How Does Adiabatic Compression Affect Entropy and Temperature in Nitrogen Gas?

I have no idea what you are doing here. Substitute P = nRT/V into PV^\gamma = A to getnRTV^{\gamma-1} = ANow substitute V' = V/2, P' = P', T' = T' (these are final values) into the above and see what happens to the equation. A is a constant. You are given the value of \gamma = 1.4. So you should be able to solve for T' in terms of T.In summary, a sample of 5 moles of nitrogen gas (γ = 1.40) is adiabatically compressed from a volume of 3.00 ×
  • #1
imy786
322
0

Homework Statement



A sample of 5 moles of nitrogen gas (γ = 1.40) occupies a volume of 3.00 × 10^−2 m3 at a pressure of 2.00 × 10^5 Pa and temperature of 280 K.
The sample is adiabatically compressed to half its original volume. Nitrogen behaves as an ideal gas under these conditions.

a)What is the change in entropy of the gas?

b)Show from the adiabatic condition and the equation of state that TV γ −1 remains constant, and hence determine the final temperature of the gas.

Homework Equations





The Attempt at a Solution



(a)

W=PV
Change in V= (3-1.5) *10^2= 1.5^10*-2

W= 2.00 × 10^5 * 1.5^10*-2
= 3*10^3 J

U=Q+W
Q= -W
Q= - 3*10^3


S= Q/T
= - 3*10^3/ 280= 10.7 JK^-1

-------------------------------------------

(b)

adiabatic condition PV^γ= A

equation of state : PV= nRT

(nRT)^gamma=A

A* V γ−1 = P

P= nRT/v

nRT/v= A* V γ−1

nRT/vA = V γ−1

TV γ −1 remains constant..
 
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  • #2
imy786 said:

Homework Statement



A sample of 5 moles of nitrogen gas (γ = 1.40) occupies a volume of 3.00 × 10^−2 m3 at a pressure of 2.00 × 10^5 Pa and temperature of 280 K.
The sample is adiabatically compressed to half its original volume. Nitrogen behaves as an ideal gas under these conditions.

a)What is the change in entropy of the gas?

b)Show from the adiabatic condition and the equation of state that TV γ −1 remains constant, and hence determine the final temperature of the gas.

Homework Equations





The Attempt at a Solution



(a)
...
S= Q/T
= - 3*10^3/ 280= 10.7 JK^-1
Assume it is compressed reversibly and adiabatically (the external pressure is slightly higher than internal pressure during compression). Is there any flow of heat into/out of the gas or into or out of the surroundings? So what is the change in entropy?

(b)

adiabatic condition [itex]PV^\gamma = A[/itex]

equation of state : PV= nRT
...
I find it a little difficult to follow your reasoning. Substitute P = nRT/V into [itex]PV^\gamma = A[/itex] to get

[tex]nRTV^{\gamma-1} = A[/tex]

AM
 
  • #3
(a) there is flow of heat to the system of temp 300K.
The change in entropy has doubled as the volume has halved.
 
  • #4
What is the adiabatic accessibility index doing? rising falling remaining constant? this will tell you what the change in entropy is.
 
  • #5
adiabatic accessibility index remains constant.
Therefore change in entropy is constant no change.
 
  • #6
imy786 said:
(a) there is flow of heat to the system of temp 300K.
The change in entropy has doubled as the volume has halved.
Careful. Adiabatic means no heat flow. Temperature changes due to internal energy increasing as a result of work being done on gas. But this does not mean there is heat flow.

AM
 

1. What is entropy?

Entropy is a measure of the disorder or randomness of a system. It is often referred to as the "arrow of time" because it tends to increase with time in natural systems.

2. How is entropy related to adiabatic processes?

In an adiabatic process, there is no exchange of heat between a system and its surroundings. This means that the change in entropy of the system is equal to zero. In other words, the entropy of an adiabatic system remains constant.

3. What is the connection between entropy and the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will tend to increase over time. This law is closely related to the concept of entropy, as entropy is a measure of the direction in which a system tends to evolve.

4. Can entropy be reversed?

In isolated systems, the total entropy can never decrease, meaning that entropy cannot be reversed. However, in open systems, such as living organisms, it is possible to decrease the entropy of a localized area by increasing the entropy of the surrounding environment.

5. How is entropy calculated?

The exact calculation of entropy requires advanced mathematical techniques, but it is essentially a measure of the number of possible microstates that a system can have. In simple terms, the more ways a system can be arranged, the higher the entropy.

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