# Thermodynamics of a Single-Component Ideal Gas

• TRB8985
In summary, the conversation discusses the use of the Sackur-Tetrode equation to calculate changes in entropy for an ideal gas in various scenarios. The first scenario involves an isobaric process with an amount of heat energy flowing into the system, where the change in entropy is determined using the specific heat at constant pressure. The second scenario involves specific values for the gas and initial state variables, where the final temperature is calculated in order to find the change in entropy. The third scenario involves an isobaric adiabatic expansion, where the final temperature is determined using the first law of thermodynamics. However, the resulting final temperature does not satisfy the ideal gas law for the final state.

## Homework Statement

In all calculations, take R = 8.31 J/m-K. Use the Sackur-Tetrode equation with N replaced by n and K replaced by R to calculate the changes in entropy. Also, assume that these processes are quasi-static so that the ideal gas law and the first law apply at all times. Consider an ideal gas in a container with initial state variables (P0, V0, T0, n0). Here, P = pressure, T = temperature in Kelvin, n = # of moles, and V = volume.

A.) Assume that an amount Q of heat energy flows into the system, which is free to expand in an isobaric process. Calculate the symbolic change in entropy of the system.

B.) Let n = 5.0, P0 = 3000 Pa, V0 = 3.0L, T0 = 206.6 K, and Q = 200 J. Calculate the magnitude of the change in entropy.

C.) Now assume that Q = 0 and the system undergoes an isobaric adiabatic expansion where the new volume is Vf = 3.23 L. Calculate the change in entropy.

## Homework Equations

Sackur-Tetrode equation:

S = Nk [ ln( V/N ( 4*pi*m*U / 3*N*h^2)^(3/2) ) + 5/2]

## The Attempt at a Solution

A.) For this part, there was some really useful detailed information regarding this situation in my textbook ("An Introduction to Thermal Physics" by Daniel V. Schroeder) which gives the following:

ΔSp = ∫(Cp/T)dT

I know that, for monatomic ideal gasses, CV = (3/2)nR. Additionally:

Cp = CV + nR = (5/2)nR

Thus, ΔSp = (5/2)nR * ∫(dT/T) from Ti to Tf. This is a pretty trivial integral which reduces to:

(5/2)nR*ln(Tf / Ti)

Which is my answer. (Slightly tentative on this one since it doesn't use the Sackur-Tetrode equation at all, but other online sources from various institutions and thermodynamics instructional videos on YouTube seem to agree with that answer.)
B.) For this part, I found myself rather stuck for a while since the Sackur-Tetrode equation contains a mass term in the expression, but my professor didn't include any mass value for this question. (Possibly an oversight on his part?)

My thought was to use a work-around using the expression I found in part A. The only thing I'd need to find is the final temperature in order to utilize that equation. Here was my attempt:

Q = (5/2)nRΔT for an isobaric process. Using the value I've been given for the heat and initial temperature:

200 J = (5/2)nR(Tf - Ti) = (5/2)nR*Tf - (5/2)nR*Ti

Solving algebraically for Tf:

Tf = (200 J * (5/2) * 5 mol * 8.31 J/mol*K * 206.6 K)/( (5/2) * 5 mol * 8.31 J/mol*K) ≅ 208.5 K

So,

ΔS = (5/2)(5 mol)(8.31 J/mol*K) * ln( 208.5K / 206.6 K) ≅ 0.96 J/K
C.) For this part, again, no mass is provided somehow so using the Sackur-Tetrode equation seemed impossible. Like in part A and B, I thought I'd continue with my work-around strategy by finding the "new" Tf once the gas had expanded to 3.23 L.

Since I was explicitly told that the first law holds at all times in these situations, I thought I'd use that to get the new temperature, which should be lower than the previous temperature due to the expansion of the gas.
i.e. dU = - dWby gas

Thus,

(3/2)nRΔT = -PΔV

(3/2)nR(Tf - Ti) = -P(Vf - Vi)

Solving algebraically for Tf:

Tf = [-3000 Pa * (3.23L - 3.0 L) + (3/2)(5 mol)(8.31 J/mol*K)(208.5 K)] / [(3/2)(5mol)(8.31 J/mol*K)]

≅ 197.5 K

Using this value:

ΔS = (5/2)(5 mol)(8.31 J/mol*K)*ln(197.5 K / 208.5 K) ≅ -5.67 J/K

I don't think this can be correct, since entropy always has to be greater than or equal to zero. I think this plays into the fact that I didn't use the Sackur-Tetrode equation, but again, there was no mass given which seems essential to calculating the entropy with that expression. Is there anyone familiar with this content that could comment on this? Thank you very much for your assistance.

Process C can't be carried out reversibly (i.e., quasi statically so that the ideal gas law applies at all times). If the volume is increasing at constant pressure, the temperature has to be increasing. This means that the internal energy is increasing. In addition, work is being done on the surroundings. So, from the first law, if the number of moles is constant, heat must be added.

As a result of all this, the answer you got for the final temperature does not satisfy the ideal gas law for the final state.

## 1. What is the "ideal gas" in thermodynamics?

The ideal gas is a theoretical concept used in thermodynamics to describe a gas that follows the ideal gas law. This means that the gas particles have no volume, do not exert forces on each other, and have perfectly elastic collisions.

## 2. What is the ideal gas law?

The ideal gas law is a mathematical equation that describes the relationship between pressure, volume, temperature, and the number of moles of an ideal gas. It is represented by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

## 3. How does temperature affect the behavior of an ideal gas?

According to the ideal gas law, as temperature increases, the volume of an ideal gas also increases. This is because the gas particles gain more kinetic energy and move faster, causing them to collide with the container walls more frequently and with more force, resulting in an increase in volume.

## 4. What is the significance of the ideal gas law in thermodynamics?

The ideal gas law is significant in thermodynamics as it provides a simple, yet accurate, way to describe the behavior of gases. It is also used to calculate important thermodynamic properties such as internal energy, enthalpy, and entropy.

## 5. Can the ideal gas law be applied to real gases?

The ideal gas law is often used as an approximation for real gases, especially at low pressures and high temperatures. However, at high pressures and low temperatures, real gases may deviate from the ideal gas behavior and other equations, such as the van der Waals equation, must be used instead.