3/2nR(T_f – T_i)

ΔU = 3/2 (0.1 mol)(8.315 J/mol.K)(500K – 300K)

ΔU = 3/2 (0.1 mol)( 8.315 J/mol.K)(200K)

ΔU = 249.45 J

ΔU = 0.39 × 2.5 × 8.31 × 26

ΔU = 210.65 J

Example 4: 2 kg of water is heated from 50°C to 80°C in a closed vessel. What will be the change in the internal energy of the water? (specific heat of water = 4184 J/kg.K)

Solution:

Given,

• Mass of Water = 2 kg
• T_i = 50°C = 323K
• Tf = 80°C = 353K
• c_v = 4184 J/kg.K

In an isochoric process, work done is 0. Hence the internal energy is only dependent on the heat absorbed. Thus,

ΔU = ΔQ = mcvΔT

ΔU = 2 kg × 4184 J/kg.K × (353K – 323K)

ΔU = 2 kg × 4184 J/kg.K × 30K

ΔU = 251.04 kJ