# Entropy with non constant specific heats

• Goatsenator
In summary, we are given the initial state of air inside a piston-cylinder device that is compressed adiabatically and then heated at constant pressure. We are asked to determine various parameters, such as pressure, temperature, work, heat, and entropy change, using data from Table A7.1 and the First Law of Thermodynamics. We cannot assume constant specific heats, so we must use the ideal gas law and solve for the unknown parameters.
Goatsenator
1. Homework Statement
Consider air inside a piston-cylinder device that is compressed adiabatically from p1 = 100kPa, T1 = 300 K, and V1 = 0.5 m^3 to V2 = 0.03 m^3 and then heated at constant pressure until T3 = 2000 K. The heat is exchanged with a heat reservoir at TH = 2000 K. Determine P2, T2, P3, V3, W1-2, W2-3, Q2-3, and ∆S1−2, without assuming constant specific heats, instead using data given in Table A7.1 in the text.

The table just gives standard entropy. 2. Homework Equations S0(T2) - S0(T1) = R*ln(V2/V1) + R*ln(T2/T1)

where S0 is the standard entropy. I derived this myself as well and understand where it comes from. 3. The Attempt at a Solution

I tried to substitute S0(T1) + R*ln(P2/P1) for S0(T2) but that doesn't seem to help.

Hello,

Thank you for sharing your attempt at a solution. I can see that you are on the right track in using the equation relating entropy to pressure and temperature. However, in this problem, we are not assuming constant specific heats, so the equation you have used may not be applicable. Instead, we can use the First Law of Thermodynamics to solve for the various parameters.

First, let's consider the adiabatic compression from state 1 to state 2. Since the process is adiabatic, we know that Q1-2 = 0. Therefore, we can use the First Law of Thermodynamics to write:

W1-2 = ΔU1-2 = m(Cv1)(T2-T1)

where m is the mass of the air, Cv1 is the specific heat at constant volume for air at state 1, and T2-T1 is the change in temperature during the adiabatic compression.

Next, let's consider the heating at constant pressure from state 2 to state 3. Since the process is at constant pressure, we know that:

W2-3 = PΔV2-3 = P(V3-V2)

where P is the constant pressure during this process and V3-V2 is the change in volume during the heating.

We can also use the First Law of Thermodynamics to write:

Q2-3 = ΔU2-3 + W2-3 = m(Cp2)(T3-T2) + P(V3-V2)

where Cp2 is the specific heat at constant pressure for air at state 2, and T3-T2 is the change in temperature during the heating process.

Now, we can use the given data in Table A7.1 to determine the specific heats at constant volume and pressure for air at states 1 and 2. From the table, we can see that:

Cv1 = 0.718 kJ/kg-K
Cp2 = 1.005 kJ/kg-K

We can also use the ideal gas law to relate pressure, temperature, and volume at each state:

P1V1/T1 = P2V2/T2 = P3V3/T3

Using this, we can solve for the unknown parameters in terms of the given parameters. I will leave it to you to plug in the numbers and solve for the values.

I hope this helps. Good luck with your problem

## 1. What is entropy with non constant specific heats?

Entropy with non constant specific heats refers to the measure of the disorder or randomness of a system when the specific heat of the system varies with temperature. It takes into account the changes in energy and temperature of the system, and helps to determine the direction of spontaneous processes.

## 2. How is entropy with non constant specific heats calculated?

The calculation of entropy with non constant specific heats involves integrating the specific heat with respect to temperature. This can be done using the formula: ∆S = ∫(Cp/T)dT, where ∆S represents the change in entropy, Cp is the specific heat at constant pressure, and T is the temperature.

## 3. What factors affect entropy with non constant specific heats?

The main factors that affect entropy with non constant specific heats are the temperature and composition of the system. The specific heat of a substance can also vary depending on its state (solid, liquid, or gas) and any phase changes that occur.

## 4. How does entropy with non constant specific heats relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of an isolated system always increases over time. Entropy with non constant specific heats is a way of quantifying this increase in disorder or randomness. It helps to explain why certain processes occur spontaneously and why others do not.

## 5. Can entropy with non constant specific heats be negative?

Yes, it is possible for entropy with non constant specific heats to be negative. This can occur when the specific heat decreases with increasing temperature, causing the integral to have a negative value. However, this is not a violation of the second law of thermodynamics as the total entropy of the universe always increases, even if a part of the system has a negative entropy change.

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