Isochoric Process: Understanding Work, Heat Transfer, and

## Example Problems

Do you ever wonder about the science behind the functioning of engines, refrigerators, and other devices that operate using gases or fluids? Thermodynamics is the branch of physics that deals with heat and work transfer, and it is essential to understand these concepts to comprehend the behavior of matter under varying conditions.

In this article, we will delve into one of the essential processes of thermodynamics- the “Isochoric Process.” We will provide a detailed explanation of the concept, discuss the work done, and explore the heat transfer mechanism involved. We will also solve example problems to demonstrate the practical application of the concept.

Isochoric Process: Definition and Examples

An isochoric process is a thermodynamic process that occurs under constant volume. This process is also referred to as an isometric process or an isovolumetric process.

The word “isochoric” comes from the Greek word “iso,” meaning “same,” and “khora,” meaning “space.”

During an isochoric process, the volume of the system remains constant, and the boundaries do not allow for any movement. Some example systems that experience this process include the Otto cycle used in four-stroke engines and refrigeration devices designed for ultra-low temperatures.

## Work Done in an Isochoric Process

Now that we have understood the definition of an isochoric process, let’s move on to the work done during this process. To calculate the work done, we use the following formula:

Work = -PV

Where,

P = Pressure

V = Change in volume

Here, as the volume is constant, V = 0; hence, the work done during an isochoric process becomes zero.

Hence, the energy added or removed from the system during this process shows up as a change in internal energy.

## Heat Transfer and Internal Energy during an Isochoric Process

The first law of thermodynamics states that energy is conserved. Thus, the change in internal energy (U) of the system during an isochoric process is a result of heat transfer (Q) and work done (W) by the system.

Therefore, we have:

U = Q – W

As discussed earlier, the work done is zero during an isochoric process; hence, the formula simplifies to:

U = Q

To calculate the amount of heat transferred during an isochoric process, we use the following equation:

Q = C_v T

Where,

C_v: Specific heat capacity at constant volume, which is the heat capacity at constant volume per unit mass

T: Change in temperature

The internal energy of an ideal gas, U, can be expressed using the ideal gas model as follows:

U = 3/2 nRT

Where,

n: number of moles of the gas

R: Gas constant

T: Temperature in Kelvin

We can obtain the change in enthalpy (H) by using the ideal gas model; in this case, H is equal to U + PV. However, the volume being constant, V = 0, which means the change in enthalpy will be equal to the change in internal energy.

## Example Problems

Problem 1: Determine the work done and change in internal energy of a monoatomic gas with a volume of 25m3, which undergoes a temperature change from 200K to 400K. The gas constant is 8.31J/K/mol.

## Solution:

## Given:

Volume, V = 25 m3

Temperature, T1 = 200K

Temperature, T2 = 400K

Gas constant, R = 8.31J/K/mol

The work done during an isochoric process is zero; hence W = 0. U = Q

Q = U = C_v T

Here, we need to find C_v, which can be determined using the following formula:

C_v = (3/2)R

## Therefore:

C_v = (3/2) x 8.31 J/K/mol = 12.5 J/K/mol

## Hence:

Q = U = C_v T = 12.5 J/K/mol x 200 K = 2500 J/mol

Problem 2: Determine the heat absorbed and change in internal energy of 4 kg water that undergoes an isochoric process with a temperature change from 298K to 333K.

The specific heat capacity of water is 4182J/kg K. Solution:

## Given:

Mass of water, m = 4 kg

Initial temperature, T1 = 298K

Final temperature, T2 = 333K

Specific heat capacity of water, C_v = 4182J/kg K

The work done during an isochoric process is zero; hence W = 0.

U = Q

Q = U = mC_v T

Here, we can use the above formula to determine the amount of heat absorbed. Therefore:

Q = U = mC_v T = 4 kg x 4182 J/kg K x (333K 298K) = 584184 J

## Conclusion:

In this article, we have covered the isochoric process, the work done, and the heat transfer mechanism at play.

The isochoric process is a crucial process in thermodynamics as it is vital in the functioning of numerous systems and devices. We have also discussed examples of practical applications and solved example problems to better understand the concept.

Understanding the isochoric process is essential in comprehending the principles that govern the behavior of matter around us. In summary, this article provides an in-depth explanation of the isochoric process, which is a crucial aspect of thermodynamics governing the behavior of matter under varying conditions.

We have discussed the work done, heat transfer, and example problems related to the topic, essential to understanding the science behind the functioning of various devices. A strong takeaway from this article is that a deep understanding of the isochoric process is necessary to comprehend the principles that govern the behavior of matter.

## FAQs:

1. What is an isochoric process?

An isochoric process is a thermodynamic process that occurs under constant volume, with the volume of the system remaining unchanged. 2.

What is the work done during an isochoric process? The work done during an isochoric process is zero since the volume remains constant.

3. How is heat transferred during an isochoric process?

Heat is transferred during an isochoric process by a change in internal energy since the work done is zero. 4.

What are some practical applications of the isochoric process? Some practical applications of the isochoric process include the Otto cycle used in four-stroke engines and refrigeration devices designed for ultra-low temperatures.

5. What is the ideal gas model that can be used for calculations related to isochoric processes?

The ideal gas model can be used to calculate the internal energy and change in enthalpy of the system, with U = (3/2) nRT and H = U + PV, where V = 0 for isochoric processes.