# Calculating Enthalpy Change in Adiabatic Compression

• roam
In summary, the conversation discusses the process of refrigeration and its effects on air temperature. The first question is answered by stating that refrigerating the gas causes it to do work, resulting in a decrease in enthalpy and temperature. The conversation then goes on to discuss the calculation process for determining the work required to adiabatically compress one kg of air at different pressures and temperatures, and the use of specific heat values for diatomic molecules. The second stage involves determining the work required to isothermally compress the air, and the relationship between pressure and volume. The conversation concludes by mentioning the absence of heat transfer in the first stage and the heat transfer being equal to the negative work done in the second stage.
roam

## Homework Statement

http://img29.imageshack.us/img29/4386/prob1p.jpg

## 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.

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

roam said:
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?
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)?

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 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

Andrew Mason said:
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)?

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?

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.

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?

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?

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.

roam said:
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?
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

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

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

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

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?
How is pressure related to volume??!. Since Ti and Tf are the same, this is rather easy.
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 work done on the gas Q=-W.
Very good.

AM

Andrew Mason said:
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

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?

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

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?

roam said:
Since Cp-Cv=Runiversal, and we have Cv=717.5 then we can find
Cv = 717.5 J/Kg K not J/mol K

Cp = R+Cv = 8.31+717.5=725.81

The ratio of molar specific heats is

$\gamma = \frac{C_p}{C_v}=\frac{725.81}{717.5}= 1.0116$
You have to be careful with units. You are mixing moles and kg.

The ratio of Cp/Cv is 7/5 = 1.4.
$P_i^{1-\gamma} T_i^\gamma = P_f^{1-\gamma} T_f^\gamma$

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

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?
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.
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?
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

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Andrew Mason said:
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.

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

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

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.

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?

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

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

Andrew Mason said:
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

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.

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roam said:
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?
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.

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.
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

Andrew Mason said:
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

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?

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

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

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

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

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

## 1. What is adiabatic compression?

Adiabatic compression is a process in which the volume of a gas is decreased without any heat transfer between the gas and its surroundings. This results in an increase in pressure and temperature of the gas.

## 2. What is the difference between adiabatic and isothermal compression?

The main difference between adiabatic and isothermal compression is that isothermal compression involves keeping the temperature constant, while adiabatic compression does not transfer any heat and thus can result in a change in temperature.

## 3. How is adiabatic compression used in industrial processes?

Adiabatic compression is often used in industrial processes such as compressors and pumps to increase the pressure of gases or liquids. It is also used in refrigeration systems to compress refrigerant gases.

## 4. What factors affect the efficiency of adiabatic compression?

The efficiency of adiabatic compression is affected by factors such as the type of gas being compressed, the speed of compression, and the design of the compression system. A higher efficiency can be achieved with a slower compression speed and a well-designed system.

## 5. What are the potential drawbacks of adiabatic compression?

One potential drawback of adiabatic compression is the increase in temperature that can occur, which may cause the gas to reach its critical temperature or even ignite. It can also result in a decrease in the density of the gas, which may affect its performance in certain applications.

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