What is the entropy change of gas during an internal combustion engine test?

[sparkle123]

Homework Statement

During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 C was compressed suddenly (and irreversibly) to 0.500L by driving in a piston. In the process the temperature of the gas increased to 28.1 C. Assume ideal behavior. What is the change in entropy of the gas?

Homework Equations

dS=nRln(V2/V1)
dS=Cln(T2/T1)
C=5R/2

The Attempt at a Solution

I think that you need the number of moles of gas to be able to solve this...
The answer provided is -14.6 J/K but how do you do this? Please help!

 

[Apphysicist]

I would use your second equation, noting that the specific heat with constant pressure = 5R/2. Since R is a constant and the two temperatures are given, you can solve that without knowing n. Don't forget to convert to Kelvin, and do note that ds in your equations is not a differential change in entropy, but rather a (delta-s) change in entropy.

 

[sparkle123]

So I don't need to take into account the change in volume?
Thanks!

 

[Apphysicist]

So I don't need to take into account the change in volume?
Thanks!

I don't think so, but then, I'm rusty with some of this thermo stuff.

 

[LeonhardEuler]

You do need to take account of both. Since entropy is a state function, you can imagine any process from state A to state B, and the entropy change will be the same. So you can imagine first an isothermal compression from 3.0L to 0.5L, followed by a change in temperature from 18.5 to 21.8C at constant volume. When do those formulas you gave apply?

 

[gneill]

I plugged the numbers in as follows:

$$c_v = 20.81 \frac{J}{K\ mol}$$

$$R = 8.314472 \frac{J}{mol\ K}$$

$$dS = c_v ln(\frac{T2}{T1}) + R\ ln(\frac{2}{V1}) = -14.22 \frac{J}{mol\ K}$$

The result is close to, but not the same as, the supposed solution value. The solution also seems to have mucked up the units.

A reasonable reference sheet for this stuff is http://www.grc.nasa.gov/WWW/K-12/airplane/entropy.html"

 

[LeonhardEuler]

I plugged in the same values and got a different answer. In your formula, you have an "ln(2/V1)". I assume you mean ln(V2/V1) which is ln(0.5/2), right?

 

[LeonhardEuler]

I think I found your error. You forgot to convert Celcius to Kelvin when you did T2/T1, didn't you? You can only use an absolute temperature any time you're dividing temperatures in thermodynamics.

 

[gneill]

I plugged in the same values and got a different answer. In your formula, you have an "ln(2/V1)". I assume you mean ln(V2/V1) which is ln(0.5/2), right?

Right.