# Homework Help: Basic thermodynamics

1. Jan 17, 2006

### KoGs

Hello everyone first time posting here.

I have a thermodynamics question that my teacher told me should be realy easy. But I am having a lot of difficulty on it. Any help would be appreciated.

For question 2 by simplying plugging in the information given I am able to get the energy of each system before equilibrium without any difficulty. And we know energy is conserved, so this total energy is the same at equilibrium. I'm kind of stuck what to do at this point.

My teacher told me to think of the forces at equilibrium. Ok so F net = 0 at equilibrium. My teacher said from there we should be able to say something about pressure. I'm not too sure exactly what that is.

I know at equilibrium we can write p(1)v(1)/n(1) = p(2)v(2)/n(2). But doesn't my n(1) and n(2) values change once we reach equilibrium? Also the volume levels of each system, as well as pressure?

Oops I guess my attachment is too big. I can't use superscripts on this keyboard so I will use T(1) to denote temperature of system 1. Ok here is the question:

Two systems of ideal gases have the following equations of state:

1 / T(1) = 3RN(1) / (2 U(1)), P(1) / T(1) = R N(1) / V(1).
1 / T(2) = 5RN(2) / (2 U(2)), P(2) / T(2) = R N(2) / V(2).

Where R = 8.314J / mol K is the universal gas constant. The mole number of the first system is N(1) = 0.5 mol and that of the second system is N(2) = 0.75 mol. The two systems are initially contained in a closed cylinder, separated by a rigid adiabatic wall. The initial temperatures are T(1) = 200 K and T(2) = 300 K, and the total volume is 20 litres. Subsequently, the wall is replaced by a freely movable diathermal piston and the total system is allowed sufficient time to come to equilibrium.
a) What is the energy and volume of each system in equilibrium?
b) What is the pressure and temperature in equilibrium?

Thanks for any help.

Last edited: Jan 17, 2006
2. Jan 17, 2006

### Mindscrape

So, wait, I want to make sure I got this right, you have two containers that are separated by a wall that (ideally) keeps the two systems completely separate, and then they come together and it wants to know the stats of the new equilibrium? Also, the equations are as follows (what is U?):
$$\frac{1}{T_1} = \frac{3RN_{1}}{2U_1}$$

$$\frac{P_1}{T_1} = \frac{RN_1}{V_1}$$

$$\frac{1}{T_2} = \frac{5RN_{2}}{2U_2}$$

$$\frac{P_2}{T_2} = \frac{RN_2}{V_2}$$

I don't think force has anything to do with it.

Last edited: Jan 17, 2006
3. Jan 17, 2006

### KoGs

Yes that is what the question is. And those are the equations. U is energy.

4. Jan 17, 2006

### KoGs

Last edited: Jan 17, 2006
5. Jan 17, 2006

### KoGs

6. Jan 18, 2006

### Mindscrape

Okay, sorry that I didn't get back to you in time, or I hope that you eventually got it. The two systems start with different temperatures and are not in thermal equilibrium. EneSo system 1 is monotomic gas, and system 2 is a diatomic gas. I'm used to E representing energy, but for you it would be that $$U_{1 i} =\frac{3}{2} n_1 R T_{1 i}$$ and $$U_{2 i} = \frac{5}{2} n_2 R T_{2 i}$$ and $$U_{tot} = U_1 + U_2$$. You seem to have gotten to this point well (part a), though I did not double check the actual number crunch.

For part b, all we need to find the heat entering or leaving each system which can be found by subtracting the final energies from the initial energies $$Q_1 = E_{1 f} - E_{1 i}$$ for system 1. (You should have an equation that would tell you how to solve for the final energies, I don't want to give you too many steps.) I'm pretty sure this qualifies as a constant-volume scenario, during the time for equilibrium at least (which is what we are focusing on and was part of the point of finding the new volume), which means that $$Q = n C_V \Delta T$$ where n is the number of moles and C_v is (3/2)R for monotomic gases and (5/2)R for diatomic gases if I remember right (though you really only need to solve one). Then manipulate the algebra and such to solve for change in temperature, and then use the change to find the final temperature.