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.