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

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SUMMARY

The discussion focuses on the adiabatic compression of 5 moles of nitrogen gas (γ = 1.40) from an initial volume of 3.00 × 10-2 m3 at a pressure of 2.00 × 105 Pa and temperature of 280 K. The change in entropy is calculated to be 10.7 J/K, confirming that adiabatic processes involve no heat transfer. The final temperature of the gas is determined using the adiabatic condition, demonstrating that the relationship TVγ−1 remains constant throughout the process.

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  • Understanding of ideal gas laws and equations of state
  • Knowledge of adiabatic processes and their characteristics
  • Familiarity with thermodynamic concepts such as entropy and internal energy
  • Basic mathematical skills for manipulating equations
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This discussion is beneficial for students in thermodynamics, physicists, and engineers involved in gas dynamics and energy systems, particularly those studying the behavior of gases under varying conditions.

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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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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 PV^\gamma = A

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

nRTV^{\gamma-1} = A

AM
 
(a) there is flow of heat to the system of temp 300K.
The change in entropy has doubled as the volume has halved.
 
What is the adiabatic accessibility index doing? rising falling remaining constant? this will tell you what the change in entropy is.
 
adiabatic accessibility index remains constant.
Therefore change in entropy is constant no change.
 
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
 

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