# Isothermal process and first Law

In an isothermal process, $$dQ=0$$, the temperature remains constant, but if the system does work, then $$\text{d}U=-dW$$, but if the internal energy reduces due to work done, doesn't that reduce the temperature of the system?

In addition, can anyone give examples for the formula $$\text{d}U=dQ-dW$$ using real life scenarios?

• Who says that, if dQ=0, the temperature remains constant? Commented Jul 25, 2021 at 13:48

## 2 Answers

It's already been pointed out that $$dQ=0$$ is for an adiabatic process and $$dT=0$$ is for an isothermal process and therefore $$dU=-dW$$ is the differential change in internal energy for an adiabatic process.

but if the internal energy reduces due to work done, doesn't that reduce the temperature of the system?

In the case of an adiabatic expansion ($$dQ=0$$) the work done is at the expense of the internal energy of the system ($$dU=-dW$$) and therefore will necessarily result in some reduction in the temperature of the system.

The amount of reduction depends on the nature of the system since, in general, a system's internal energy is comprised of both microscopic kinetic energy (which is related to temperature) and microscopic potential energy (which is related to intermolecular forces). In the case of an ideal gas, all its internal energy is considered kinetic and thus all of the decrease in internal energy results in a decrease in temperature.

In addition, can anyone give examples for the formula $$\text{d}U=dQ-dW$$ using real life scenarios?

The first law is basically a statement of the law of conservation of energy. The following contains daily life examples of the application of conservation of energy:

https://lawofthermodynamicsinfo.com/examples-of-first-law-of-thermodynamics/

Hope this helps.

In an isothermal process $$dT=0$$, it's an adiabatic process where $$dQ=0$$.

In the adiabatic process it's true that $$dU = -dW$$ and the temperature could reduce.

An example would be a gas under high pressure in a cylinder. If it could quickly expand and do work against the outside pressure, $$W=PdV$$, then the temperature of the gas would reduce (if it's quick there is no time for heat energy to enter or leave the system).

An aerosol deodorant spray does this, the spray comes out cold.