# Don't have point B (cycle process)

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## Homework Statement

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.

http://img179.imageshack.us/img179/4843/graphnc4.th.jpg [Broken]

## Homework Equations

$$\Delta U= Q + W$$

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

$$Q= nC_v \Delta T$$

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

$$C_v= 3/2R$$

$$C_p= 5/2R$$

## 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.

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Doc Al
Mentor
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.
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.

Gold Member
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.
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.

Doc Al
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.

Gold Member
Treat the helium as an ideal monoatomic gas--that should tell you what gamma is.
okay so $$\gamma= 1.67$$

Relating point A to point B (isothermal):
$$P_a V_a = P_b V_b$$
so I have the $$P_a= 8.00x10^4Pa$$ and $$V_a= 2.00m^3$$
so $$P_aV_a= P_bV_b= 1.6x10^5Pa*m^3$$

Relating point C to point B (adiabatic):
$$P_c V_c^{\gamma} = P_b V_b^{\gamma}$$
well for this $$P_cV_c^{\gamma} = (3.00x10^4 Pa)(5.00m^3)^{1.67}= 4.40x10^5 Pa*m^3= P_b V_b^{\gamma}$$

Two equations; two unknowns.
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?

Doc Al
Mentor
I don't think I can substitute one into the other and or etc.
Why not? Use one equation to write $P_b$ in terms of $V_b$, then plug that into the second.

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

Doc Al
Mentor
Recall this property of exponents:

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

Gold Member
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?

Doc Al
Mentor
is this correct?
No.

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

Gold Member
No.

$$\frac{X^a}{X} = \frac{X^a}{X^1} = X^{a-1}$$
so it would be this...

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

Doc Al
Mentor
so it would be this...

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

Gold Member
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!).
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...

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

Gold Member
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.

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Gold Member
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

Gold Member
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

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

Gold Member
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

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

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

Doc Al
Mentor
For CD which is Isobaric
P= constant
$$W= -P(V_F-V_i)$$ for the internal E
This is work, not internal energy (of course).

$$W= -(3.00x10^4Pa)(5.00x10^{-3}- 2.00x10^{-3})= -9x10^8 J$$
Redo this calculation. Also, since the gas is being compressed, the work done on it will be positive.

$$Q= nCP \Delta T$$
$$Q= (1mol)(20.8J/mol*K)= 20.8 J$$
Redo this. What's delta T?

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

$$\Delta U= 20.8J + -9x10^8J= -8.99x10^8 J$$
You'll have to redo this, given the above.

Doc Al
Mentor
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.
Yes, you'll have to integrate:
$$W = -\int p dV$$

Hint: Write p as a function of V, taking advantage of $pV^{\gamma} = C$ (where C is a constant).

Gold Member
Looks good! Be careful with terminology: On a P-V diagram only lines can be isotherms or adiabats, not points.
oh okay.
Your methodology is correct, but redo your calculation of delta T and what follows. (You forgot to multiply by the volume.)
$$\Delta T= \frac{(8.00x10^4-3.00x10^4)(2x10^-3)} {(1mol)(8.314472 m3·Pa·K·mol)} = 12.027 K$$

$$\Delta U= nC_v \Delta T = 1mol(12.5)(12.027K)= J$$

$$\Delta U= Q= 150.3375 J$$

This is work, not internal energy (of course).
um yes
Redo this calculation. Also, since the gas is being compressed, the work done on it will be positive.
Oh yeah, I forgot about that.

$$W= -(3.00x10^4Pa)(2.00x10^{-3}-5.00x10^{-3} )= 90 J$$

Redo this. What's delta T?
$$W=nR \Delta T$$

$$90J= 1mol (8.314472 m3·Pa·K·1·mol-1 )\Delta T$$

$$\frac{90J} {1mol (8.314472 m3·Pa·K·1·mol-1)}= 10.82K$$

$$Q= nCP \Delta T$$

$$Q= (1mol)(20.8J/mol*K)(10.82K)= 225.056 J$$

you'll have to redo this, given the above.
$$\Delta U= Q + W$$

$$\Delta U= 225.056J + 90J= 315.056 J$$

Yes, you'll have to integrate:
$$W = -\int p dV$$

Hint: Write p as a function of V, taking advantage of $pV^{\gamma} = C$ (where C is a constant).
hm..my teacher gave us a hint on this part and he said to use PV=nRT though. I would say that relates to $pV^{\gamma} = C$ but it doesn't include the gamma on the V so is it related?

I have issues with integration.

Thank you very much Doc Al

Doc Al
Mentor
hm..my teacher gave us a hint on this part and he said to use PV=nRT though. I would say that relates to $pV^{\gamma} = C$ but it doesn't include the gamma on the V so is it related?
As far as finding the work done in an adiabatic process, I don't quite see the point of your teacher's hint. PV=nRT applies to all ideal gas processes, not just adiabatic ones.
I have issues with integration.
Give my previous suggestion a shot and try to set it up properly. Then we'll worry about doing the integration. (It will turn out to be an easy integral. )

Gold Member

As far as finding the work done in an adiabatic process, I don't quite see the point of your teacher's hint. PV=nRT applies to all ideal gas processes, not just adiabatic ones.

Give my previous suggestion a shot and try to set it up properly. Then we'll worry about doing the integration. (It will turn out to be an easy integral. )
okay. (sorry about the late post-few tests this week)

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

and I think that's weird looking. Is it? (Okay I found this online but I'm not sure how they did the integration to get=> $$W= \frac{C(V_f^{1-\gamma}- V_i^{1-\gamma})} {1-\gamma}$$

Thank you Doc Al

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