Thermodynamics problem help

In summary, the entropy changes during this process are: (a) ∆s = R ln(P 2 /P 1 ) (b) ∆s = R ln(P 2 /P 1 )- q/T (c) ∆s = R ln(P 1 /P 2 ) (d) ∆s = R ln(P 1 /P 2 )-q/T
  • #1
Deathcrush
40
0

Homework Statement



A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure
P 1 to pressure P 2 while loosing heat to the surroundings at temperature T in the amount of q. If the gas
constant of the gas is R, the entropy change of the gas ∆s during this process is
(a) ∆s =R ln(P 2 /P 1 ) (b) ∆s = R ln(P 2 /P 1 )- q/T (c) ∆s =R ln(P 1 /P 2 ) (d) ∆s =R ln(P 1 /P 2 )-q/T
(e) ∆s= 0

Homework Equations



PV=nRT (1)
u=q+w (2)
h=u+PV (3)
dw=-PdV (4)
ds=dq/T (5)

The Attempt at a Solution



first I considered h=0, since it is an isothermic process (I don't know if this is correct, this is one of my doubts)

I derived 2 and 3

du=dq+dw and du=-VdP-PdV

now, using 4 in this last derivative and comparing it with the first one, I found that dq=-VdP

and I used it in 5, also, I used 1 in V and integrated the whole thingS=R [tex]\int-dp/P)[/tex] from state 1 to state two

I integrated from state two to one, and changed the negative sign

this would be something like
S=R Ln(P1/P2)
after using the properties of the logarithms

My biggest doubt is that of the enthalpy, if it is not 0, the whole thing is wrong. I chequed the answers in the book and mine is right, but I don't know if the process is ok, would anybody help me on this please?
 
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  • #2
It may be of use to know that the enthalpy H of an ideal gas is [itex]H=C_P\Delta T+H_0[/itex], where [itex]C_P[/itex] is the heat capacity, [itex]\Delta T[/itex] is the temperature change, and [itex]H_0[/itex] is an arbitrary reference value.
 
  • #3
Your have the right answer but it is simpler than you have done.

Since dT = 0, dU=nCvdT = 0. So dQ = PdV (Use the first law dQ = dU + dW where dW is the work done BY the gas = PdV)

dV/dP = d/dP(nRT/P) = -nRT/P^2

So:

[tex]\Delta S = \int ds = \int dQ/T = \frac{1}{T}\int_{P_1}^{P_2} PdV = -nR\int_{P_1}^{P_2}\frac{dP}{P}[/tex]

AM
 
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  • #4
Mapes said:
It may be of use to know that the enthalpy H of an ideal gas is [itex]H=C_P\Delta T+H_0[/itex], where [itex]C_P[/itex] is the heat capacity, [itex]\Delta T[/itex] is the temperature change, and [itex]H_0[/itex] is an arbitrary reference value.

thanks a lot both of you, now, just another doubt, is [itex]H=C_P\Delta T+H_0[/itex] valid only for ideal gases? because I am working out another problem now, and I was planning to use the equation I got in this problem, but now it is not an ideal gas, it is air at about 25atm, is the equation valid?
 
  • #5
It is valid only for ideal gases. Depending on the nonideality, you may need to add a correction term (aka an enthalpy departure function).
 
  • #6
Deathcrush said:
thanks a lot both of you, now, just another doubt, is [itex]H=C_P\Delta T+H_0[/itex] valid only for ideal gases? because I am working out another problem now, and I was planning to use the equation I got in this problem, but now it is not an ideal gas, it is air at about 25atm, is the equation valid?
This is true only for constant pressure processes. It is simply saying that for constant pressure processes [itex]\Delta Q = H - H_0 = C_p\Delta T[/itex], ie. just a special case of applying the first law. In this case, pressure is not constant. Temperature is. So you just apply the first law: [itex]\Delta Q = \int PdV[/itex]. There is positive heat flow into the gas in this case.

AM
 
  • #7
If you want to determine the change in enthalpy in this process, start with the definition:

H = U + PV

This means that:

dH = dU + PdV + VdP

In this case, dU = 0 so dH = PdV + VdP:

[tex]\Delta H = \int_{V_i}^{V_f} PdV + \int_{P_i}^{P_f} VdP = nRT( \int_{V_i}^{V_f}
\frac{dV}{V} + \int_{P_i}^{P_f} \frac{dP}{P}) = nRT\left(\ln{\frac{V_f}{V_i}} + \ln{\frac{P_f}{P_i}}\right)[/tex]

But since T is constant, Vf/Vi = Pi/Pf, so

[tex]\Delta H = 0[/tex]

AM
 
  • #8
No, I am actually trying to find the intermediate pressure for a 2 step compression so that the work needed is minimum
 

1. What is thermodynamics?

Thermodynamics is the branch of science that deals with the study of energy and its transformations in a system, including heat and work.

2. What are the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern energy and its transformations. The first law states that energy cannot be created or destroyed, only transferred or converted. The second law states that in any energy transfer or conversion, some energy will be lost as heat. The third law states that the entropy of a pure crystal at absolute zero temperature is zero.

3. What is an example of a thermodynamics problem?

An example of a thermodynamics problem could be calculating the efficiency of a heat engine, where heat is converted into work. This involves using the first and second laws of thermodynamics to determine the relationship between the heat input, work output, and heat loss.

4. How can I solve a thermodynamics problem?

To solve a thermodynamics problem, you will need to understand the fundamental principles of thermodynamics, such as the laws and equations. You will also need to have a good understanding of the system being studied, including its properties and boundary conditions. It is also helpful to have a strong foundation in mathematics and problem-solving skills.

5. What are some common applications of thermodynamics?

Thermodynamics has many practical applications in various fields, such as engineering, chemistry, biology, and physics. Some common applications include power generation, refrigeration and air conditioning systems, chemical reactions, and biological processes. Thermodynamics also plays a crucial role in understanding and predicting the behavior of complex systems, such as weather patterns and ecosystems.

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