Solving Isothermal Problem: Pressure vs. Volume

[sahil_time]

Isothermal problem!

Imagine an idead cylinder and piston and gas filled in it with the Pgas=Patm.
Ideal gas Equation for initial state is PV=nRT for the gas....(1)
Now is dQ heat is imparted to the gas. Pressure of the gas at that instant increases by dP so it expands by dV to attain Mechanical Equilibrium with the surrounding!
Now the Ideal gas equation for final state is P(V+dV)=nRT...(2)
Since Initial and Final P are equal and Initial and Final T are equal
BUT eqn 1 is not equal to eqn 2. How?

 

[Curl]

The pressure cannot be the same as the initial pressure Patm if you have "isothermal" heat addition. It's impossible.

 

[cjl]

Why do you say that initial and final T are equal? If you have constant pressure heat addition, with expanding volume, the final temperature will be higher than the initial temperature.

 

[sahil_time]

What i meant was if dQ heat is imparted for just once and not continuously then what will be the final state of the isothermal system?

 

[Curl]

Pressure must decrease if you want to expand it in a way such that T is constant.

 

[cjl]

What i meant was if dQ heat is imparted for just once and not continuously then what will be the final state of the isothermal system?

If you impart heat, and the system is allowed to maintain mechanical (pressure) equilibrium with its surroundings, then the final state will be larger in volume, and higher in temperature than the initial state.

It is not possible to have a system maintain both constant pressure and constant temperature while adding heat.

 

[sahil_time]

If you impart heat, and the system is allowed to maintain mechanical (pressure) equilibrium with its surroundings, then the final state will be larger in volume, and higher in temperature than the initial state.

It is not possible to have a system maintain both constant pressure and constant temperature while adding heat.

But let's say what u said happens!..now if temp is higher than surroundings then The system will also want thermal equilibrium!wont it?

 

[Q_Goest]

But let's say what u said happens!..now if temp is higher than surroundings then The system will also want thermal equilibrium!wont it?

I agree with cjl and Curl. If you add heat, temperature and volume will increase as the gas moves along an isobar (the process is constant pressure). If you cool it back down, the heat is being removed and the gas follows the same isobar right back to where it started.

 

[sahil_time]

I agree with cjl and Curl. If you add heat, temperature and volume will increase as the gas moves along an isobar (the process is constant pressure). If you cool it back down, the heat is being removed and the gas follows the same isobar right back to where it started.

How can it be isobaric if the internal pressure varies!

 

[Q_Goest]

How can it be isobaric if the internal pressure varies!

Ok, maybe this is where the misunderstanding is. In the OP you said:

Imagine an idead cylinder and piston and gas filled in it with the Pgas=Patm.

I think people are interpeting that as the piston has Pgas on one side and Patm on the other side and otherwise, it is unrestrained. In that case, as heat is added or removed, the piston is in equilibrium with atmospheric pressure so Pgas is always equal to Patm.

If there is some other additional restraint on the piston such as a spring or some force that changes as the piston moves, then the pressure in the cylinder can vary also. With just atmospheric pressure on the one side and no other force such as a spring, the pressures are always equal.

 

[cjl]

But let's say what u said happens!..now if temp is higher than surroundings then The system will also want thermal equilibrium!wont it?

Then the heat will flow out of the system, and it will return to where it started. That's kind of useless though, because then the state will never really change. You specified a piston at mechanical equilibrium with its surroundings (which implies an isobaric system). Then, you specified that heat was added. Either the heat must be immediately lost with no change in the system's state, or the temperature must increase with the volume also increasing, or the temperature must increase with the pressure increasing, or the volume must increase with the pressure decreasing. Out of these, the only isobaric solution (in other words, the only one which agrees with the problem statement) is the one involving a temperature and volume increase at constant pressure.

 

[sahil_time]

So what exactly happens when we add dQ heat to the system whose
1)Intial pressure is P which is equal to the atmospheric pressure Patm ,
2)Inital volume is V and
3)Inital temp is T.
4)Hence PV=nRT.
Provided the system is capable of exchanging heat with surroundings.
What will be its final state?

 

[Q_Goest]

Hi sahil. It depends on how you constrain the gas in the system. Let's take a look at two extremes:
1) We lock the piston in place so the volume remains constant and can't change. PV = nRT always applies. In this case, the V remains constant so P and T vary according to how much heat is added to the cylinder. To determine how they vary, you apply the first law of thermodynamics and find that dU = dQ. In this case, no work is being done on or by the gas in the cylinder because the volume hasn't changed. This is called an isochoric process. Note I'll use the convention that energy added to the gas is positive, so if heat is added, dU increases. If heat is removed, dU decreases.
This is a special case in which dQ= c_v m dT

2) We allow the piston to slide so that the pressure remains constant. This is called an isobaric process, and work is being done by heating the gas. Again, we can apply the first law of thermodynamics and find dU = Q + W. This equation is written assuming heat added to the gas is positive and work done ON the gas is positive. Sometimes the equation is written dU = Q - W such that heat added to the gas is positive and work done BY the gas is positive such that it is subtracted from dU. This seems to generate a lot of confusion, but simply put, energy added to the gas increases the internal energy and energy removed from the gas decreases the internal energy as shown https://www.physicsforums.com/library.php?do=view_item&itemid=32". This is the special case in which the gas is doing work on the atmosphere equal to the pressure times area times distance the piston moves through, which equals the pressure times the change in volume.
This is a special case in which dQ= c_p m dT

Note that in both these cases, dQ can be either positive or negative depending on heat flux. If we add heat, the temperature will increase in both cases. But that heat can also leave the cylinder and put the system right back to where it was originally. The process always follows the equation PV = nRT.

There are a wide variety of equations that can be applied to the above 2 cases. In addition, we could imagine an infinite number of different cases simply by applying a spring load of varying spring constant to the piston. Depending on how pressure changes in the volume then, the pressure and temperature would not be the same as other cases, but the gas would still follow the equation PV = nRT.

 

[cjl]

So what exactly happens when we add dQ heat to the system whose
1)Intial pressure is P which is equal to the atmospheric pressure Patm ,
2)Inital volume is V and
3)Inital temp is T.
4)Hence PV=nRT.
Provided the system is capable of exchanging heat with surroundings.
What will be its final state?

The same as its initial state. If the system is capable of exchanging heat with its surroundings, and you add some heat, the system will simply undergo some transient during which it conducts that heat to its surroundings. After the transient, it will be back to the initial state.

(This is assuming that no work is done on or by the system, so the heat is the only energy transfer)

 

[sahil_time]

What if work is done?

 

[cjl]

Then you'll end up with an isothermal expansion or compression process, with the final pressure and density either lower or higher than the initial.