Irreversible isothermal Process Work External pressure not provided

In summary, an irreversible isothermal process occurs when a system exchanges heat with its surroundings at a constant temperature, and the external pressure is not specified. In such a process, the work done by the system can be calculated using the ideal gas law and the concept of pressure-volume work. The absence of external pressure information implies that the process deviates from equilibrium, leading to inefficiencies and potential energy losses. Understanding this process is crucial for applications in thermodynamics and engineering.
  • #1
Aurelius120
251
24
Homework Statement
An ideal gas is irreversibly isothermally expanded from ##(8bar ,4L)##to##(2bar,16L)##to##(1bar,32L) ##Find heat.
Relevant Equations
NA
Screenshot_20231231_031131_Chrome.jpg

It is clear that the process is isothermal else it is not possible to find heat absorbed.
$$W=-P_{ext}(\Delta V)$$

However ##P_{ext}## is not given. How do I proceed?
I tried taking ##W=-(P_2V_2-P_1V_1+P_3V_3-P_2V_2)=\Delta(PV)## but it is wrong for obvious reasons.
 
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  • #2
The one solution I found uses
##W_1=-(2)(16-4)## and ##W_2=-(1)(32-16)##
##W=(W_1+W_2) bar.Litre=-4000J##
And ##Q=4000J##

How external pressure becomes ##2\ bar## and ##1\ bar## is beyond me. It also seems wrong that ##P_{ext}## should change in an irreversible process? That is like the only thing that is good about irreversible calculations.
 
  • #3
Sketch a PV diagram of the process. Do you know what an isotherm looks like for an ideal gas?
 
  • #4
Mister T said:
Sketch a PV diagram of the process. Do you know what an isotherm looks like for an ideal gas?
images.jpeg

Something like this correct?
 
  • #5
Looks like you've found the way!
Aurelius120 said:
How external pressure becomes ##2\ bar## and ##1\ bar## is beyond me.
A slow compression expansion.

Aurelius120 said:
It also seems wrong that ##P_{ext}## should change in an irreversible process?
A dramatic example would be an explosion.
 
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FAQ: Irreversible isothermal Process Work External pressure not provided

What is an irreversible isothermal process?

An irreversible isothermal process is a thermodynamic process where the temperature remains constant, but the process is not reversible. This means that the system does not return to its initial state without leaving changes in the surroundings, often due to factors like friction, rapid compression or expansion, and non-equilibrium states.

How do you calculate work done in an irreversible isothermal process without external pressure?

In an irreversible isothermal process, if the external pressure is not provided, you can still calculate the work done using the internal pressure of the gas. The work done (W) can be calculated using the formula W = P_internal * ΔV, where P_internal is the internal pressure of the gas and ΔV is the change in volume.

Why is external pressure not provided in some irreversible isothermal process problems?

External pressure is not provided in some problems to emphasize the focus on the internal properties of the system or to simplify the problem by considering only the state variables of the gas. In real-world applications, external pressure can vary and may not be constant, making it more practical to use internal pressure for calculations.

What role does temperature play in an isothermal process?

In an isothermal process, temperature remains constant throughout the entire process. This constancy ensures that the internal energy of an ideal gas does not change, as internal energy is a function of temperature. Therefore, any heat added to the system is used to do work on the surroundings or vice versa, without changing the temperature.

Can you provide an example of an irreversible isothermal process?

An example of an irreversible isothermal process is the rapid expansion of a gas into a vacuum (free expansion). In this scenario, the gas expands quickly and irreversibly, and because it is an isothermal process, the temperature of the gas remains constant throughout the expansion. However, no work is done on the surroundings because there is no external pressure opposing the expansion.

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