Quasi-static, adiabatic problem

In summary, two moles of a monatomic ideal gas are at a temperature of 0°C and a quasi-static volume of 45 liters. The gas is expanded adiabatically until its temperature falls to - 50°C. Its initial pressure is 100,956 Pa (0.1MPa) and its final pressure is still unknown.
  • #1
mateomy
307
0
Two moles of a monatomic ideal gas are at a temperature of 0°C and a
(dQ = 0) and quasi-static volume of 45 liters. The gas is expanded adiabatically
until its temperature falls to - 50°C. What are its initial and final pressures and its final volume?

I've been beating my head against a wall for a while now on this one.

I know how to find the initial pressure. That's just an easy PV=nRT problem. I get an answer of 100,956 Pa (0.1MPa). I can't find the final pressures though, totally stuck.

So what I've done thus far is to utilize the first law in that I know dQ = 0. So I know that dU = -dW. Solving the U = cNRT for initial and final temperatures and taking the difference, I get an energy which I know must be attributed to the work done. This is where I get stuck.

Not sure what steps to take next. Need some assistance, thanks.

I thought I could try
[tex]
W = - \int PdV
[/tex]
which I could do some substitution to get to (eventually),
[tex]
W = -nRTln\left(\frac{V_f}{V_i}\right)
[/tex]
But when I do that and I set W to the difference in energy I found earlier, I get an answer that doesn't match my book's.
 
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  • #2
No, but you almost have it. Going back to your equation for the first law:

dU=cNRdT=-PdV=-(NRT/V)dV

where c = 3/2

Cancel NR from both sides, divide both sides by T, and integrate.

Chet
 
  • #3
mateomy said:
I thought I could try
[tex]
W = - \int PdV
[/tex]
which I could do some substitution to get to (eventually),
[tex]
W = -nRTln\left(\frac{V_f}{V_i}\right)
[/tex]
But when I do that and I set W to the difference in energy I found earlier, I get an answer that doesn't match my book's.
This doesn't work because that expression for the work done is for isothermal processes. That's not what you have here.
 
  • #4
Chestermiller said:
dU=cNRdT=-PdV=-(NRT/V)dV

where c = 3/2

Cancel NR from both sides, divide both sides by T, and integrate.

Chet

So this is what I've done on your suggestion (asking for my work to be checked because my answer still isn't correct):

[tex]
cNRdT = -\frac{NRTdV}{V}
[/tex]

[tex]
\frac{cdT}{T} = \frac{dV}{V}
[/tex]

[tex]
c \int_{T_i}^{T_f} \frac{dT}{T} = - \int_{V_i}^{V_f} \frac{dV}{V}
[/tex]

[tex]
c\,ln\left(\frac{T_f}{T_i}\right) = - ln\left(\frac{V_f}{V_i}\right)
[/tex]

[tex]
e^{c} (T_f - T_i) = -(V_f - V_i)\,\,\,\,\,\,negatives\,cancel\,here
[/tex]

from which I get a result of 224.04. However, the answer is [itex]61E3[/itex]. Still unsure...
 
  • #5
mateomy said:
[tex]
c\,ln\left(\frac{T_f}{T_i}\right) = - ln\left(\frac{V_f}{V_i}\right)
[/tex]

[tex]
e^{c} (T_f - T_i) = -(V_f - V_i)\,\,\,\,\,\,negatives\,cancel\,here
[/tex]
You exponentiated the LHS wrongly. eab is not eaeb.
 
  • #6
I'm insanely confused then. Is it,
[tex]
\left(\frac{T_f}{T_i}\right)^{c}
[/tex]
 
  • #7
mateomy said:
I'm insanely confused then. Is it,
[tex]
\left(\frac{T_f}{T_i}\right)^{c}
[/tex]
Now you're cookin'. Of course, you follow the same recipe on the RHS of the equation.

Chet
 
  • #8
Now I'm becoming embarrassed...
[tex]
\left(\frac{T_f}{T_i}\right)^{c} = \left(\frac{V_f}{V_i}\right)^{-1}
[/tex]
[tex]
\left(\frac{T_f}{T_i}\right)^{c} = \left(\frac{V_i}{V_f}\right)
[/tex]
[tex]
V_i \left(\frac{T_i}{T_f}\right)^{c} = V_f
[/tex]
But I'm still not getting the correct answer.
 
  • #10
The given answer (from Callen) is [itex]61 x 10^{3} m^{3}[/itex]. My answer is 0.06095. It almost seems like there's a conversion error but I don't see it. From the initial values: 45 liters = 0.045 cubic meters, 0 C = 273K, and -50 C = 223K.
 
  • #11
Although...

I'm looking at my PDF (don't have the 'real' book) and it's entirely possible that the negative is smudged out. I've noticed a lot of cosmetic errors in this PDF. That seems reasonable if my equation is correct and I am off by that magnitude. Maybe someone could check my numbers to confirm?
 
  • #12
mateomy said:
The given answer (from Callen) is [itex]61 x 10^{3} m^{3}[/itex]. My answer is 0.06095. It almost seems like there's a conversion error but I don't see it. From the initial values: 45 liters = 0.045 cubic meters, 0 C = 273K, and -50 C = 223K.
No need to check the calculation in detail. It's quite clear that the volume would not expand by more than a factor of 10, probably much less. So 61 litres is reasonable.
 
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  • #13
mateomy said:
The given answer (from Callen) is [itex]61 x 10^{3} m^{3}[/itex]. My answer is 0.06095. It almost seems like there's a conversion error but I don't see it. From the initial values: 45 liters = 0.045 cubic meters, 0 C = 273K, and -50 C = 223K.

Yeah. That should be 10-3. You did a really nice job on this problem. Too bad the book had that typo.
 
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  • #14
Ugh. I can't believe it was a typo. At least I got some lessons out of it. Thank you, both.
 

1. What is a quasi-static process?

A quasi-static process is a type of thermodynamic process in which the system changes slowly enough that it can be considered in equilibrium at all times. This means that the system is always close to its internal and external equilibrium states, and the process can be described by a series of equilibrium states.

2. What is an adiabatic process?

An adiabatic process is a thermodynamic process in which there is no heat exchange between the system and its surroundings. This means that the system is thermally isolated and there is no transfer of energy in the form of heat during the process.

3. What is the difference between a quasi-static and adiabatic process?

The main difference between a quasi-static and adiabatic process is that in a quasi-static process, the system is always close to equilibrium and the process is slow enough to be described by a series of equilibrium states. In an adiabatic process, there is no heat exchange between the system and its surroundings, but the process does not necessarily have to be slow or in equilibrium.

4. What are some real-life examples of quasi-static, adiabatic processes?

Some examples of quasi-static, adiabatic processes include the compression or expansion of a gas in a cylinder with a slow-moving piston, the slow melting of ice in a thermally insulated container, and the slow heating or cooling of a substance in a well-insulated container.

5. What is the importance of studying quasi-static, adiabatic processes?

Studying quasi-static, adiabatic processes is important in understanding how different systems behave and how energy is transferred within them. These processes are also commonly used in various industries, such as engineering and chemistry, and can help in the design and optimization of systems and processes.

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