Thermodynamics, Heat transfer question

AI Thread Summary
The discussion revolves around a thermodynamics problem involving heat transfer, where a specific volume of air is heated at constant pressure and then allowed to expand. The user seeks guidance on calculating the final temperature, work transfer, and change in internal energy, expressing uncertainty about the initial and final parameters. Clarifications on the adiabatic process and the ideal gas law are provided, emphasizing the need to derive the final temperature using known equations. The conversation highlights a typo regarding the gamma coefficient, which is discussed in the context of its theoretical limits. Overall, the user is encouraged to apply familiar thermodynamic principles to solve the problem effectively.
Mathy
Messages
7
Reaction score
0
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!
 
Physics news on Phys.org
Could you clarify on the "law pv power of 12 = a constant" part? Does this mean you have

PV^1^2 = constant
 
Yes indeed

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

PV^1^2 = constant



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):

\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.
 
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"

\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.
 
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 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
 
Back
Top