## Isothermal reversible of ideal gas

isochoric/isometric: $\Delta V = 0$
isothermal: $\Delta T = 0$
isobaric: $\Delta P = 0$

aphung1E
Posts: 101
Joined: Sat Aug 24, 2019 12:15 am

### Isothermal reversible of ideal gas

Can someone tell me what values such as heat, pressure, etc are equals to zero in an ideal gas of an isothermal reversible system?

Ashley Tran 2I
Posts: 108
Joined: Thu Jul 11, 2019 12:17 am

### Re: Isothermal reversible of ideal gas

Isothermal reversible just means the temp stayed constant so q=-w and you can use the equation nrlnv2/v1 to calculate entropy of the system.

Ally Huang- 1F
Posts: 103
Joined: Thu Jul 25, 2019 12:16 am

### Re: Isothermal reversible of ideal gas

Delta U is equal to zero for an isothermal reversible system. The same amount of heat enters the system as the work that leaves the system so the temperature stays constant and delta U will also equal zero based on DeltaU=q+w.

Pegah Nasseri 1K
Posts: 100
Joined: Wed Feb 27, 2019 12:15 am

### Re: Isothermal reversible of ideal gas

An ideal gas in an isothermal, reversible reaction will have an internal energy of 0. This means that the heat will equal the negative energy of the work done:

Delta U = q + w = 0
This indicates that: q = -w

In addition, the total change in entropy will be zero:
$\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings}$=0
This indicates that: $\Delta S_{system} = - \Delta S_{surroundings}$

Use the equation $\Delta S = nRln(V_{2}/V_{1})$ to find the change in entropy of the system.
Because the initial and final states of the system are the same since the reaction is returning to its initial state, both the internal energy and total change in entropy is zero.

Helen Struble 2F
Posts: 97
Joined: Sat Aug 24, 2019 12:17 am

### Re: Isothermal reversible of ideal gas

Delta U will equal 0, because q = -w. This simply means that any energy transferred from system to surroundings as work will immediately be replaced by a flow of heat from surroundings to system (or vice versa, depending on if it's an expansion or compression). Because there is no net change in energy, there is no temperature change, making it isothermal. Pressure and volume both change variably depending on the problem.