# Homework Help: Thermodynamics, Heat transfer question

1. Nov 22, 2005

### Mathy

I have done a series of thermodynamics questions covering heat transfer, internal energy, temperature pressure etc. I have a new one but im unsure how to start it, its unlclear whether i know certain things. I can do the question form looking back at pervious questions, if i knew how to start.

0.36m cubed of air at a pressure of 1.1MN/m2 and 339k is given an energy of 3.4MJ by means of heating at a constant pressure. The air is then allowed to expand to a volume of 1.44m3 according to the law pv power of 12= a constant.

For each process calculate the final temperature, the work tarnsfer and the change in internal energy.

I am assuming the question is two parts, and therefore i dont know
temperature two/final temp or volume 2. thats why im unsure how to get the final volume

how would i find the final temperature in order to carry out the rest of the question?

any help whould be great!

2. Nov 22, 2005

### mezarashi

Could you clarify on the "law pv power of 12 = a constant" part? Does this mean you have

$$PV^1^2 = constant$$

3. Nov 23, 2005

### Mathy

Yes indeed

Yes thats it. but i also made a mistake with my phrasing, i want to know the final temperature, not the final volume. The question asks for the final temperature. Im not sure because i dont know temp final but i also dont seem to know v2 either.

4. Nov 23, 2005

### mezarashi

I'd just like to note that I've never seen such a high gamma coefficient before :P
Assume the ideal gas law holds at all times.
For the first process, which is isobaric, we can write (as we always can):

$$\Delta U = Q - W$$

Q is given, and $$W = \int P dV$$
where P is constant making it quite easy. The following calorimetric equation also holds at constant pressure $$Q = nC_p \Delta T$$

Using the ideal gas law, you can then find all the other parameters.

Implicitly stated in the second process is that it is adiabatic, so that:

$$PV^\gamma = constant$$

Combining with the ideal gas law, you can derive:

$$T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$$

From the previous step you should have the initial temperature and volume. The final volume is given.

5. Nov 23, 2005

### Mathy

Thanks should help

Yes those all seem familiar, i have used them in questions but i am still not familiar with the thermodynamics physics enough to tackle any question straight off. i know enough once i get started, so thanks i think that will be fine. I let you know how i get on

6. Nov 24, 2005

### Gokul43201

Staff Emeritus
I'm almost certain that "12" is a typo for "1.2"

$\gamma=12$ is theoretically impossible.

$$\gamma = C_p/C_v = 1 + \frac{R}{C_v}$$

But the equipartition theorem tells us that $C_v = nR/2$, where n is an integer denoting the number of degrees of freedom that contribute to the internal energy. So, $R/C_v = 2/n$ and can be at most 2, so $\gamma$ can be no larger than 3.

7. Nov 30, 2005

### Mathy

I now know that for the first part i need to get final temperature then put it into u=mCVdt dt=(t2-t1) cv=0.718 then Q=U+W I think w is simply the enrgy stated at the start 3.4 MJ = 3400KJ so Q=U+3400KJ then i find change in internal enrgy with:
can anynone tell me whether this is right. I am still unsure how to get the final temperature for process one too. I am struggling as it seems i dont know v2 for process one or t2, are either of them just the same as the v1 or v2?

8. Nov 30, 2005

### Mathy

ps yes its right there was a typo its was pv 1.2 not 12