Temperature change in a gas expansion

In summary, the ideal gas law can be used to express δP_V and δT_S as functions of δV, δT, and γ, and the first law can be used to express δP_V and δT_s as functions of δV, δT, and γ.
  • #1
yamata1
61
1
Homework Statement
The fluid is a perfect gas. Constant pressure heating is broken down into two stages:
- an elementary transformation AB which is a heating at constant volume, during which
temperature and pressure vary by ##\delta T_V## and ##\delta P_V##.

-an elementary transformation BC which is a reversible adiabatic relaxation, during
which the temperature and the pressure vary respectively from ##\delta T_S## and ##\delta P=-\delta P_V##

1-Express ##\delta P_V ## as a function of ##\delta T_V##,T and P

2-Express ##\delta T_S## as a function of ##\delta P_V##,T , P and ##\gamma=\frac{C_p}{C_v}## then as a function of ##\delta T_V## and ##\gamma##.
Relevant Equations
##PV^{\gamma}=cst\; \; \; \; \;TV^{\gamma -1}=cst## and ##TP^{\frac{1-\gamma }{\gamma}}=cst##

PV=nRT , dU=TdS-PdV ,##\delta Q=C_V\deltaT## , ##\delta V =(\frac{\partial V}{\partial T})_P \delta T+(\frac{\partial V}{\partial P})_T \delta P##
1- ##\delta P_V =(\frac{\partial P}{\partial T} )_V \delta T_V##

2-##\delta V =(\frac{\partial V}{\partial T})_P \delta T+(\frac{\partial V}{\partial P})_T \delta P## so ##C_v \delta T=-P\delta V=-P((\frac{\partial V}{\partial T})_P \delta T+(\frac{\partial V}{\partial P})_T \delta P)## I can replace ##-P((\frac{\partial V}{\partial T})_P \delta T)=nR\delta T## since we have an ideal gas and make ##\gamma## appear that way.

Is there some other equation I am forgetting ?

Thank you
 
Last edited:
Physics news on Phys.org
  • #2
Using the ideal gas law, what is your actual answer for part 1 (not in terms of partial derivatives)?

For part 2, you don't need to use the first law if you use ##TP^{\frac{1-\gamma }{\gamma}}=cst##.
 
  • #3
Chestermiller said:
Using the ideal gas law, what is your actual answer for part 1 (not in terms of partial derivatives)?

For part 2, you don't need to use the first law if you use ##TP^{\frac{1-\gamma }{\gamma}}=cst##.
##\delta P_V =(\frac{\partial P}{\partial T} )_V \delta T_V =\frac{nR}{V}\delta T_V##.
I don't know how to use ##TP^{\frac{1-\gamma }{\gamma}}=cst## to make ##\delta T_S## a function of ##\delta P_V##,T,P and ##\gamma##.
 
  • #4
yamata1 said:
##\delta P_V =(\frac{\partial P}{\partial T} )_V \delta T_V =\frac{nR}{V}\delta T_V##.
I get $$\delta T_V=\frac{\delta P}{P}T$$
I don't know how to use ##TP^{\frac{1-\gamma }{\gamma}}=cst## to make ##\delta T_S## a function of ##\delta P_V##,T,P and ##\gamma##.
I would write algebraically, $$(T+\delta T_V+\delta T_S)P^{\frac{1-\gamma }{\gamma}}=(T+\delta T_V)(P+\delta P)^{\frac{1-\gamma }{\gamma}}$$
 
  • Like
Likes yamata1
  • #5
Chestermiller said:
I would write algebraically, $$(T+\delta T_V+\delta T_S)P^{\frac{1-\gamma }{\gamma}}=(T+\delta T_V)(P+\delta P)^{\frac{1-\gamma }{\gamma}}$$
Thank you but now I don't see a way to remove T and have ##\delta T_S## as a function of only ##\delta T_V## and ##\gamma ##.
 
  • #6
yamata1 said:
Thank you but now I don't see a way to remove T and have ##\delta T_S## as a function of only ##\delta T_V## and ##\gamma ##.
What does "Express δ##P_V## as a function of δ##T_V##,T and P" and "Express ##\delta T_S## as a function of ##\delta P_V##,T , P and ##\gamma##" mean to you?
 
  • #7
Chestermiller said:
What does "Express δ##P_V## as a function of δ##T_V##,T and P" and "Express ##\delta T_S## as a function of ##\delta P_V##,T , P and ##\gamma##" mean to you?
Since δ##T_V## is a function of T and P there should be a way to change variables P and T in the formula for ##\delta T_S## and express it as a function of only δ##T_V## and ##\gamma##.
 
  • #8
yamata1 said:
Since δ##T_V## is a function of T and P there should be a way to change variables P and T in the formula for ##\delta T_S## and express it as a function of only δ##T_V## and ##\gamma##.
Who says?
 
  • #9
Chestermiller said:
Who says?
Question 2 :
Express ##\delta T_S## as a function of ##\delta P_V##,T , P and ##\gamma## first then as a function of ##\delta T_v## and ##\gamma##
 
  • #10
yamata1 said:
Question 2 :
Express ##\delta T_S## as a function of ##\delta P_V##,T , P and ##\gamma## first then as a function of ##\delta T_v## and ##\gamma##
The "then" part can't be done unless T is also included. This is just an omission from the problem statement.
 
  • Like
Likes yamata1
  • #11
Chestermiller said:
The "then" part can't be done unless T is also included. This is just an omission from the problem statement.
I guess so.Thank you for these clarifications.
 

What is temperature change in a gas expansion?

Temperature change in a gas expansion refers to the change in temperature that occurs when a gas expands or contracts. This change in temperature is due to the transfer of energy between the gas and its surroundings.

What factors affect temperature change in a gas expansion?

The temperature change in a gas expansion is affected by several factors, including the initial temperature of the gas, the pressure of the gas, and the type of gas being expanded. The volume of the gas and the surrounding environment also play a role in the temperature change.

How does temperature change during gas expansion?

During gas expansion, the temperature of the gas decreases due to the decrease in pressure and the expansion of the gas molecules. This decrease in temperature is known as the Joule-Thomson effect.

Why is temperature change important in gas expansion?

Temperature change in gas expansion is important because it affects the behavior and properties of the gas. It also plays a crucial role in various industrial processes, such as refrigeration and air conditioning.

What is the relationship between temperature change and work done in gas expansion?

The relationship between temperature change and work done in gas expansion is described by the first law of thermodynamics, which states that the change in internal energy of a gas is equal to the work done on the gas plus the heat added to the gas. This means that the temperature change is directly proportional to the work done during gas expansion.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
600
  • Introductory Physics Homework Help
Replies
2
Views
925
  • Introductory Physics Homework Help
Replies
4
Views
969
  • Introductory Physics Homework Help
Replies
11
Views
307
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Replies
1
Views
634
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
941
  • Introductory Physics Homework Help
Replies
1
Views
966
Back
Top