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