Calculation of work (P.dV) in an isothermal system

In summary, the textbook way of calculating the work done by an ideal gas in an isothermal case is to use PV=nRT, P.dV=(nRT/V).dV, and V(x)=A.x. Differentiating both sides, d[P(x)V(x)]=0, which results in P(x).dV= -V(x).dP.
  • #1
metalrose
113
0
Here's the textbook way of calculating the work done by an ideal gas in an isothermal case.

PV=nRT
P.dV=(nRT/V).dV
∴ ∫P.dV=nRT∫dV/V
→ W2-W1=nRT*ln(V2/V1)

My question.
Consider a cylindrical (or of any other shape) container of surface area A and a frictionless movable piston attached.
Let the coordinate x run along the length of this container, with origin at the bottom of the container.
Let the piston be at some x at some point of time.

At this x, P.V= constant = nRT
This holds for all x.
Now since V(x)=A.x
P(x)=constant/A.x=K/x ... for some contant k (=nRT/A)

Now P and V are both functions of x.
While calculating the infinitesimal work P.dV, how can we treat P(x) constant over the range dx?

Instead, since P(x)V(x)=constant
differentiating both sides, d[P(x)V(x)]=0
→ V(x).dP + P(x).dV=0
∴ P(x).dV= -V(x).dP

Shouldn't this relation hold?

I ask this because I was solving a similar problem, although which wasn't isothermal, involved P and V that depended on x and the approach used there was the second one.
 
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  • #2
The relationship holds and if you integrate -VdP instead of pdV you get the same result, don't you?
I am not sure what is the question...
 
  • #3
Usually, in calculus, functions of x are taken as being constant along any differential displacement dx.
 
  • #4
DaTario said:
Usually, in calculus, functions of x are taken as being constant along any differential displacement dx.

Any rigorous mathematical explanation you could point me to? I know that if dx is infinitesimally small, f(x) will not change much in dx, but it will change neverthless. So i am looking for a rigorous mathematical explanation of why we can treat it as constant.
 
  • #5


I appreciate your question and the effort you have put into thinking about this problem. It is always important to question and analyze our methods in scientific calculations to ensure accuracy and understanding.

In the first method, the calculation of work is based on the ideal gas law, where pressure and volume are directly proportional at a constant temperature. This is a simplified approach and assumes that the pressure remains constant throughout the change in volume. This is a valid assumption for an isothermal system, where the temperature is constant and the gas follows the ideal gas law.

However, in your example, the pressure and volume are not constant as they are both functions of x. In this case, the second method you have proposed is more accurate as it takes into account the changing pressure and volume at different points along the container. This is known as the generalized work equation and is valid for systems where the pressure and volume are not constant.

In summary, both methods can be used to calculate work in an isothermal system, but the second method is more accurate for systems where pressure and volume are not constant. Thank you for bringing this up and for your critical thinking.
 

1. What is the formula for calculating work in an isothermal system?

The formula for calculating work (P.dV) in an isothermal system is W = PΔV, where W represents work, P represents pressure, and ΔV represents the change in volume.

2. How is work calculated in an isothermal system?

Work in an isothermal system is calculated by multiplying the pressure and change in volume. This is because in an isothermal system, the temperature remains constant, so the work done is equal to the change in internal energy, which is equal to the product of pressure and volume.

3. What is the significance of calculating work in an isothermal system?

Calculating work in an isothermal system helps to understand the amount of energy that is transferred during a process. It also provides insight into the efficiency of the system and can be used to make predictions about the behavior of the system.

4. Can work be negative in an isothermal system?

Yes, work can be negative in an isothermal system. This can happen when the system is compressed, causing the volume to decrease. In this case, the work done by the system is negative because the energy is being transferred from the system to the surroundings.

5. How does the ideal gas law relate to calculating work in an isothermal system?

The ideal gas law (PV=nRT) can be used to calculate work in an isothermal system because it describes the relationship between pressure, volume, and temperature. By rearranging the formula to solve for work (W = PΔV), we can use it to calculate the work done in an isothermal process.

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