Find the equation of state of each gas

Click For Summary

Homework Help Overview

The discussion revolves around finding the equation of state for different gases as presented in a thermodynamics context. Participants are exploring the relationships between pressure, volume, and temperature for various gases, referencing established equations of state.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the validity of assuming the right-hand side of the equation of state to be simply ##nRT## and question how to proceed using thermodynamic laws. There is also consideration of whether the relationships must hold across all temperatures and the implications of equating different gas equations.

Discussion Status

The discussion is active, with participants questioning assumptions and exploring different interpretations of the equations. Some guidance has been offered regarding the nature of the functions involved, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants note potential constraints, such as the requirement to find each gas's equation of state separately at different temperatures and the possibility of more complex terms appearing in the equations beyond simple functions of temperature.

curious_mind
Messages
48
Reaction score
9
Homework Statement
Consider three gases with ##(P_1,V_1),(P_2,V_2)## and ##(P_3,V_3)##. It is found that when the first two are in equilibrium the following condition is satisfied: ##P_1V_1 =\left(P_2 +\frac{a}{V_2^2}\right)(V_2 −b)##, while the equation satisfied when the first and the last are in equilibrium is ##P_3(V_3 −c)=P_1V_1 e^{\frac{−d}{V_3P_1V_1}}##. Find the respective equations of state and identify them.
Relevant Equations
Equation of states of gas at temperature T##f(P,V,T)=0##
The problem is from the book "The Principles of Thermodynamics" by ND Hari dass.

It looks trivial problem, but I am not able to form logical arguments for going into next step.

For example, It seems like first gas has equation of state ##PV =nRT## and second has ## \left( P_2 +\frac{a}{V_2^2} \right) (V_2 −b) = nRT ##
But I cannot straightforward assume Right hand side of equation of state to be simply ##nRT## in general right ? So what could be valid way to proceed from thermodynamical laws ?

Thanks.
 
Physics news on Phys.org
For the relationships to hold across all T, each relationship must be of the form (first expression =second expression = some function of T).
 
  • Like
Likes   Reactions: curious_mind
If we equate all three relations, then it will be valid only if all three gases in equilibrium, which is not required to be found. We require to find all three gases equation of state separately, at different temperatures.

Also, how can we say that it individual ##f(P,V)## is some function of ##T## ONLY?. Right hand side of equation of state might containt terms like ##\cos (TVe^P)## etc etc or something, in general - right m? Or am I missing something very fundamental?
 
Last edited:
curious_mind said:
If we equate all three relations
I did not say that. The "some function of T" does not have to be the same for each.
curious_mind said:
Also, how can we say that it individual ##f(P,V)## is some function of ##T## ONLY?.
It is the same principle as "separation of variables" in PDEs.
We know that ##T_1=g_1(P_1,V_1)## and ##T_2=g_2(P_2,V_12)## for some functions ##g_1, g_2##. So at any given temperature T we know ##g_1(P_1,V_1)=g_2(P_2,V_12)##. And these are the forms you are given.
 
  • Like
Likes   Reactions: curious_mind
It seems to me your original approach was correct, except I would assume the V is molar volume rather than volume itself, so you would get rid of the n's in the equations.
 
  • Like
Likes   Reactions: curious_mind

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K