Thermo: Need help finding T2 and V2 in an adiabat

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In summary: I'm assuming that the constant K will remain the same for p1*v1 and p2*v2If it didn't, it would not be constant!
  • #1
Courtknee
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Homework Statement



1 mole of ideal gas initially at 294 K is compressed adiabatically and
reversibly from 0.83 atm to 10.0 atm. Calculate the initial and …final volumes,
the final temperature, [itex]\Delta[/itex]U; [itex]\Delta[/itex]H; Q; and W: Assume that Cv = (5/2)R

Given:

n = 1 mol
T1 = 294 K
P1 = 0.83 atm
P2 = 10 atm
Q = 0

Find
V1, V2, [itex]\Delta[/itex]U, Q, [itex]\Delta[/itex]H, T2, W

Homework Equations



1. The professor gave that specific heat at constant volume is (5/2)R, however, the volume of the problem changes. Is this equation completely irrelevant to the problem?

2. How to you find the final temperature and final volume?


The Attempt at a Solution



I started off with PV = nRT for the initial givens and obtained that V1 = 29.1 L

Now, I'm stuck. I've tried using the ideal gas equation again, but I have 2 unknowns. I tried rearranging the equation and setting it equal the initial equations. I know how to find the other unknowns, but I'm stuck at this part for some reason.
 
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  • #2
Courtknee said:

Homework Statement



1 mole of ideal gas initially at 294 K is compressed adiabatically and
reversibly from 0.83 atm to 10.0 atm. Calculate the initial and …final volumes,
the final temperature, [itex]\Delta[/itex]U; [itex]\Delta[/itex]H; Q; and W: Assume that Cv = (5/2)R

Given:

n = 1 mol
T1 = 294 K
P1 = 0.83 atm
P2 = 10 atm
Q = 0

Find
V1, V2, [itex]\Delta[/itex]U, Q, [itex]\Delta[/itex]H, T2, W

Homework Equations



1. The professor gave that specific heat at constant volume is (5/2)R, however, the volume of the problem changes. Is this equation completely irrelevant to the problem?

2. How to you find the final temperature and final volume?

The Attempt at a Solution



I started off with PV = nRT for the initial givens and obtained that V1 = 29.1 L

Now, I'm stuck. I've tried using the ideal gas equation again, but I have 2 unknowns. I tried rearranging the equation and setting it equal the initial equations. I know how to find the other unknowns, but I'm stuck at this part for some reason.
You need to use the adiabatic condition: [itex]PV^\gamma = constant = K[/itex] which can be rewritten: [itex]TV^{\gamma-1} = K/nR = constant[/itex]

AM
 
  • #3
I'm assuming that the constant K will remain the same for p1*v1 and p2*v2

If I set P1V1γ = K, then I can use that constant for P2V2γ and obtain the volume, correct?
 
  • #4
Courtknee said:
I'm assuming that the constant K will remain the same for p1*v1 and p2*v2
If it didn't, it would not be constant!

If I set P1V1γ = K, then I can use that constant for P2V2γ and obtain the volume, correct?
Exactly.

AM
 
  • #5




I would suggest approaching this problem by using the adiabatic process equation, P1V1^γ = P2V2^γ, where γ is the ratio of specific heats (Cp/Cv) for an ideal gas. In this case, since Cv = (5/2)R, then γ = 1.4. From this equation, you can solve for V2 and then use the ideal gas law to solve for T2. Additionally, since the process is adiabatic, Q = 0, so you can use the first law of thermodynamics, ΔU = Q - W, to solve for ΔU and then use the relationship ΔU = ΔH - Δ(PV) to solve for ΔH. Lastly, you can use the work done equation, W = -PΔV, to solve for W. I hope this helps guide you in finding the final temperature and volume.
 

FAQ: Thermo: Need help finding T2 and V2 in an adiabat

How do I find T2 and V2 in an adiabat?

In an adiabatic process, there is no heat exchange with the surroundings, so the change in internal energy is equal to the work done by the system. This can be represented by the equation U2 - U1 = W. Since temperature is directly proportional to the internal energy of a gas, we can use the ideal gas law (PV = nRT) to find T2. Rearranging the equation to solve for V2, we get V2 = nRT2/P2.

What is the ideal gas law?

The ideal gas law is a fundamental equation in thermodynamics that describes the relationship between pressure, volume, temperature, and the number of moles of an ideal gas. It is represented by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

How does an adiabatic process differ from an isothermal process?

An adiabatic process is one in which there is no heat exchange with the surroundings, resulting in a change in temperature. In contrast, an isothermal process is one in which there is no change in temperature, so the heat exchange with the surroundings is equal to the work done by the system. In an adiabatic process, the internal energy of the gas changes, while in an isothermal process, the internal energy remains constant.

What is the significance of finding T2 and V2 in an adiabat?

Finding T2 and V2 in an adiabatic process allows us to calculate the final state of the gas based on its initial state. This information is important in understanding the behavior of gases and can be used to make predictions about the system.

How is the adiabatic process used in real-world applications?

The adiabatic process is used in various real-world applications, such as in refrigeration and air conditioning systems, where gases are compressed adiabatically to decrease their volume and increase their temperature, creating a cooling effect. It is also used in the design of engines, where an adiabatic expansion of hot gases is used to generate power.

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