1. Mar 9, 2012

### roam

1. The problem statement, all variables and given/known data

http://img29.imageshack.us/img29/4386/prob1p.jpg [Broken]

3. The attempt at a solution

To answer the first question as to why the air needs to be refrigerated, I think it's because refrigerating the gas cause it to do work, so its enthalpy will decrease resulting in a decrease in temperature. Is the correct?

This is my attempt at the calculation part: Since this is an Adiabatic process we consider the case, Q = 0 (when the heating or cooling is zero). So du = cvdT=-pdα. To use this equation we need to know the density since α=1/ρ.

I first found the initial volume using the equation PV/T = nR, and since nR is constant (no gas escapes):

$\frac{P_iV_i}{T_i} = \frac{P_fV_f}{T_f}$

$\frac{(300 \times 10^2)V_i}{-40} = \frac{(800 \times 10^2) \times 0.0005 \ m^2}{20}$

$\therefore V_i = - 0.00266 m^3$

I don't understand the -ve in volume but:

$\rho = \frac{PV}{RT} = \frac{(800 \ hPa -300 \ hPa) \times (0.0005+0.00266)}{287 \times (-40+20)} = -0.0275 \ Kg/m^3$

$du = -(800 \times 10^2) \frac{1}{-0.0275} = 2906329.114 \ J$

I don't think my calculation is correct because it didn't take into account the specific heat of air that they provided. If I used du = cvdT = 1005 x (20+40) = 60300, which is a completely different answer. Any help or hints are greatly appreciated.

Last edited by a moderator: May 5, 2017
2. Mar 9, 2012

### Staff: Mentor

Most equations involving temperature require that it be in Kelvin (i.e., degrees absolute).

3. Mar 9, 2012

### Andrew Mason

Apply the first law: ΔU = Q + W where W is the work done ON the gas. What happens to the temperature (which is proportional to internal energy) in adiabatically compressing the air from 30 to 80 hPa (Q = 0)?

I suggest you break it down into two stages:

1. First you have to determine the amount of work required to adiabatically compress one kg of air at a pressure of 30 hPa and temperature of -40C to a temperature of 20C.

2. Then you determine the amount of work required to isothermally compress it at 20 C until you reach 80 hPa.

When you have done that, apply the first law to each stage. Does any heat flow in 1.? (hint: it is adiabatic). What is the heat flow in 2?

AM

4. Mar 10, 2012

### roam

So the equation ΔU = Q + W will become ΔU = W. And as it is compressed, work is done on the gas, therefore ΔU increases and this increases the temprature (as temperature is proportional to the kinetic energy part of the Uint). So the point would be to increase the temprature by this refrigeration process?

Thank you!

Step1: To find the work done we want to use the equation:

$W=n C_v \Delta T$

And I know the number of moles:

$M= \frac{R_{universal}}{R_{specific}} = \frac{8.314}{287} = 0.02896 \ Kg/mol$

$\therefore n= \frac{m}{M} = \frac{1}{0.02896}=34.52 \ moles$

Because again air is mostly composed od diatomic molecules, do I need to use Cv=5R/2 instead of Cv=3R/2? If so

$C_v = \frac{5}{2} R_{specific} = 2.5 \times 287 = 717.5 \ J/K$

$\therefore \ W=n C_v (T_f-T_i) = 34.52 \times 717.5 \times (293.15 \ K-233.15 \ K) = 1486086 \ J$

(I used Kelvins but the final result was the same).

Step 2: the formula for work done during an isothermal process is

$W=nRT \ ln \left( \frac{V_i}{V_f} \right)$

What other formula can I use so it is in terms of pressure not volume?

There is no heat transferrerd between the system and its surroundings in 1. And since 2 is isothermal the heat transfer is equal to the -ve of the woek done on the gas Q=-W.

5. Mar 10, 2012

### Andrew Mason

No. The point is that the air would be too hot if you adiabatically compressed it. You have to somehow cool that air before you breathe it.

How hot? Apply the adiabatic condition $PV^\gamma = K$ and substitute nRT/P for V : $P^{1-\gamma}T^\gamma = constant$ to determine Tf if Pf/Pi = 8/3

Looks good.

How is pressure related to volume??!!!. Since Ti and Tf are the same, this is rather easy.
Very good.

AM

6. Mar 10, 2012

### roam

Since Cp-Cv=Runiversal, and we have Cv=717.5 then we can find

Cp = R+Cv = 8.31+717.5=725.81

The ratio of molar specific heats is

$\gamma = \frac{C_p}{C_v}=\frac{725.81}{717.5}= 1.0116$

$P_i^{1-\gamma} T_i^\gamma = P_f^{1-\gamma} T_f^\gamma$

$T_f^\gamma = \frac{P_i^{1-\gamma}}{P_f^{1-\gamma}} T_i^\gamma = \frac{3}{8} (-40)^{1.0116}$

Not sure if this is right but if I take γ=1, then Tf = -15.65. But this is not a very hot temprature, why cool it?

