Change in entropy in a polytropic process

In summary, the conversation discusses a problem in which a piston cylinder device contains 1.2kg of nitrogen at 120kPa and 300K, and gas is compressed slowly in a polytropic process. The process ends when the volume is reduced by one half, and the goal is to find the entropy change. The discussion centers around using the ideal gas law, PV = nRT, to calculate the number of moles of nitrogen (represented by n) and using the relation dQ/T = dS to calculate the entropy change. The final answer is found to be 0.026 kJ/K.
  • #1
ricof
14
0

Homework Statement



A piston cylinder device contains 1.2kg of nitrogen at 120kPa and 300K. Gas is compressed slowly in a polytropic process during which PV^1.3 = constant. The proces ends when the volume is reduced by one half. What is the entropy change?

Homework Equations



(P2/P1) = (V1/V2)^n


The Attempt at a Solution



Assuming V1 = 1 and therefore V2 = 0.5, I have worked out P2 as 295.5kPa but I am stuck on what to do next.
 
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  • #2
I should add that I have absolutely no idea where or how the 1.2kg of nitrogen comes into the equation, nor if V1 is in fact 1.

Please help me I've been staring at this one for hours!
 
  • #3
anyone?
 
  • #4
How many moles is 1.2kg of nitrogen?
 
  • #5
ricof said:

Homework Statement



A piston cylinder device contains 1.2kg of nitrogen at 120kPa and 300K. Gas is compressed slowly in a polytropic process during which PV^1.3 = constant. The proces ends when the volume is reduced by one half. What is the entropy change?

Homework Equations



(P2/P1) = (V1/V2)^n


The Attempt at a Solution



Assuming V1 = 1 and therefore V2 = 0.5, I have worked out P2 as 295.5kPa but I am stuck on what to do next.

ricof said:
I should add that I have absolutely no idea where or how the 1.2kg of nitrogen comes into the equation, nor if V1 is in fact 1.

You can get V1 from the ideal gas equation. The 1.2 kg will come in handy there.

For entropy change, a useful relation is dQ/T = dS.
 
  • #6
Oh right ok so

120kPa x V = 1.2 x R x 300K

As the gas is Nitrogen do I still use R=8.314?
 
  • #7
ricof said:
Oh right ok so

120kPa x V = 1.2 x R x 300K
This looks wrong. Where does this equation come from, and what does the 1.2 represent here? Include units with all quantities, and ask yourself if the units are what the should be.

As the gas is Nitrogen do I still use R=8.314?
No, use R = 8.314 Pa*m^3 / (mole*K). "8.314" is not the same thing.
 
  • #8
In your previous post you mentioned the ideal gas law which is PV=nRT no?

n is the number of moles so I think you use the mass of the nitrogen (1.2kg) in there somewhere?
 
  • #9
Yes, you use PV = nRT.

n is the number of moles of nitrogen, not the mass of nitrogen. You have to figure out how many moles there are. See Mapes's post #4
 
  • #10
The atomic weight of nitrogen is 7 but as it is diatomic, the molecular weight is 7x2 = 14.

So 1.2kgs represents 1200/14= 85.7 moles = n?
 
  • #11
That's the right idea, but you should look up the atomic weight of nitrogen.
 
  • #12
Thanks, have now got the answer
 

Related to Change in entropy in a polytropic process

1. What is entropy?

Entropy is a measure of the disorder or randomness in a system. It is a thermodynamic quantity that describes the distribution of energy within a system.

2. How is entropy related to a polytropic process?

In a polytropic process, the entropy of a system remains constant as the system undergoes a specific type of expansion or compression. This means that the disorder or randomness within the system remains unchanged.

3. What factors affect the change in entropy in a polytropic process?

The change in entropy in a polytropic process is affected by the heat transfer into or out of the system, the work done on or by the system, and the specific heat ratio of the gas involved.

4. What is the equation for calculating change in entropy in a polytropic process?

The equation for calculating change in entropy in a polytropic process is ΔS = Cpln(T2/T1) - Rln(P2/P1), where Cp is the specific heat at constant pressure, T1 and T2 are the initial and final temperatures, P1 and P2 are the initial and final pressures, and R is the universal gas constant.

5. How does a change in entropy affect the efficiency of a polytropic process?

A change in entropy can affect the efficiency of a polytropic process by influencing the amount of work that can be extracted from the system. Generally, a higher change in entropy results in a decrease in efficiency due to the decrease in available work. However, this relationship can vary depending on the specific conditions and processes involved.

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