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

  • Thread starter Thread starter imy786
  • Start date Start date
  • Tags Tags
    Adiabatic Entropy
AI Thread Summary
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.
imy786
Messages
321
Reaction score
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..
 
Physics news on Phys.org
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
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top