Thermodynamics, Heat transfer question

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Mathy
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I have done a series of thermodynamics questions covering heat transfer, internal energy, temperature pressure etc. I have a new one but I am 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 don't know
temperature two/final temp or volume 2. that's why I am 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!
 
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Yes indeed

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

[tex]PV^1^2 = constant[/tex]



Yes that's 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. I am not sure because i don't know temp final but i also don't seem to know v2 either.
 
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):

[tex]\Delta U = Q - W[/tex]

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

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:

[tex]PV^\gamma = constant[/tex]

Combining with the ideal gas law, you can derive:

[tex]T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}[/tex]

From the previous step you should have the initial temperature and volume. The final volume is given.
 
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
 
I'm almost certain that "12" is a typo for "1.2"

[itex]\gamma=12[/itex] is theoretically impossible.

[tex]\gamma = C_p/C_v = 1 + \frac{R}{C_v}[/tex]

But the equipartition theorem tells us that [itex]C_v = nR/2[/itex], where n is an integer denoting the number of degrees of freedom that contribute to the internal energy. So, [itex]R/C_v = 2/n[/itex] and can be at most 2, so [itex]\gamma[/itex] can be no larger than 3.
 
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 energy stated at the start 3.4 MJ = 3400KJ so Q=U+3400KJ then i find change in internal energy 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 don't know v2 for process one or t2, are either of them just the same as the v1 or v2?
 
ps yes its right there was a typo its was pv 1.2 not 12