Heat Engines, pV-Diagrams, & Adiabatic Changes

Today was Wednesday.... QuizDay!!!! Yay!!

Our quiz focused on analyzing a pV-diagram of a Heat Engine.

• Find the work done in the isobaric process from A to B (part 1).

In an isobaric (constant pressure) process, the work done by the system is given by this equation:

W=p(V_2 - V_1)

The work can also be found by finding the area under the curve (or in this case, line).

Since the change in volume was a positive one (i.e. V_2 > V_1) the work was positive.

• Find the work done in the isochoric process from B to C (part 2).

In an isochoric (constant volume) process, there is no work done. This is because Work=Force*Distance, and since the distance is 0, so is the work.

This is also apparent when you look at the graph. Since this portion has constant volume, the graph shows a vertical line, with the area under the line = 0. Therefore the Work = 0.

• Find the work done in the isobaric process from C to D (part 3).
  

We found work using the same formula we used in part 1. Except for in part 3, V_f <>

• Find the work done in the isochoric process from D to C (part 4).
  

This constant volume segment (same as part 2) of the process also produced no work.

• Find the net work done in the whole system.

The trick to this part was making sure the signs were correct in terms of positive and negative work being done.

Remember, in thermodynamics:

Work is (+) when work is done by the system against its surroundings (the system expands), and energy is leaving the system

Work is (-) when work is done on the gas by its surroundings (during compression of a gas), and energy is entering the system.

After the quiz we gathered in groups of three based on alphabetized elements from the periodic table. We then observed a heat engine first-hand. This was a cylinder and piston (both made of glass) connected by a tube (to allow air flow) to a metal spherical ball (metal in general is an excellent heat conductor).

First the ball was placed in a cold reservoir (an ice box) so the air inside the chamber and the tube cooled, reducing the pressure. Then the ball was placed in a large beaker of hot water. We watched in amazement as the glass piston was mysteriously moved from it's initial position as the volume inside the chamber expanded. This proved to us that work was done by the system due to the changing temperature and volume.

After the experiment, we analyzed a pV-diagram with a cycle that was similar, if not identical to the one on the quiz. We determined which part of the cycle is work done on the gas and by the gas. We also determined which parts of the cycle was heat transferred to the gas and from the gas.
Then we used the equation E_int = (3/2)pV to calculate the internal energy at each of the four points and then the change in internal energy for each part of the cycle. We then calculated the work done using the equation W=p(V_2-V_1) for the isobaric portions and 0 work for the isochoric portions. From this we calculated Q (heat energy) using the equation Q = (change in)E_int + W.

In the next activity, we derived several relationships for adiabatic expansions.

First, we used pV=nRT and the relationship C_p-C_v=R. With some algebraic manipulation, we showed 

n(delta)T= [P(delta)V+V(delta)P] /C_p - C_v

Then we used the fact that Q=0 in an adiabatic expansion and the equation (delta)Eint=C_vn(delta)T to show

n(delta)T=-P(delta)v/C_v

We then used these two new equations to show that

(delta)P/P+(C_p/C_v)((delta)V/V)=0

After this, we showed that the limit of very small changes could be integrated into this equation:

ln(P_i)+(C_p/C_v)ln(V_i)=ln(P_f)+(C_p/C_v)*ln(V_f)

With some more mathematical manipulations, we showed that

(P_f)*(V_f)^(gamma)=(P_i)(V_i)^(gamma)

Then we used this in conjuction with the ideal gas lawto show that

(T_f)(V_f)^(gamma - 1)=(T_i)(V_i)^(gamma - 1)

The last derivation of the day gave us the Work in an Adiabatic Expansion:

W_adiabatic={[(P_i)(V_i)^(gamma)][(V_f)^(1 - gamma) - (V_i)^(1 - gamma)]/(1 - gamma)

Then we used this equation to calculate the work done in an adiabatic expansion given the initial and final values of pressure and volumes.

BAM!!!