# Don't have point B (cycle process)

1. Mar 29, 2008

### ~christina~

1. The problem statement, all variables and given/known data
One mole of helium is enclosed in a cylinder with a movable piston. By placing the cylinder in contact with various reservoirs and also insulating it at proper times, the helium performs a cycle. Compute the internal energy change, heat transferred, and work perfomed for each segment of the cyle and the total amount of each of these quantities for the entire cyle. Assume helium to be an ideal gas.

2. Relevant equations
$$\Delta U= Q + W$$

PV= nRT
$$Q= nC_p \Delta T$$

$$Q= nC_v \Delta T$$

$$\gamma= C_p / C_v = 1.67$$

$$C_v= 3/2R$$

$$C_p= 5/2R$$

3. The attempt at a solution

B is unknown but not sure how to find it...

I do have Parts:

A=> P= 8.00x10^4Pa, V= 2.00m^3

B=> P= ? V=?

C=> P= 3.00x10^4 Pa, V= 5.00m^3

D=> P= 3.00x10^4 Pa, V= 2.00m^3

For DA which is isovolumetric
w= 0

$$\Delta U= Q$$ for U

$$\Delta Q= nCv (T_D-T_A)$$ for finding heat (Q)

$$\Delta E_{int}= C_v \Delta T$$ => however I'm not given T so what do I do?

do I use PV= nRT and then solve for T since I have P and V ? so
$$\Delta P V /nR = \Delta T$$

For CD which is Isobaric
P= constant
$$W= -P(V_F-V_i)$$ for the internal E

$$\E_{int}= Q + W$$ for the work

$$PV= nRT$$

$$Q= nCP \Delta T$$

$$\E _{int}= ?$$ not sure about this

For the part AB It is isothermic

T= constant
$$T_A= T_B$$

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

so based on above $$Q= -W$$ to find the heat

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

$$W= nRT ( /frac{V_A} {V_B})$$ => don't have V for the final VB

For part BC

It's adiabatic so Q= 0 thus
$$\Delta U= W$$

point B is not known (P or V) initial point

C=> P= 3.00x10^4 Pa, V= 5.00m^3

$$P_iV_i^{\gamma} = P_iV_f^{\gamma}$$

not sure how to find the $$\Delta U$$ and that would also = W but if I don't find those then how can I find the W?

Point B

I don't have the P or V for point B so I'm not exactly sure how to find the info I need.

I think that for the $$\Delta U= 0$$

$$P_iV_i^{\gamma} = P_fV_f^{\gamma}$$

so I think I can find the V and P. But don't you have to be given $$\gamma$$ ? or is it a constant?

Well after finding the Pf I was thinking of using the fact that the AB is isothermic and plugging into this..

$$Q= W= P_AV_A ln (V_B/V_A)$$

Is this it..I think so but I think I'm missing some things here and there on actually solving this with numbers.

Could someone check it for me?

Thank you.

2. Mar 29, 2008

### Staff: Mentor

Here's a tip about finding point B. Point A and B are connected via an isothermal. Similarly, point C and B are connected via an adiabatic expansion. That should allow you to find P and V for point B.

3. Mar 30, 2008

### ~christina~

I don't know but I tried to find the volume for b by the equation that I used above but I'm not sure what gamma is and I still don't know how to find the pressure for b.

4. Mar 30, 2008

### Staff: Mentor

Treat the helium as an ideal monoatomic gas--that should tell you what gamma is.

Relating point A to point B (isothermal):
$$P_a V_a = P_b V_b$$

Relating point C to point B (adiabatic):
$$P_c V_c^{\gamma} = P_b V_b^{\gamma}$$

Two equations; two unknowns.

5. Mar 30, 2008

### ~christina~

okay so $$\gamma= 1.67$$

so I have the $$P_a= 8.00x10^4Pa$$ and $$V_a= 2.00m^3$$
so $$P_aV_a= P_bV_b= 1.6x10^5Pa*m^3$$

well for this $$P_cV_c^{\gamma} = (3.00x10^4 Pa)(5.00m^3)^{1.67}= 4.40x10^5 Pa*m^3= P_b V_b^{\gamma}$$

um.. what do I do now? I don't think I can substitute one into the other and or etc. Am I missing something else, Doc?

6. Mar 30, 2008

### Staff: Mentor

Why not? Use one equation to write $P_b$ in terms of $V_b$, then plug that into the second.

7. Mar 30, 2008

### ~christina~

okay, but I guess my algebra's :yuck: then

this is what I did.

PaVa=PbVb then

$$P_b =1.6x10^5Pa*m^3 /V_b$$ then I plug that into the other equation...

$$P_c V_c^{\gamma} = P_b V_b^{\gamma}$$

$$4.40x10^5 Pa*m^3= \frac{1.6x10^5Pa*m^3} {V_b} V_b^{\gamma}$$

and that would cancel the Vb? (I don't think so)

8. Mar 30, 2008

### Staff: Mentor

Recall this property of exponents:

$$\frac{X^a}{X^b} = X^{a-b}$$

9. Mar 30, 2008

### ~christina~

unless you mean that

it would be..
$$4.40x10^5 Pa*m^3= \frac{1.6x10^5Pa*m^3} {V_b} V_b^{\gamma}$$

and so based on the rule that I forgot...

$$4.40x10^5 Pa*m^3= (1.6x10^5Pa*m^3) (V_b)^{\gamma-0}$$

is this correct?

