Thermodynamics, isothermal irreversible

In summary, the system does work on the surroundings (transferring heat) in STATE 1, and the entropy of the surroundings decreases in STATE 1. The system does no work on the surroundings in STATE 2, and the entropy of the surroundings increases in STATE 2. The internal energy of the system (in terms of Joules) decreases in STATE 1, and increases in STATE 2.
  • #1
Faiq
348
16

Homework Statement


Can you please show how to solve this question (I am not asking it for homework. I am asking because it would help me understand better). Determine the entropy change in sys, surr, uni, when a sample of helium has of mass M grams at 298K and 1 bar doubles its volume in isothermal irreversible expansion against external pressure =p

The complete question can be found on https://chemistry.stackexchange.com/q/84590/29265

Homework Equations

The Attempt at a Solution

 
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  • #2
Faiq said:

Homework Statement


Can you please show how to solve this question (I am not asking it for homework. I am asking because it would help me understand better). Determine the entropy change in sys, surr, uni, when a sample of helium has of mass M grams at 298K and 1 bar doubles its volume in isothermal irreversible expansion against external pressure =p

The complete question can be found on https://chemistry.stackexchange.com/q/84590/29265

Homework Equations

The Attempt at a Solution

Based on your effort in the link you cited, we are permitted by PF rules to continue without further effort.

Let's assume temporarily that the constant external pressure p = 0.5 bars so that, if the gas pressure is suddenly dropped from 1 bar to p at time zero and the gas subsequently allowed to expand irreversibly while it re-equilibrates with the (ideal reservoir) at 298 K, the final volume will be double the initial volume. (Later we can talk about how to solve if the constant external pressure p is made less than 0.5 bars, and the gas expansion is forced to stop after doubling).

Let's begin by using the ideal gas law to determine the initial volume in terms of M. Using the value of ##R=0.08314\ \frac{liter.bar}{K.mole}##, what is the initial volume of helium in terms of M? In terms of M, what is the final volume at 0.5 bars and 298 K?

Chet
 
  • #3
Initial volume = ##M*0.08314*298/4##= 6.19M liters
Final Volume = ##M*0.08314*298/2##= 12.39M liters
 
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  • #4
Faiq said:
Initial volume = ##M*0.08314*298/4##= 6.19M liters
Final Volume = ##M*0.08314*298/2##= 12.39M liters
OK. So STATE 1 is:
0.25M moles, 298 K, 6.19M liters

STATE 2 is:
0.25M moles, 298 K, 12.39M liters

Before we calculate ##\Delta S## for the system, let's first apply the first law of thermodynamics to determine the change in entropy of the surroundings in this irreversible process.

For an expansion process, the equation for the work W done by the system on the surroundings is:
$$W=\int{p_{ext}dV}$$ where ##p_{ext}## is the externally applied pressure. In our irreversible process, at time zero, we suddenly drop the external pressure from 1 bar to 0.5 bars, and then hold the external pressure constant while the gas expands, until the system re-equilibrates both mechanically and thermally. So, in our process, ##p_{ext}## is constant at 0.5 bars. Based on this, what do you get for the work W done by the system on the surroundings (in terms of M, expressed in liter-bars)?

What is the change in internal energy for this "isothermal" change between STATES 1 and 2? Based on the first law, what is the amount of heat Q transferred from the surroundings to the system in our irreversible process?
 
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What is thermodynamics?

Thermodynamics is a branch of physics that deals with the relationships between heat, energy, and work. It studies the behavior of systems at a macroscopic level and how they change and interact with their surroundings.

What is an isothermal process?

An isothermal process is one in which the temperature of a system remains constant throughout. This means that the energy entering the system is equal to the energy leaving the system, resulting in no net change in temperature.

What does it mean for a process to be irreversible?

An irreversible process is one in which the system and its surroundings cannot be returned to their original state. This can be due to factors such as friction, heat loss, or irreversibility in the chemical reactions involved.

What is the significance of isothermal irreversible processes?

Isothermal irreversible processes are important in thermodynamics because they allow us to study the relationship between heat, energy, and work in real-world situations. They also help us understand how systems behave when there is a constant temperature change.

How do isothermal irreversible processes relate to the second law of thermodynamics?

The second law of thermodynamics states that in any natural process, the total entropy (measure of disorder) of a closed system will always increase. Isothermal irreversible processes demonstrate this law by showing how energy is constantly being dissipated and lost to the surroundings, resulting in an increase in entropy.

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