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