How to Determine the Change in Entropy of a Monoatomic Ideal Gas?

[chocolatepie]

Homework Statement

A monoatomic ideal gas is initially trapped in a 0.500L container at 298K and 100kPa. The sample is then allowed to expand to 1.00L and the temperature is increased to 373K. Determine change of the entropy (ΔS)

Homework Equations

ΔS = ∫(T1 to T2) dqrev/T

The Attempt at a Solution

Is the pressure constant?

 

[I like Serena]

Hi chocolatepie!

I'm drawing the conclusion that, yes, pressure is constant.

And I think you need better relevant equations than you have.
Perhaps you can find some on:
http://en.wikipedia.org/wiki/Table_of_thermodynamic_equations

 

[chocolatepie]

Thank you for start replying my question. I REALLY appreciate it!
Ok. what I would do is to assume that pressure is constant, so I would use the formula I posted earlier.

C p,m = 3/2R
C p = 3/2R x change in mole (determined by PΔV=ΔnRΔT)
which would give me 0.9989 J/K

Then I would find the change of the entropy by integrating and it gave me:
ΔS = 3/2R (ln373-ln298)
= 2.8 J/K

Yes?

 

[I like Serena]

Hmm, you've got C p,m wrong (check the table).

I'm afraid there is no change in mole, but you do need to calculate the number of moles.
How did you get that formula?

 

[chocolatepie]

Hmm, you've got C p,m wrong (check the table).

I'm afraid there is no change in mole, but you do need to calculate the number of moles.
How did you get that formula?

Uh.. I wrote 3/2R because I was using the equipartition theorem. Since the monoatomic ideal gas does not have neither rotational nor transitional energy, 3/2RT. Then I sub this into heat capacity formula where T would get cancelled.

I am not even sure if I am doing this question right.. confused..

 

[I like Serena]

Uh.. I wrote 3/2R because I was using the equipartition theorem. Since the monoatomic ideal gas does not have neither rotational nor transitional energy, 3/2RT. Then I sub this into heat capacity formula where T would get cancelled.

I am not even sure if I am doing this question right.. confused..

Oh, you're doing the question right.
There's just some minor stuff to take care of...

Let's start with the proper relevant formulas, which you seem to have left out.

You should have:
1. PV=nRT for an ideal gas.
2. Cv,m = (3/2)R for a monatomic ideal gas (molar heat capacity at constant volume).
3. Cp,m = (5/2)R for a monatomic ideal gas (molar heat capacity at constant pressure).
4. ΔS = Cp,m ln(T2 / T1) for 1 mole of an ideal gas at constant pressure.

See what I mean?
You should be able to find these on the wiki page I mentioned.

From (1) you can calculate the number of moles n...

 

[chocolatepie]

Oh OK. I see what you mean now

But when determining the number of mole using PV=nRT, how do I know which T or which V I need to use? I have 2 values (for example, V1 and V2, etc). This is actually why I used the change (Δ).

 

[I like Serena]

Hmm, let's organize this a bit:

Darn! I'm just noticing that the process is probably not at constant pressure!
I'll have to think about that.

\begin{array}{| c | c | c |}<br /> \hline \\<br /> &amp; \textrm{Before (1)} &amp; \textrm{After (2)} \\<br /> \hline \\<br /> V &amp; 0.500 \textrm{ L} &amp; 1.00 \textrm{ L} \\<br /> T &amp; 298 \textrm{ K} &amp; 373 \textrm{ K} \\<br /> P &amp; 100 \textrm{ kPa} &amp; ? \\<br /> n &amp; ? &amp; ? \\<br /> S &amp; S_1 &amp; S_2, \Delta S = ? \\<br /> \hline<br /> \end{array}

Can you fill in the question marks?

 

[I like Serena]

And I'll bring in a new formula (always true for an ideal gas):

\Delta S = n C_{v,m} \ln{T_2 \over T_1} + n R \ln{V_2 \over V_1}

 

[chocolatepie]

And I'll bring in a new formula (always true for an ideal gas):

\Delta S = n C_{v,m} \ln{T_2 \over T_1} + n R \ln{V_2 \over V_1}

Whoa!
I got:
n= 0.02 for both initial and final
P2 = 62 kPa
ΔS = 0.12 J/K

However, I did not learn that formula.. I've seen the first and second part of the equation separately but not together like that. Shouldn't I use the second part of the equation for the reversible isothermal expansion? This was not isothermal though?

 

[I like Serena]

Whoa!
I got:
n= 0.02 for both initial and final
P2 = 62 kPa
ΔS = 0.12 J/K

Looks good!

However, I did not learn that formula.. I've seen the first and second part of the equation separately but not together like that. Shouldn't I use the second part of the equation for the reversible isothermal expansion? This was not isothermal though?

Ah well, I was looking for a general equation, but couldn't find it quickly enough, so I derived it myself.
It's probably on the internet somewhere.

Here's the derivation if you're interested.

\def\d{\textrm{ d}}
From \d U=T \d S - P \d V (always true!) we find:
\d S = {\d U \over T} + {P \d V \over T}

With U = n C_{v,m} T:
\d S = {n C_{v,m} \d T \over T} + {P \d V \over T}

With P V = n R T we also have {P \over T} = {n R \over V} implying:
\d S = {n C_{v,m} \d T \over T} + {n R \d V \over V}

Integrate, and we get:
\Delta S = n C_{v,m} \ln {T_2 \over T_1} + n R \ln {V_2 \over V_1}

 

[chocolatepie]

Thank you so much for your help!

I looked through my notes, and I realized that I actually had that formula LOL.
Anyway, thank you again!

Take care!