Work and Heat in Gas Cycle (formatted: may load slowly)

In summary, the problem involves a cycle of an ideal diatomic gas with a mole of gas initially at 20.6°C and 1.00 atm pressure. The cycle consists of three paths: (1) an isochoric increase in pressure until the temperature reaches 159°C, (2) an adiabatic expansion until the pressure returns to 1.00 atm, and (3) an isobaric compression until the gas returns to its original volume. The original volume of the gas at the beginning and end of the cycle is 0.0241 m^3. The pressure of the gas at the completion of path (1) is 142149 Pa. The volume of the gas at the completion
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Problem:
One (1.00) mole of an ideal diatomic gas (with = 1.40) initially at 20.6°C and 1.00 atm pressure is taken around the following cycle:
Path (1): an isochoric increase in pressure until the temperature of the gas is 159°C and the pressure is P;
Path (2): an adiabatic expansion until the pressure returns to 1.00 atm;
Path (3): an isobaric compression until the gas reaches its original volume.
[[FIGURE ATTACHED]]


(a) What is the original volume of the gas at the beginning and end of the cycle?

(b) What is the pressure of the gas at the completion of path (1)?

(c) What is the volume of the gas at the completion of path (2)?

(d) What is the temperature of the gas at the completion of path (2)?

(e) Calculate the work done by the gas during each path of the cycle and the total work done by the gas.
1. (along path (1))
2. (along path (2))
3. (along path (3))
4. (total work)
(f) Calculate the heat transfer to the gas during each path of the cycle and the total heat transfer to the gas over the cycle.
1. (along path (1))
2. (along path (2))
3. (along path (3))
4. (total heat transfer)
(g) Find the efficiency of the cycle.

(h) Calculate the maximum efficiency that a heat engine could have if it operated between the hottest and coldest temperatures encountered by this gas in this cycle.


Attempt at a Solution:
**Let the point at the beginning of Path 1=a, end of path 1=b, end of path 2=c

a)[tex]V_{a}[/tex]=[tex]\frac{nRT}{P}[/tex] (where P=[tex]P_{a}[/tex] and T=[tex]T_{a}[/tex]) --->This gives the correct answer .0241 m^3

b)[tex]P_{b}[/tex]=[tex]\frac{nRT}{V}[/tex] --> [tex]V_{b}[/tex]=[tex]V_{a}[/tex]; this gives the correct answer 142149 Pa

c) [tex]P_{c}[/tex][tex]V^{\gamma}_{c}[/tex]=[tex]P_{b}[/tex][tex]V^{\gamma}_{b}[/tex] ----->this gives the correct answer .0307 m^3

d) [tex]T_{c}[/tex]=[tex]\frac{P_{c}V_{c}}{nR}[/tex] ----->this gives the correct answer 382 K.

...NOW, to where I'm having trouble.

e) 1. W=0 because [tex]\int[/tex]PdV = [tex]W_{AB}[/tex] ---->dv=0 this gives the correct answer

e2. [tex]W_{BC}[/tex] = -[tex]\Delta[/tex]U +Q, adiabatic process--> Q=0, so [tex]W_{BC}[/tex]=-[tex]\Delta[/tex]U
=-n[tex]C_{v}[/tex][tex]\Delta[/tex]T
=-1.00(20.8)(432K-382K) = -1040 but apparently this is wrong. I tried the positive value just to be safe, but that was incorrect as well.

e3. [tex]W_{CA}[/tex]=[tex]\int[/tex]PdV ---> pressure is constant, so

[tex]W_{CA}[/tex]=P[tex]\Delta[/tex]V = (1.013 x [tex]10^{5}[/tex] Pa)(.0307-.0241) = 669, but I also tried the negative and both were incorrect.

e4. Just the sum of all e1+e2+e3, I assume

f) 1. Q=n[tex]C_{v}[/tex][tex]\Delta[/tex]T ---> this gives correct answer2879 J

f2. Q=0, adiabatic process-->correct answer

f3. Q= n[tex]C_{p}[/tex][tex]\Delta[/tex]T ---->this gives correct answer -2572 J

f4. First, I tried adding them all up, which was wrong, so I tried the negative, which was wrong and then realized it should be the opposite value of each number, which gave the wrong answer so I tried the negative of that, which was also wrong...

I got g (and I'm getting tired of using this formatting :smile:) but if you want me to show how I did it just to see my thought process, I will. I'm assuming h shouldn't be hard either, once I have the correct values for everything.


THANK YOU FOR ANY HELP YOU CAN GIVE!
 

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  • #2
I confirm your results for e2. The answer key must be wrong.
 

1. What is the difference between work and heat in a gas cycle?

In a gas cycle, work is the energy transferred to or from the gas due to a change in volume, while heat is the energy transferred to or from the gas due to a temperature difference. Work and heat are both forms of energy, but they differ in how they are transferred to or from the gas.

2. What is the ideal gas law and how does it relate to work and heat in a gas cycle?

The ideal gas law is a mathematical relationship between pressure, volume, and temperature of a gas. In a gas cycle, changes in pressure, volume, and temperature affect the work and heat transferred to or from the gas. The ideal gas law can be used to calculate these changes and understand how they impact the gas cycle.

3. How is the efficiency of a gas cycle determined?

The efficiency of a gas cycle is determined by the ratio of the work output to the heat input. This is known as the thermal efficiency and it can be calculated using the ideal gas law and the first law of thermodynamics. The higher the thermal efficiency, the more efficient the gas cycle is at converting heat into work.

4. What are the different types of gas cycles?

There are several types of gas cycles, including the Carnot cycle, Otto cycle, Diesel cycle, and Brayton cycle. Each type of cycle has its own unique characteristics and is used in different applications. The efficiency and performance of each cycle also vary depending on the specific conditions and processes involved.

5. How does the concept of entropy relate to work and heat in a gas cycle?

Entropy is a measure of the disorder or randomness of a system. In a gas cycle, entropy plays a crucial role in determining the direction and magnitude of heat transfer. Heat will naturally flow from a higher temperature region to a lower temperature region due to the increase in entropy. This process is essential in understanding the thermodynamics of a gas cycle and its efficiency.

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