What Is the Net Entropy Change in a Two-Step Gas Process?

[Faux Carnival]

Homework Statement

One mole of a monatomic gas first expands isothermally, then contracts adiabatically. The final pressure and volume of the gas are twice its initial pressure and volume. (P_f=2P_i and V_f=2V_i)

Find the net change in the entropy of the gas.

Homework Equations

The Attempt at a Solution

I found the entropy change for the first step. Let the volume between the two steps be V_m.

TV^γ-1 = constant, so V_m = 2^2.5 V_i

ΔS = nR ln(V_m/V_i) = 1.73 nR

But I have no idea about the entropy change of the second step. What is the entropy change for an adiabatic process? Isn't dQ=0 ?

 

[Andrew Mason]

Homework Statement

One mole of a monatomic gas first expands isothermally, then contracts adiabatically. The final pressure and volume of the gas are twice its initial pressure and volume. (P_f=2P_i and V_f=2V_i)

Find the net change in the entropy of the gas.

Homework Equations

The Attempt at a Solution

I found the entropy change for the first step. Let the volume between the two steps be V_m.

TV^γ-1 = constant, so V_m = 2^2.5 V_i

ΔS = nR ln(V_m/V_i) = 1.73 nR

But I have no idea about the entropy change of the second step. What is the entropy change for an adiabatic process? Isn't dQ=0 ?

There is no entropy change in a reversible adiabatic process, so you just have to determine the change in entropy of the isothermal process. You appear to have tried doing this by applying the adiabatic condition (after finding the final temperature) to get the volume at the end of the isothermal process, but I can't figure out how you get your answer. If you apply the adiabatic condition:

$$\left(\frac{V_m}{V_f}\right)^{\gamma-1} = \frac{T_f}{T_m} = 4$$

AM

 

[Faux Carnival]

STEP 1 (ISOTHERMAL)

P_i V_i T_i ---> P_unknown V_m T_i

STEP 2 (ADIABATIC)

P_unknown V_m T_i ---> 2P_i 2V_i 2T_i

So, for the adiabatic process I applied TV^γ-1 = constant.

T_i V_m^2/3 = 2T_i (2V_i)^2/3

Isn't this correct?

Another thing, is the change in entropy ΔS zero for adiabatic processes?

 

[Andrew Mason]

STEP 1 (ISOTHERMAL)

P_i V_i T_i ---> P_unknown V_m T_i

STEP 2 (ADIABATIC)

P_unknown V_m T_i ---> 2P_i 2V_i 2T_i

So, for the adiabatic process I applied TV^γ-1 = constant.

T_i V_m^2/3 = 2T_i (2V_i)^2/3

Isn't this correct?

It is not correct. What is the final temperature? (hint: apply the ideal gas law). The final temperature is not twice the initial temperature.

Another thing, is the change in entropy ΔS zero for adiabatic processes?

Yes. See my previous post.

AM

 

[rwisz]

I would approach this question by first understanding the concept of entropy and its relationship to temperature and energy. Entropy is a measure of the disorder or randomness of a system and is related to the amount of energy that is unavailable for work. In this case, we are dealing with a monatomic gas, which means that it consists of single atoms and has a simple relationship between temperature and energy.

To find the net change in entropy, we need to consider the two steps of the process separately and then combine them to find the overall change. The first step is an isothermal process, which means that the temperature remains constant. Therefore, the change in entropy for this step can be calculated using the formula ΔS = nR ln(Vf/Vi), where n is the number of moles of gas, R is the gas constant, and Vf and Vi are the final and initial volumes, respectively. This is the formula that was used in the attempt at a solution provided.

The second step is an adiabatic process, which means that there is no exchange of heat between the gas and its surroundings. This does not mean that there is no change in entropy, as there can still be changes in temperature and volume. The formula for the change in entropy for an adiabatic process is ΔS = nCv ln(Tf/Ti), where Cv is the heat capacity at constant volume and Tf and Ti are the final and initial temperatures, respectively. However, in this case, we are not given any information about the temperature change, so we cannot calculate the entropy change for this step.

To find the net change in entropy, we can use the principle of entropy additivity, which states that the total change in entropy for a process is equal to the sum of the changes in entropy for each step. Therefore, the net change in entropy for this process would be ΔS = ΔS1 + ΔS2, where ΔS1 is the change in entropy for the first step (isothermal expansion) and ΔS2 is the change in entropy for the second step (adiabatic contraction). However, since we do not have enough information to calculate ΔS2, we cannot determine the net change in entropy for this process.

In conclusion, as a scientist, I would explain that the net change in entropy for this process cannot be calculated without knowing the temperature change during the adiabatic step. However, we can still calculate the