Understanding Quasi-Static Process: Work and Temperature Changes

[spaghetti3451]

Consider a quasi-static expansion of a gas. If you change the external force by dF_ext, then the system will do work on the surroundings until the internal pressure equals the external pressure, right?

Now, how does the temperature of the system and the surroundings chnage in the process?

Thanks for any help

 

[Andrew Mason]

Consider a quasi-static expansion of a gas. If you change the external force by dF_ext, then the system will do work on the surroundings until the internal pressure equals the external pressure, right?

Now, how does the temperature of the system and the surroundings chnage in the process?

This is an adiabatic expansion. So the adiabatic condition applies. If it is an ideal gas, then:

P_fV_f^\gamma = P_iV_i^\gamma and

T_fV_f^{\gamma - 1} =T_iV_i^{\gamma - 1}

where \gamma is the ratio of specific heats: Cp/Cv

As far as the surroundings are concerned, it depends on the surroundings. Work is done on the surroundings. That may or may not change the temperature of the surroundings. For example, it might lift a weight in which case no temperature change occurrs. Or it may run a heating coil in an insulated container, in which case T increases.

AM

 

[spaghetti3451]

This is an adiabatic expansion. So the adiabatic condition applies. If it is an ideal gas, then:

P_fV_f^\gamma = P_iV_i^\gamma and

T_fV_f^{\gamma - 1} =T_iV_i^{\gamma - 1}

where \gamma is the ratio of specific heats: Cp/Cv

I see! So, if the pressure increases, the temperature increases and vice-versa.

But that's for an ideal gas only. What happens in the most general case? Is there any way to predict?

Also, is there a general adiabatic condition?

As far as the surroundings are concerned, it depends on the surroundings. Work is done on the surroundings. That may or may not change the temperature of the surroundings. For example, it might lift a weight in which case no temperature change occurrs. Or it may run a heating coil in an insulated container, in which case T increases.

I see! So you are saying that whether the temperature changes depends on the way the energy is used in the surroundings. But what if the energy is simply stored?

 

[Andrew Mason]

I see! So, if the pressure increases, the temperature increases and vice-versa.

But that's for an ideal gas only. What happens in the most general case? Is there any way to predict?

It depends on the equation of state of the gas. It will be close.

I see! So you are saying that whether the temperature changes depends on the way the energy is used in the surroundings. But what if the energy is simply stored?

If the work output is stored (say by lifting a weight) then would there be heat flow to the surroundings?

AM