Thermo dynamics, entropy based problem

[MightyG]

Homework Statement

A rigid vessel of total volume 0.5m3 contains 1kg of air at 15°C initially
constrained in a volume of 0.06m3 by a diaphragm. The remaining volume is
completely evacuated (pressure = 0). If the diaphragm is burst, allowing free
adiabatic expansion of the air, determine the change of entropy of the air and
of the surroundings. If instead of a free expansion, the process had been a
reversible expansion between the two states, what would then be the change
of entropy of the air, and of the surroundings? Compare the net change of
entropy for the system and surroundings in the two cases. Surrounding
temperature 15°C .
Answer: +0.609, 0, +0.609 – 0.609 kJ/K

Homework Equations

[tex]\Delta[/tex]S=C_v.Ln(P2/P1) + C_p.Ln(V2/V1) (1)

[tex]\Delta[/tex]S=C_v.Ln(T2/T1) + R.Ln(V2/V1) (2)

[tex]\Delta[/tex]S=C_p.Ln(T2/T1) - R.Ln(P2/P1) (3)

[tex]\Delta[/tex]S=C_v.Ln(P2/P1) - C_p.Ln(V2/V1) (4)

The Attempt at a Solution

I can work out the first part of the problem easy enough taking the expansion as being isothermic so (2) becomes [tex]\Delta[/tex]S=R.Ln(V2/V1) but I am stumpped by the second part.

im told that the Eq becomes;

Q = [tex]\Delta[/tex]u + W since there is no heat transfer to the system it becomes Q=W=m.R.t.Ln(V2/V1) but I've got no idea where this Eq comes from.#

EDIT:

Is this just a case of W=[tex]\int[/tex]PdV and since Pv=mRt so P=(mRt)/v

then W=[tex]\int[/tex][tex]\frac{mRt}{V}[/tex] dV

= mRt[tex]\int[/tex][tex]\frac{1}{V}[/tex] dV

 

[MightyG]

ok, just worked that through and I get the right magnitude for the entropy change but I don't understand why the answer is negative.

my working:

[tex]W = mRt\int^{2}_{1}\frac{1}{v}dv = mRt [Ln v]^{2}_{1} = 1*0.287*288*(Ln0.5 - Ln0.06) =175.253[/tex]

so,

[tex]\Delta S = \frac{Q}{T} = \frac{175.253}{288} = 0.609Kj[/tex]

but acording to the answers it should be -0.609Kj?

 

[piercas]

= mRt.Ln(V) + C

= mRt.Ln(V2/V1) since C would be 0 in this case.

And since Q=W and Q=0 then W=0 so \DeltaS=0.

In the first part of the problem, the change in entropy of the air and surroundings is calculated using the equation for isothermal expansion (2). This assumes that the expansion is reversible, meaning that the pressure inside the vessel is always equal to the external pressure. However, in the second part of the problem, the expansion is not isothermal and is instead adiabatic, meaning that there is no heat transfer between the system (air) and surroundings. This changes the equation for the change in entropy, which now includes the work done on the system (air) during the expansion.

The equation for this change in entropy is given by:

\DeltaS = \frac{Q}{T} + \frac{W}{T}

Since there is no heat transfer (Q=0) and the temperature remains constant (T=15°C), the equation simplifies to:

\DeltaS = \frac{W}{T}

The work done (W) in this case is given by the integral:

W=\int PdV

Since the pressure inside the vessel is initially 0 and then increases as the air expands, the work done on the system is in the form of compression work. This is given by:

W = -\int_{V1}^{V2} PdV

Using the ideal gas law, P=(mRt)/V, we can substitute this into the equation for work:

W = -\int_{V1}^{V2} \frac{mRt}{V}dV

Integrating this equation gives:

W = mRt\ln\left(\frac{V2}{V1}\right)

Substituting this into the equation for change in entropy gives:

\DeltaS = \frac{mRt\ln\left(\frac{V2}{V1}\right)}{T}

Since T=15°C and R is a constant, we can simplify this equation to:

\DeltaS = m\ln\left(\frac{V2}{V1}\right)

This is the change in entropy of the air during the adiabatic expansion. To calculate the change in entropy of the surroundings, we can use