# Homework Help: Thermodynamics (Adiabatic PV Diagram)

1. Apr 8, 2010

### alexgmcm

1. The problem statement, all variables and given/known data

Give a definition of an adiabatic process. Derive an expression for the work performed by n mol of an ideal gas undergoing an adiabatic expansion in terms of the initial and the final temperature of the gas. [Hint: you may find it helpful to apply the First Law to the process.]

Give a definition of an isothermal process. Derive an expression for the work performed by n mol of an ideal gas undergoing a quasistatic isothermal expansion in
terms of the initial, and the final, volume of the gas.

One mol of an ideal monatomic gas is contained in a cylinder at an initial temperature of 300 K. The gas is subjected to a quasistatic two-step process. In the first step, the gas expands adiabatically until its volume reaches 5 times the original volume. During the second step the gas is compressed isothermally back to its original volume.

(a) Sketch this two-step process on a pressure-volume diagram.
(b) Calculate the temperature at the end of the first (adiabatic) step.
(c) Calculate the work performed by the gas during the first (adiabatic) step.
(d)Calculate the work performed by the gas during the second (isothermal) step.
(e) Calculate the change in the internal energy for each step of the process.

[The ratio of the molar heat capacities at constant pressure and constant volume for an ideal monatomic gas is 5/3]

2. Relevant equations
Adiabatic condition: PV$$^{\gamma}$$ = K (a constant) I seriously can't remember the derivation for this.
Therefore via more magical derivation: Work done during adiabatic expansion:
$$W = \frac{1}{1 - \gamma} \left(P_{1}V_{1} - P_{2}V_{2} \right)$$

And for Isothermal, $$W = nRT ln \left(\frac{V_{2}}{V_{1}} \right)$$ which I can remember the derivation for :)

3. The attempt at a solution

This is an old exam question I am doing for practice purposes for which only numerical solutions are provided which really doesn't help me actually understand the problem. I understand that an adiabatic process is one in which Q = 0 as there is no heat transfer (e.g. the process occurs instaneously so no heat can transfer or something similar) but I have trouble deriving an expression for the work performed by it in terms of temperature.

The Isothermal one is just a process at constant temperature and I can do the derivation for the work easily. It's just adiabatic processes seem to be the spawn of Satan.

For the sketch I drew the Isothermal process as following the familiar isotherm shape and the Adiabatic process as following a curved line that isn't as steeply curved as the isotherm if that makes sense.

I can't do b) or even see any possible way one might approach it, which probably stems from my difficulty with deriving the equation at the start and with adiabatic processes in general. I also can't do c) although it seems easy enough if you just replace the PV in the work equation with nRT and use the T found in b)? (which I don't have as I can't do b).

d) seems easy enough but once again you need T from b) :(

I guess e) would just use the fact that a closed loop has no change in internal energy?

It's really my problem with b) that is stopping me here, any advice on adiabatic processes generally would be greatly appreciated though. Especially as I have seen the derivation of the adiabatic condition asked for on previous papers.

b) T = 102.6 K
c) W = 2461.8 J
d) W = -1372.9 J
e) Adiabatic dU = -2461.8 J , Isothermal dU = 0.

tl;dr: Any help on b) would be greatly appreciated.

2. Apr 10, 2010

### Andrew Mason

Substitute nRT/V for P in $PV^{\gamma} = K$. That will give you the adiabatic condition in terms of temperature and volume.

Use the above expression for adiabatic condition in terms of temperature and volume to find the temperature after an adiabatic expansion to 5 x initial V.

That is one way to do it. But all you really have to do is apply the first law. That shows that $W = -\Delta U$. What is $\Delta U$ in terms of $\Delta T$?

AM