10. Mar 30, 2008

### Staff: Mentor

No.

$$\frac{X^a}{X} = \frac{X^a}{X^1} = X^{a-1}$$

11. Mar 30, 2008

### ~christina~

so it would be this...

$$4.40x10^5 Pa*m^3= (1.6x10^5Pa*m^3) (V_b)^{\gamma-1}$$

12. Mar 30, 2008

### Staff: Mentor

Now you're cooking. (Although I didn't check your numbers.)

Now you should be able to solve for $V_b$ (using a calculator, of course!).

13. Mar 30, 2008

### ~christina~

Okay so if that's like that then.

$$4.40x10^5 Pa*m^3= (1.6x10^5Pa*m^3) (V_b)^{\gamma-1}$$

$$2.75= Vb^{.67}$$
so
Vb= 4.52 m^3

and then since I have Vb

PaVa=PbVb

$$1.6x10^5 Pa m^3= Pb(4.52m^3)$$

$$Pb= 1.017x10^5Pa$$

um that looks funny though since the graph has Pa is 10^4...

14. Mar 30, 2008

### Staff: Mentor

Yeah, something's not right. Check your values for volume: Realize that the values on the diagram need to be muliplied by $10^{-3}$.

15. Mar 30, 2008

### ~christina~

so $$P_aV_a= P_bV_b= 1.6x10^2Pam^3$$

so then $$Pb= 1.6x10^2 Pam^3/Vb$$

$$P_cV_c^{\gamma} = (3.00x10^4 Pa)(5.00x10^-3)^{1.67}= 4.30 Pa*m^3= P_b V_b^{\gamma}$$

$$4.30 Pa*m^3= (1.6x10^2Pa m^3) (V_b)^{\gamma-1}$$

$$2.68x10^-2= V_b^{.67}$$

$$V_b= 4.52x10^-3 m^3$$

$$P_aV_a= P_bV_b$$

$$\frac{1.6x10^2Pam^3} {4.52x10^-3m^3} = P_b$$

$$P_b= 3.54x10^4 Pa$$

I think I got it.

Now to find the work what do I do? since I know that the point b) is both an isotherm and a adiabat.

Last edited: Mar 30, 2008
16. Mar 30, 2008

### ~christina~

you didn't say whether this was right or not but assuming it is.

For DA which is isovolumetric
w= 0

$$\Delta U= Q$$ for U

$$\Delta Q= nCv (T_D-T_A)$$ for finding heat (Q)

$$\Delta U= C_v \Delta T$$ => however I'm not given T so what do I do?

$$\Delta P V /nR = \Delta T$$

$$\Delta T= \frac{(8.00x10^4-3.00x10^4)(2x10^-3)} {(1mol)(8.314472 m3·Pa·K·mol)} = 6.013x10^3 K$$

and I think I take that and plug this below to find the $$E_{int}$$
and since Cv= 12.5 for He
$$\Delta U= nC_v \Delta T = 1mol(12.5)(6.013x10^3 K)= 7.516x10^4 J$$

and $$\Delta U= Q= 7.516x10^4 J$$ right?

and W= 0

17. Mar 30, 2008

### ~christina~

For CD which is Isobaric
P= constant
$$W= -P(V_F-V_i)$$ for the internal E

$$W= -(3.00x10^4Pa)(5.00x10^{-3}- 2.00x10^{-3})= -9x10^8 J$$

$$Q= nCP \Delta T$$
$$Q= (1mol)(20.8J/mol*K)= 20.8 J$$

$$\Delta U= Q + W$$ for the work
so it would be

$$\Delta U= 20.8J + -9x10^8J= -8.99x10^8 J$$

18. Mar 30, 2008

### ~christina~

Last part.

For part BC

It's adiabatic so Q= 0 thus
$$\Delta U= W$$

$$Point C=> P_f= 3.00x10^4 Pa, V_f= 5.00m^3$$

$$P_iV_i^{\gamma} = P_iV_f^{\gamma}$$

I found the Vi and Pi though before (assuming their correct)

not sure how I find the work though...and it's equal to the U

so I need help on that. I think you have to take take the integral of the area under the curve but not sure how that would be set up unfortunately.

Thanks Doc Al

19. Mar 31, 2008

### Staff: Mentor

Looks good! Be careful with terminology: On a P-V diagram only lines can be isotherms or adiabats, not points.

20. Mar 31, 2008

### Staff: Mentor

Your methodology is correct, but redo your calculation of delta T and what follows. (You forgot to multiply by the volume.)