Carnot Engine cycle on PV Diagram

[Sorelle]

I posted this in the engineer / comp science thread, but I've had no one reply or help. I really could use some guidance and I don't know where else to post.

Homework Statement

I am tasked to create a PV Diagram of a Carnot Engine Cycle. I must find pressure, volume, Q, W, ΔU, and ΔS on all four points. This is what has been given to me by my teacher:

https://lh6.googleusercontent.com/LupjMh6SJxv2M4loxPxaSIMEaU6wtMeMfz0_jjY6dgyyEewHwdgzplahrV0Xidfkjpn4dfu3h0wcjoc=w1920-h1021
T_Cold = 300 K
T_Hot = 1700 K
p_c = 1.01*105 Pa
v_c = 0.01 m3
Q_{a to b} = 300 J
γ (gamma) = 1.40

Homework Equations

(1) p₁v₁=p₂v₂
(2) W=nRT ln(V₂/V₁)
(3) p₁v₁^γ=p₂v₂^γ
(4) W= p1v1-p2v2 / (γ-1)
(5) T₁V1^(γ-1) = T2V2(γ-1)
(6) W = nRT ln(V2/V1)
(7) pv = nRT

The Attempt at a Solution

Using the above equations I managed to get point b's pressure and volume. What I got for point b:
vb = 0.028 m3
pb = 2.389*104 Pa
First, I used the gas law equation (7) to get moles. This came out to n = 0.405 moles. Then, I used equation (5) to get volume, solving for V_b^(y-1). I then used equation (3) to get pressure, solving for pb. I then got W = -852.7 J for the path b to c using equation (4). This seems kind of odd to me and I'm not sure if its correct because it's doing more work than the amount of heat it is providing. I assume Q_{a to b} = is Q_H ? It seems so low though.

I'm trying to figure out how am I going to get points d and a without knowing pressure or volume on those points. A class mate had suggested I used equation (5) for a to b and c to d, but those paths are isothermal. Isn't equation (5) adiabatic only? I don't know if I can even use the gas law because I'd need pressure or volume and I don't have that.

There is also the idea that I have to use equation (6) and solve for volume that way. The problem is I don't know where to start. I'm a little rusty on my calculus (it's been about 4 years). I tried to break it down to W = nRT(ln (V2) - ln (V1))= nRT( 1/V2 - 1/V1), but this doesn't seem to help and I may have done it wrong.

I'd be grateful for any kind of help.

 

[vela]

Did you mean $Q_{A \to B} = 300\text{ J}$? 300 K doesn't make sense.

Try figure out what you easily can from the given information. For example, what's the efficiency of the engine? If you know that, you can figure out $Q_{C to D}$ and W and so on.

 

[Sorelle]

Did you mean $Q_{A \to B} = 300\text{ J}$? 300 K doesn't make sense.

Try figure out what you easily can from the given information. For example, what's the efficiency of the engine? If you know that, you can figure out $Q_{C to D}$ and W and so on.

Yeah, sorry I had the wrong unit there (fixed it).

I could try using e = T_h - T_c / T_H to get the efficiency and then use e = W / Q_H in some way to get volume or pressure perhaps.

I wonder if perhaps I have the temperatures mixed up. and T_c should be c → d ?

 

[vela]

You can figure out Q_C from e and Q_H.

On each leg, either Q_{i to f}=0 (adiabatic) or ΔU_{i to f}=0 (isothermal). Take C to D. It's an isothermal process, so you know the internal energy doesn't change. The first law then tells you that W(C to D) = ±Q_C (I'm not sure what sign convention your class follows). Well you found Q_C earlier, so you know W(C to D), so you can figure out V_B using (6). Then use the ideal gas law to find p_B. Just keep going like that around the cycle.

 

[Sorelle]

Thank you you've been very helpful.

I have just 1 last question though. How can I get V_B from ln (V₂/V₁) ? would e^W/nRT / V₂ = V₁ work?

 

[vela]

Close. The units don't work out there. You have $\rm{m^{-3}}$ on one side and $\rm{m^3}$ on the other. $V_1$ should be some multiple of $V_2$. Other than the algebra mistake, though, you have the right idea.