Ideal Gas Under A (Complex) Reversible Process

[memnoch3434]

Homework Statement

An ideal gas undergoes a reversible process 1-2-3-4-5. The stage from 1-2 as well as isobaric 2-3 and isochoric 4-5 stages are represented by linear segments. The Stage 2-3 is isothermal. The following relations are true:

V_2=2*V_1
p_2=1.5*p_1
V_3=4*V_1
V_4=6*V_1
T_5=T_1

The initial states are V_1=3m^3 and p_1 = 2.5 atm.
Assuming that the internal energy of an ideal gas depends only on temperature, find the total heat received by the gas during the entire process also find change in enthalpy (H_1-H_5).

Homework Equations

pv=nRT

dw=p(v)dv

dw=pdv (isobaric)

dw=nRT*ln(v_f-v_i) isothermic

Q=U+W

H=U+pV

U=n*c_v*T+n*u_0 *****

The Attempt at a Solution

So far I have calculated all the pressures and volumes for all points (in atm and m^3):
P_1=3
V_1=2.5
p_2=3.75
V_2=6
p_3=1.875
V_3=12
p_4=1.875
V_4=18
p_5=5/12
V_5=18

After that I started trying to find either n or T, but I am guessing the point of the problem is to solve it without those. So moving on to find Q total, I noticed that without T, this problem involves at least 4 unknowns. So my troubles come in calculating the work from 1-2 since pressure is non constant and not any special type of process, and from eliminating n and T from the equation for q total.

****This is also the chapter in the book before we talk about the energy of an ideal gas, so I am assuming that we are not to use that equation, but just in case there it is.

 

[rude man]

You need to clarify what stages are isobaric, isochoric, isothermal, adiabatic, etc. better. I'm royally confused.

Linear segments on what type of plot?

 

[memnoch3434]

Sorry about that just tried to copy the problem word for word. But looks like I made a mistake typing, sorry about that.

Isochoric- 4-5
Isobaric- 3-4
Isothermal- 2-3
No specified adiabatic processes.

The plot is a P-V diagram.

 

[rude man]

Homework Statement

An ideal gas undergoes a reversible process 1-2-3-4-5. The stage from 1-2 as well as isobaric 2-3 and isochoric 4-5 stages are represented by linear segments. The Stage 2-3 is isothermal. The following relations are true:

V_2=2*V_1
p_2=1.5*p_1
V_3=4*V_1
V_4=6*V_1
T_5=T_1

The initial states are V_1=3m^3 and p_1 = 2.5 atm.
Assuming that the internal energy of an ideal gas depends only on temperature, find the total heat received by the gas during the entire process also find change in enthalpy (H_1-H_5).

Homework Equations

pv=nRT

dw=p(v)dv

dw=pdv (isobaric)

dw=nRT*ln(v_f-v_i) isothermic

Q=U+W

H=U+pV

U=n*c_v*T+n*u_0 *****

The Attempt at a Solution

So far I have calculated all the pressures and volumes for all points (in atm and m^3):
P_1=3
V_1=2.5
p_2=3.75
V_2=6
p_3=1.875
V_3=12
p_4=1.875
V_4=18
p_5=5/12
V_5=18

After that I started trying to find either n or T, but I am guessing the point of the problem is to solve it without those. So moving on to find Q total, I noticed that without T, this problem involves at least 4 unknowns. So my troubles come in calculating the work from 1-2 since pressure is non constant and not any special type of process, and from eliminating n and T from the equation for q total.

****This is also the chapter in the book before we talk about the energy of an ideal gas, so I am assuming that we are not to use that equation, but just in case there it is.

By the end of your last chapter, I assume it discussed Cv and Cp. Because unless it avoided the adiabatic equation pV^γ = constant, it had to mention that γ = Cp/Cv.

By definition, Cv =∂U/∂T|V but since U = U(T only), then Cv = dU/dT.


My first suggestion is to plot all your p's and V's on your p-V diagram.
Then: what is ∆W on a p-V diagram when any two states are plotted?
Can you readily see what ∫pdV and ∫Vdp are on your diagram?

Combine the plot with dU = dQ - dW and pV = nRT and you can proceed step-by-step thru all 5 states & arrive at ∆U.

Then of course ∆H is immediately available since you now know p5 and V5.

 

[paranormal]

I would approach this problem by first acknowledging the complexity of the reversible process and the limitations of the given information. While the equations provided may be useful in certain scenarios, it is important to consider that an ideal gas may not always follow these exact relationships and that there may be other factors at play.

With that being said, I would start by using the ideal gas law (pv=nRT) to calculate the number of moles (n) of gas present in the initial state (1). From there, I would use the given relationships to calculate the pressures and volumes at each stage, keeping in mind that the internal energy of an ideal gas depends only on temperature.

To find the total heat received by the gas during the entire process, I would use the equation Q=U+W, where U is the change in internal energy and W is the work done on the gas. The work can be calculated by subtracting the area under the pressure-volume curve for each segment (1-2, 2-3, 4-5) from the area under the temperature-volume curve for segment 2-3. This will give the work done on the gas during each stage, which can then be added together to find the total work.

To find the change in enthalpy (H_1-H_5), I would use the equation H=U+pV and calculate the enthalpy at each stage by adding the internal energy (calculated from the given temperature) to the product of pressure and volume.

In conclusion, while the given equations may be helpful in some cases, it is important to approach this problem with a critical mindset and consider all possible factors that may affect the behavior of an ideal gas.