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## Homework Statement

Consider an ideal monatomic gas that undergoes an adiabatic free expansion starting from equilibrium state A with volume 500 cm

^{3}, pressure 40 kPa and temperature 300K to state B, which has a final equilibrium volume of 1000 cm

^{3}.

- Construct an reversible isothermal path that joins the initial and final states and show it on a schematic p-V diagram.
- Calculate the total entropy change between the two states using the reversible path.
- Would you expect this to be the same as the entropy change for the irreversible free expansion?

## Homework Equations

1) P

_{1}*V

_{1}= P

_{2}*V

_{2}

ΔT for free adiabatic expansion = 0

2) ΔS = nRln(V

_{f}/V

_{i})

3) Ideal Gas Law pV=nRt

## The Attempt at a Solution

1) 40*500 = P

_{2}* 1000

P

_{2}= 20 kPa

I've attached a P-V diagram. What I'm failing to understand is how i draw a reversible isothermal path that joins the states. If ΔT in an adiabatic free expansion = 0, and then I'm asked to draw a reversible isothermal path that joins the initial and final states together then wouldn't this be the exact same path because ΔT for isothermal path = 0?

2) P

_{1}= 40,000Pa

V

_{1}= 0.0005m

^{3}

T = 300 K

R (gas constant) = 8.3144

∴ = PV / RT = n = 0.008018217

ΔS = nRln(V

_{f}/V

_{i})

ΔS = (0.008018217)*(8.3144)* ln(0.0005/0.001)

= - 0.046209810 J/K

3) The entropy change in the adiabatic free expansion is the same as in question two, only it is positive because entropy is increasing.