# Isochoric process: formulas and calculations, daily examples

An **isochoric process** is any thermodynamic process in which the volume remains constant. These processes are also called isometric or isovolumetric. In general, a thermodynamic process can occur at constant pressure and is called an isobaric process.

When it takes place at a constant temperature, then it is considered an isothermal process. If there is no heat exchange between the system and the environment, then it is called adiabatic. On the other hand, when there is a constant volume, the generated process is called isochoric.

In the case of the isochoric process, it can be said that in these processes the pressure-volume work is zero, as this results from the multiplication of pressure by the increase in volume.

Furthermore, in a pressure-volume thermodynamic diagram, isochoric processes are represented in the form of a straight vertical line.

__Formulas and Calculation__

__Formulas and Calculation__

**The first principle of thermodynamics**

In thermodynamics, work is calculated from the following expression:

W = P ∆ ∆ V

In this expression, W is the work measured in Joules, P is the pressure measured in Newton per square meter and ∆ V is the change or increase in volume measured in cubic meters.

Likewise, what is known as the first principle of thermodynamics states that:

∆ U = Q – W

In the aforementioned formula W is the work performed by the system or in the system, Q is the heat received or emitted by the system and *U* is the variation of the system’s internal energy. On this occasion, the three magnitudes are measured in Joules.

As in an isochoric process the work is null, it happens that:

∆ U = Q _{V} (since, ∆ V = 0 and therefore W = 0)

In other words, the system’s internal energy variation is solely due to the heat exchange between the system and the environment. In this case, the heat transferred is called the constant volume of heat.

The heat capacity of a body or system results from the division of the amount of energy in the form of heat transferred to a body or system in a given process and the change in temperature experienced by that body.

When the process is carried out at a constant volume, it refers to a heat capacity at a constant volume and is denoted by C _{v} (molar heat capacity).

In this case, it will be fulfilled:

Q _{v} = n ∙ C _{v} ∙ ∆T

In this situation, n is the number of moles, C _{v} is the aforementioned molar heat capacity at constant volume, and ∆T is the temperature increase experienced by the body or system.

__everyday examples__

__everyday examples__

It’s easy to imagine an isochoric process, just think of a process that takes place in constant volume; that is, in which the container containing the material or material system does not change the volume.

An example might be the case of an (ideal) gas closed in a closed container whose volume cannot be changed by any means to which heat is supplied. Suppose the case of a gas in a bottle.

When transferring heat to the gas, as explained above, it will eventually result in an increase or increase in its internal energy.

The reverse process would be that of a gas closed in a container whose volume cannot be changed. If the gas cools and heats the environment, the pressure of the gas will be reduced and the internal value of energy of the gas will decrease.

**The ideal Otto cycle**

The Otto cycle is an ideal case of the cycle used by gasoline machines. However, its initial use was in machines that used natural gas or other fuels in a gaseous state.

Anyway, the ideal Otto cycle is an interesting example of an isochoric process. It occurs when combustion of the mixture of gasoline and air takes place instantly in an internal combustion car.

In this case, there is an increase in the temperature and pressure of the gas inside the cylinder, the volume remains constant.

**Practical examples**

**first example**

Given an (ideal) gas enclosed in a cylinder with a piston, indicate whether the following cases are examples of isochoric processes.

– 500 J work is performed on the gas.

In this case, it would not be an isochoric process, since to carry out work with gas it is necessary to compress it and, therefore, change its volume.

– Gas expands by moving the piston horizontally.

Again, this would not be an isochoric process, as the expansion of gas implies a change in its volume.

– The cylinder piston is fixed so that it cannot move and the gas cools down.

This time, it would be an isochoric process, since there would be no variation in volume.

**second example**

Determine the variation of internal energy that a gas contained in a vessel with a volume of 10 L under a pressure of 1 atm will experience, if its temperature rises from 34 °C to 60 °C in an isochoric process, known for its specific molar heat *C _{v}* = 2.5 ·

*R*(where

*R*= 8.31 J / mol · K).

As it is a constant volume process, the internal variation of energy will only occur as a result of the heat supplied to the gas. This is determined with the following formula:

Q _{v} = n ∙ C _{v} ∙ ∆T

To calculate the heat supplied, it is first necessary to calculate the moles of gas contained in the container. For this, it is necessary to resort to the ideal gas equation:

P ∙ V = n ∙ R ∙ T

In this equation n is the number of moles, R is a constant whose value is 8.31 J / mol · K, T is the temperature, P is the pressure to which the gas measured in the atmosphere is subjected and T is the temperature measured in Kelvin.

It’s clean and you get:

n = R ∙ T / (P ∙ V) = 0.39 moles

So that:

∆ U = Q _{V} = n ∙ C _{v} ∆T = 0.39 ∙ 2.5 ∙ 8.31 ∙ 26 = 210.65 J