# Isothermal process?

In an isothermal process, does the temperature of an ideal gas remain constant?
And if so does the heat capacity at constant volume = zero?

temperature remains constant. Why do you think heat capacity is 0 at constant volume?

Heat capacity is the ability of a system to store heat as the temperature changes. Ah... I think I see what you mean... Just because a systems temperature doesnt change, doesn't mean it doesnt have a heat capacity right? So I need to calculate the heat capacity at constant volume?

Andrew Mason
Homework Helper
Heat capacity is the ability of a system to store heat as the temperature changes. Ah... I think I see what you mean... Just because a systems temperature doesnt change, doesn't mean it doesnt have a heat capacity right? So I need to calculate the heat capacity at constant volume?
Unfortunately, thermodynamics was developed before everyone really understood what heat was. Heat, in thermodynamics, is really heat flow so the concept of storing "heat" doesn't really work.

A system can store or release energy as a change in internal energy or it can perform work or have work done on it. The sum of those has to be the net heat flowing into/from the system. The relationship between the temperature of a system and the heat flow into/from the system determines the heat capacity.

Now, if the process is isothermal what does that tell you about the change in internal energy of the system? What does that tell you, then, about the relationship between heat flow and work? If work is done, can the volume remain constant? So if there is heat flow, can the volume remain constant if temperature is constant?

AM

If a system is isothermal it tells me that there is no change in temperature and therefore there is no change in the internal energy of the system.

I know that a change in internal energy = heat transfer + work. So if heat flow is applied to the system, then under normal conditions, the temperature would rise i.e, it would have greater internal energy. To prevent this, the volume would have to increase… Is this correct?

Andrew Mason