Is Internal Energy Always Dependent Only on Temperature in an Ideal Gas?

[AznBoi]

Does the internal energy always depend only on temperature in an ideal gas? Would the equation $$U=\frac{3}{2}nRT$$ tell us that? are there any other equations?

I'm also confused with the area within a cyclic curve on a PV graph. Does the area within the curve represent the work done ON the system/gas?

My text says, for heat engines, that: "The work done by the engine for a cyclic process is the area enclosed by the curve representing the process on a PV diagram" --which would equal the work done by the system/gass.

I know that the W(engine)= -W(system) because the system is doing work on the engine. So... I'm kind of confused here, my guess would be that the signs would change?

 

[Hootenanny]

Does the internal energy always depend only on temperature in an ideal gas? Would the equation $$U=\frac{3}{2}nRT$$ tell us that? are there any other equations?

Internal energy is a function of all the state variables, and therefore, can be represented by any or a combination of the state variables (pressure, temperature, volume, entropy), but temperature is usually the simplest

I'm also confused with the area within a cyclic curve on a PV graph. Does the area within the curve represent the work done ON the system/gas?

My text says, for heat engines, that: "The work done by the engine for a cyclic process is the area enclosed by the curve representing the process on a PV diagram" --which would equal the work done by the system/gass.

I know that the W(engine)= -W(system) because the system is doing work on the engine. So... I'm kind of confused here, my guess would be that the signs would change?

The work done by the system is the area enclosed by the curve;
$$W = \int^{f}_{i}\vec{P}\cdot d\vec{V}$$
The work done on the system is the negative area enclosed by the curve;
$$W = -\int^{f}_{i}\vec{P}\cdot d\vec{V}$$

 

[AznBoi]

Okay I get it now =]

 

[AznBoi]

What the difference between: 1) Compute the maximum possible efficiency of a heat engine operating between two given temperatures. 2) Comput the actualy efficiency of a heat engine.

I've only learned this equation so far: $$e= 1- \frac{W_{eng}}{Q_{h}}$$

Also, what are the easiest ways of calculating temperature?? Do you need to use PV=nRT or are there other equations? Thanks.

 

[AznBoi]

Another question: So if Cv=3/2R, then Cp=5/2R right?

So can you express heat as:$$Q=nC_{p}\Delta T=5/2nRT$$ ??

So Q=3/2nRT when the process is isochoric/isovolumetric and Q=5/2nRT when the process is isobaric, or when the pressure is constant?

 

[tim_lou]

the concept of heat is ambiguous, and you cannot have a "function" of heat unless the type of process is specified (I assumed you mean heat flow). For your question, i suppose you mean the change in internal energy.

at constant pressure, yes
$$\Delta U=C_p \Delta T$$

and for ideal gas,
$$C_p=C_v+nR$$

for isochoric process, the change in internal energy=heat flow.

but one should never talk about a function for "heat". it simply doesn't make sense. Heat is a just a quantity that is added to force the conservation of energy.

for example, suppose I push my car on the road, from mechanics, I know that the work done by me must go into the kinetic energy of the car (let me denote it by U)
$$\Delta U=\Delta W$$
but when I take friction into account, that equation simply isn't true. So people just at a Q there to force it to be true, ie.
$$\Delta U=\Delta W +Q$$

suppose after my hard labor, the car still doesn't move. What is the "heat" function?

I can say I wasn't pushing really hard, so Q is quite small. Or I can say I burnt all my body fat trying to push that car, then Q is really big. At the end of the day, Q really depends on W itself.

 

[Hootenanny]

The [ideal gas] Carnot engine is the theoretic limit on the effiency of a heat engine. The efficiency of the Carnot engine operating between to temperatures is;

$$\xi_C = 1-\frac{T_H}{T_C}$$

As for what is the easiest way of calculating temperature, that depends on the data your given. As for your other questions, I'm afraid I haven't got time to take a proper look at them now, but I'm sure someone else will step upto the plate.

 

[AznBoi]

Thanks for the help guys. I'm on the Carnot engine right now and it all seems confusing to me. Should I just skip this section? All I know is that it is the most efficient engine and that the processes are: 1.isothermal,2.adiabatic,3.isothermal,4.adiabatic. Is there anything else important that I need to know?

I'm really confused and don't know how to find the temperature in graphs. I mean a PV diagram only shows pressure and volume right? How would you know if the curve shows if the temperature increased or decreased or stayed the same?? Same for heat. I really appreciate your help! =]

 

[tim_lou]

also, when you try to calculate the efficient of a real engine, you need to have a PV diagram. You basically have to understand the work done, and the energy that flows in the system for various processes. There is no easy way out.

The ideal efficiency is the maximum that can be achieved by a reversible engine. Real engines have a lower efficiency since otherwise, the real engine can be attached to the idea engine and energy would flow from a hot source to a cold course, which doesn't make sense.

 

[tim_lou]

as for temperature:

for most engines, the working substance is an ideal gas (make sure that it is first), so
PV=nRT,

in other words, the curve with constant temperature is T=const x*y in the PV diagram.
that curve looks like a hyperbola (actually, it is a hyperbola). so you can sort of see which hyperbola a point in the PV diagram belongs to, and the higher up the hyperbola, the higher the temperature.

 

[AznBoi]

Thanks Tim! What is this objective asking?:
Relate the heats exchanged at each thermal reservoir in a Carnot cycle to the temperatures of the reservoirs.

Do you use the first law of thermo to relate heat and temp?

 

[Andrew Mason]

Another question: So if Cv=3/2R, then Cp=5/2R right?

So can you express heat as:$$Q=nC_{p}\Delta T=5/2nRT$$ ??

Only if P is constant. If V is constant, then:

$$\Delta Q = nC_v\Delta T$$

If neither P nor V is constant, you must go back to the first law.

$$\Delta Q = \Delta U + W= nC_v\Delta T + PdV$$

AM