Don't have point B (cycle process)

[~christina~]

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]

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

Almost. You're missing a natural log.

 

[~christina~]

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.

 

[Doc Al]

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.

 

[~christina~]

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.

 

[Doc Al]

\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.

 

[~christina~]

\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.

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.