Need help finding temperature-volume relation in solving Adiabatic problem

In summary, the conversation discusses a monatomic ideal gas contained within a perfectly insulated cylinder with a movable piston. The initial pressure of the gas is 1.31 × 105 Pa and the Kelvin temperature doubles due to the piston being pushed to compress the gas. Using the equation P_i*(V_i)^γ = P_f*(V_f)^γ, the final pressure of the gas is determined to be P_f = P_i*(V_i/V_f)^γ. The conversation also discusses using logarithms to solve for the final pressure in terms of temperature and volume.
  • #1
defmar
8
0
A monatomic ideal gas (γ = 5/3) is contained within a perfectly insulated cylinder that is fitted with a movable piston. The initial pressure of the gas is 1.31 × 105 Pa. The piston is pushed so as to compress the gas, with the result that the Kelvin temperature doubles. What is the final pressure of the gas?

I know the equation I will need to solve this with is P_i*(V_i)^γ = P_f*(V_f)^γ
I'm stuck at figuring out how to plug that T_f = 2*T_i to find the volume. Any pointers?
 
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  • #2
defmar said:
A monatomic ideal gas (γ = 5/3) is contained within a perfectly insulated cylinder that is fitted with a movable piston. The initial pressure of the gas is 1.31 × 105 Pa. The piston is pushed so as to compress the gas, with the result that the Kelvin temperature doubles. What is the final pressure of the gas?

I know the equation I will need to solve this with is P_i*(V_i)^γ = P_f*(V_f)^γ
I'm stuck at figuring out how to plug that T_f = 2*T_i to find the volume. Any pointers?
Substitute nRT/V for P into the equation.

AM
 
  • #3
Andrew Mason said:
Substitute nRT/V for P into the equation.

AM

In that case I'm getting an equation that reduces to (V_f/V_i)^γ = V_i/(2*V_f).

I don't know the initial or final volume or how to solve from here.
 
  • #4
defmar said:
In that case I'm getting an equation that reduces to (V_f/V_i)^γ = V_i/(2*V_f).

I don't know the initial or final volume or how to solve from here.

Substituting nRT/V for P gives:

[tex]PV^\gamma = nRTV^{(\gamma-1)} = K[/tex]

So if:

[tex]\frac{T_f}{T_i} = 2[/tex]

then what is

[tex]\left(\frac{V_i}{V_f}\right)^{(\gamma-1)} = ?[/tex]

AM
 
  • #5
I don't know. I would say either 2 or 1/2, but that'd just be me guessing. Or I can say it's (V_i/V_f)^(2/3) - but that doesn't help me either. I don't know how to make sense of this because the problem is none of constant volume, temperature or pressure.
 
  • #6
defmar said:
I don't know. I would say either 2 or 1/2, but that'd just be me guessing. Or I can say it's (V_i/V_f)^(2/3) - but that doesn't help me either. I don't know how to make sense of this because the problem is none of constant volume, temperature or pressure.
You are making it harder than it is. It is just algebra. I gave you the equation!

Since:

[tex]\frac{T_f}{T_i} = \left(\frac{V_i}{V_f}\right)^{(\gamma-1)}[/tex]

and since:

[tex]\frac{T_f}{T_i} = 2[/tex]

then

[tex]\left(\frac{V_i}{V_f}\right)^{(\gamma-1)} = 2[/tex]

Solve that for Vi/Vf. How do you determine Pf/Pi from Tf/Ti and Vi/Vf?

AM
 
  • #7
Andrew Mason said:
You are making it harder than it is. It is just algebra. I gave you the equation!

Solve that for Vi/Vf. How do you determine Pf/Pi from Tf/Ti and Vi/Vf?

AM

Thank you so far. I took (V_i/V_f)^(2/3) = 2 ==> (V_i/V_f) = 2*SQRT(2).
I then attempted to apply both Charles and Gay-Lussac's law to come up with:
P_f/P_i = 2*SQRT(2) ==> P_f = P_i*2*SQRT(2), but this is not correct. I'm sure I'm missing something obvious and simple, but I'm oblivious to it yet.
 
  • #8
defmar said:
Thank you so far. I took (V_i/V_f)^(2/3) = 2 ==> (V_i/V_f) = 2*SQRT(2).
I then attempted to apply both Charles and Gay-Lussac's law to come up with:
P_f/P_i = 2*SQRT(2) ==> P_f = P_i*2*SQRT(2), but this is not correct. I'm sure I'm missing something obvious and simple, but I'm oblivious to it yet.
You may need to brush up a bit on algebra involving logarithms.

[tex]\left(\frac{V_i}{V_f}\right)^{\gamma-1} = 2[/tex]

Take the log of both sides:

[tex](\gamma-1)\ln{\left(\frac{V_i}{V_f}\right)} = \ln{2}[/tex]

Since [itex]\gamma = 5/3[/itex]

[tex]\ln{\left(\frac{V_i}{V_f}\right)} = 3\ln{2}/2[/tex]

Take the antilog of both sides:

[tex]\left(\frac{V_i}{V_f}\right) = e^{1.04} = 2.83[/tex]

Use PV = nRT to determine the final pressure.

AM

(P.S. One could also express the adiabatic condition in terms of T and P and work it out - just avoids the last step)
 
  • #9
Thank you. I knew I brain-farted something obvious. At this point I found it easier to plug the V_i/V_f value back into the P_i*(V_i)^γ = P_f*(V_f)^γ equation and solve for:
P_f = P_i*(V_i/V_f)^γ.

Thank you again
 

Related to Need help finding temperature-volume relation in solving Adiabatic problem

1. What is the Adiabatic process?

The Adiabatic process is a thermodynamic process in which there is no exchange of heat between the system and its surroundings. This means that the system is insulated and there is no transfer of thermal energy.

2. How do you find the temperature-volume relation in solving an Adiabatic problem?

The temperature-volume relation in an Adiabatic problem can be found using the equation T1V1^(γ-1) = T2V2^(γ-1), where T1 and T2 are the initial and final temperatures, V1 and V2 are the initial and final volumes, and γ is the adiabatic index. This equation is derived from the First Law of Thermodynamics.

3. What is the adiabatic index?

The adiabatic index, also known as the heat capacity ratio, is a thermodynamic property that relates the change in temperature of a gas to the change in its internal energy. It is denoted by the symbol γ and is equal to the ratio of the specific heats at constant pressure and constant volume.

4. Can the temperature-volume relation be applied to any type of gas?

Yes, the temperature-volume relation for Adiabatic problems can be applied to all types of gases, as long as the process is adiabatic. This includes ideal gases, real gases, and even mixtures of gases.

5. How is the temperature-volume relation used in solving real-world problems?

The temperature-volume relation for Adiabatic problems is often used in real-world applications, such as in the design of engines, refrigeration systems, and air compressors. It is also used in weather forecasting and studying the behavior of gases in the Earth's atmosphere.

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