So, W = n x Cv x 0 = 0? How does this help to find the total amount of energy that needs to be removed to produce 0.5 L at 20 degrees?

7. Mar 10, 2012

### Andrew Mason

Cv = 717.5 J/Kg K not J/mol K

You have to be careful with units. You are mixing moles and kg.

The ratio of Cp/Cv is 7/5 = 1.4.
First of all, you are not using absolute temperatures. You have to convert Celsius to Kelvin!

Second, you have to use the right value of gamma which is 1.4.
Heat flow out of the gas occurs as it is isothermally compressed at 20 C (293K). In that case, as you found before:

$$Q = W = nRT\ln(P_f/P_i)$$

The question asks you to find the heat removed from a quantity of air that occupies .5 l at 20C and 800hPa. So you have to find n for that quantity and calculate Q.

AM

Last edited: Mar 10, 2012
8. Mar 11, 2012

### roam

Oops, thank you. But I still got a cold temprature:

$T_f^{1.4}= \frac{3}{8}(233.15)^{1.4} \implies T_f = 116 \ K$

I wasn't sure how to find n but here's my attempt. Density in 20 degrees and 800 hPa is:

$\rho = \frac{MP}{RT} = \frac{0.02896 \times (800 \times 10^2)}{8.314 \times 293.15} = 0.9505 \ Kg/m^3$

The molar volume is:

$v_m = \frac{M}{\rho} = \frac{0.02896 \ Kg/mol}{0.9505} = 0.0304 \ m^3/mol$

$n= \frac{v}{v_m}=\frac{0.0005 \ m^3}{0.0304}= 0.0164 \ moles$

Is this the correct calculation for n?

9. Mar 11, 2012

### Andrew Mason

The adiabatic condition is $PV^\gamma = K$ or $P^{1-\gamma}T^{\gamma} = k$. So:

$$P_f^{1-\gamma}T_f^{\gamma} = P_i^{1-\gamma}T_i^{\gamma}$$

$$\left(\frac{T_f}{T_i}\right)^\gamma = \left(\frac{P_i}{P_f}\right)^{(1-\gamma)}$$

and taking the log of both sides:

$$\gamma\ln\left(\frac{T_f}{T_i}\right) = (1-\gamma)\ln\left(\frac{P_i}{P_f}\right)$$

$$\ln\left(\frac{T_f}{T_i}\right) = \frac{(\gamma-1)}{\gamma}\ln\left(\frac{P_f}{P_i}\right)$$

That works out to Tf = 308K which is 35C.
Yes. But it is simpler to just use n = PV/RT. V = .0005 m^3; P = 8x10^4Pa and T = 293K, R = 8.314 J/mol K. So n = 40/8.3x293 = .0164 mol.

AM

10. Mar 12, 2012

### roam

Using that equation the heat that must be removed would be:

$Q = 0.0164 \times 8.314 \times 293.15 \ln \left(\frac{800}{300}\right)= 39.2 \ J$

Is this right? And what was the point of calculating the work done during the adiabatic process? Because the Q that I just calculated is the amount of energy removed (this occurs during the isothermal process (step 2)), and I think this is what the question really asks for.

Last edited: Mar 12, 2012
11. Mar 12, 2012

### Andrew Mason

No. You have to use the pressure at the beginning of the isothermal compression ie. at the end of the adiabatic compression that gets the air to 20C.

You calculated the temperature change of an adiabatic compression to 800 hPa. That shows the need for cooling.

To answer the question though, you have to break the compression into an adiabatic and an isothermal part. The adiabatic part brings the temperature to 20C.

What is the pressure when the air reaches 20C? I'll call that P20. hint: use the adiabatic condition for pressure and temperature. Be sure to convert all temperatures to Kelvin.

You have determined that we are dealing with n = .0164 moles of air. What is the heat that is removed when that amount of air is compressed from P20 to 800 hPa? That is the answer to the question.

AM

12. Mar 12, 2012

### roam

I used the adiabatic condition to find P20:

$\ln \frac{293.15}{308} = \frac{1.4-1}{1.4} \ln \left( \frac{P_{20}}{300 \times 10^2} \right)$

$P_{20} = 252.36 \ hPa$

Is this correct this time?

$Q = 0.0164 \times 8.314 \times 293.15 \times \ln \left( \frac{800}{252.36} \right) = 46.12$

13. Mar 12, 2012

### Andrew Mason

Not quite. The initial temp. is -40C. It should be:

$\ln \frac{293.15}{233.15} = \frac{1.4-1}{1.4} \ln \left( \frac{P_{20}}{300 \times 10^2} \right)$

That is the right idea. But you have to use the correct value for P20.

The question asks you to assume that the air is compressed to 800 hPa and then cooled to 293K.

If you do that you should get the same result if you do the adiabatic compression to P20 followed by isothermal compression to 800hPa.

AM

Last edited: Mar 12, 2012
14. Mar 13, 2012

### roam

Okay, I get it. Thank you very much for your help.