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

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The discussion focuses on the effects of adiabatic compression on nitrogen gas, specifically regarding changes in entropy and temperature. When 5 moles of nitrogen gas are adiabatically compressed to half its volume, the change in entropy is calculated to be approximately 10.7 J/K, indicating no heat flow occurs during the process. The relationship TV^(γ-1) remains constant under adiabatic conditions, confirming that temperature increases as volume decreases. The final temperature can be determined using the ideal gas law and the adiabatic condition. Overall, the principles of thermodynamics are applied to understand the behavior of nitrogen gas during adiabatic compression.
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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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