# Don't have point B (cycle process)

Doc Al
Mentor
Well, what's the integral of $x^n$? Of $x^{-n}$?

Gold Member
Well, what's the integral of $x^n$? Of $x^{-n}$?
$$\int x^n= \frac {x^{n+1}} {n+1}$$ n not = -1

$$\int x^{-n}= \frac {x^{-n+1}} {-n+1}$$ as long as x not = 1 but I think this one is wrong though.

Thanks Doc

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Doc Al
Mentor
All perfectly correct! If you think the second one is wrong, check it by taking the derivative of the answer.

Once you are convinced that both are correct, compare the second one with the integral you need for this problem.

Gold Member
All perfectly correct! If you think the second one is wrong, check it by taking the derivative of the answer.
okay I did and the other with the negative is fine.

Once you are convinced that both are correct, compare the second one with the integral you need for this problem.
$$W= -\int p dV= C \int \frac{dV} {V^{\gamma}}$$
I was wondering if there was supposed to be a (-) in the second part above. (the example I saw on the net didn't but why is that?)

$$\int x^{-n}= \frac {x^{-n+1}} {-n+1}$$

$$\int x^{-n}= \frac {x^{1-n}} {1-n}$$ $rearranged$

$$W= \frac{C(V_f^{1-\gamma}- V_i^{1-\gamma})} {1-\gamma}$$
well the last thing is that I wasn't getting how they had Vf and Vi but I figured it out that it's because of the initial and final Volumes BUT if the pressure is not constant then why don't we include Pi and Pf??

Thanks Doc

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Doc Al
Mentor
Since $PV^\gamma = C$, P and V are not independent variables. We chose to evaluate $\int Pdv$ using the variable V. But we could easily have changed variables and wrote it in term of P, in which case we would have used Pi & Pf.

Gold Member
Since , P and V are not independent variables. We chose to evaluate using the variable But we could easily have changed variables and wrote it in term of P, in which case we would have used Pi & Pf.

I forgot AB for isothermal process

For the part AB It is isothermic

T= constant
$$T_A= T_B$$

$$\Delta U= nR\Delta T= 0$$ internal Energy which is = 0

$$Q=-W$$

$$P_A= 8.00x10^4 Pa$$
$$P_A= 2.00 m^3$$

$$W= nRT ( \frac{V_A} {V_B})$$

thanks

Doc Al
Mentor
$$W= nRT ( \frac{V_A} {V_B})$$
Almost. You're missing a natural log.

Gold Member
Since $PV^\gamma = C$, P and V are not independent variables. We chose to evaluate $\int Pdv$ using the variable V. But we could easily have changed variables and wrote it in term of P, in which case we would have used Pi & Pf.
Okay I think I get it.

Almost. You're missing a natural log.
$$W= nRT ln ( \frac{V_A} {V_B})$$

$$W= -\int p dV= C \int \frac{dV} {V^{\gamma}}$$

I think I get how it gets to this $$W= \frac{C(V_f^{1-\gamma}- V_i^{1-\gamma})} {1-\gamma}$$

is it negative because you are taking the integral of $$\frac{1} {V^{\gamma}}$$ so it is actually $$V^{-\gamma}$$ ?

if it is then why did you put a negative in the integral at the begining?

I know it's like the neverending thread, so I promise that that was my last question.

Thank You Doc Al.

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Doc Al
Mentor
$$W= -\int p dV$$
If you're asking about the negative sign in this equation, it's there because you want the work done on the gas. $\int p dV$ by itself is the work done by the gas.

Gold Member
If you're asking about the negative sign in this equation, it's there because you want the work done on the gas. $\int p dV$ by itself is the work done by the gas.
but if there is work that is done by the gas (vol increasing) then it should be + then ? (in this case)

I understand it as if the vol increases then there is work doen by the gas.

If the volume decreases then there is work on the gas.

Thanks.

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Doc Al
Mentor
$\int p dv$ is the work done by the gas. When the volume increases, it's positive.

$-\int p dv$ is the work done on the gas. When the volume increases, it's negative.

Either one is fine, depending upon the sign convention used in your course. In your first post you wrote:
$$\Delta U= Q + W$$
That tells me you want the second version, the work done on the gas.

Gold Member
$\int p dv$ is the work done by the gas. When the volume increases, it's positive.

$-\int p dv$ is the work done on the gas. When the volume increases, it's negative.

Either one is fine, depending upon the sign convention used in your course. In your first post you wrote:

$$\Delta U= Q + W$$

That tells me you want the second version, the work done on the gas.
:uhh: I've gotten myself confused...I looked up this equation online and my teacher gave the class another equation and my book has the same one that I first posted.

so the only difference would be that it would be +/- in the end so if I made the calculations with in on the gas then the result would be (-).

but if the volume decreases then the work would be possitive

sooo, I've decided to use the one done on the gas.

Thanks very much for your explanation.