Thermodynamics - Ideal Diatomic Gas

In summary, the question involves an ideal diatomic gas with a specific heat at constant volume of (3/5)*R. The gas initially occupies a volume of 2 meters cubed at a pressure of 5 atm and a temperature of 293.15 K, and is compressed to a final pressure of 12 atm. The final volume, temperature, work done, heat transferred, and change in internal energy are to be calculated for both a reversible isothermal and adiabatic compression. For the isothermal process, the final volume is 0.8333333 meters cubed and the work done on the system is -2.134 * 10^6 J. For the adiabatic process, the gas follows
  • #1
ChronicQuantumAddict
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The question is as follows:
An ideal diatomic gas for which cv = (3/5)*R, where R is the ideal gas constant, and occupies a volume V= 2 meters^3, at a Pressure Pi= 5 atm and temperature Ti = 20 degrees Celcius (= 293 K). The gas is compressed to a final pressure of Pf = 12 atm. Calculate the final volume Vf, the final temperature Tf, the work done dW, the heat transferred dQ, and the change in Internal energy dU for the following:
a). a reversible isothermal compression
b). a reversible adiabatic compression
i know in the adiabatic process, dQ = 0, and in the isothermal dT = 0. Where do i take it from here? :confused:
 
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  • #2
anyone please?
 
  • #3
ChronicQuantumAddict said:
The question is as follows:
An ideal diatomic gas for which cv = (3/5)*R, where R is the ideal gas constant, and occupies a volume V= 2 meters^3, at a Pressure Pi= 5 atm and temperature Ti = 20 degrees Celcius (= 293 K). The gas is compressed to a final pressure of Pf = 12 atm. Calculate the final volume Vf, the final temperature Tf, the work done dW, the heat transferred dQ, and the change in Internal energy dU for the following:
a). a reversible isothermal compression
b). a reversible adiabatic compression
i know in the adiabatic process, dQ = 0, and in the isothermal dT = 0. Where do i take it from here? :confused:
The first one is fairly straightforward. Think PV=nRT. If T is constant, what does that tell youi about PV at all times? That should give you Vf. What is U at intial and final volumes? What is W in terms of P and [itex]\Delta V[/itex]? That should give you W. What is the relationship between U, Q and W? That will give you Q.

For the second part, it is more complicated. You have to use the adiabatic relationship:

[tex]PV^\gamma = constant[/tex]
where [itex]\gamma = C_p/C_v[/itex]

What is [itex]\gamma[/itex] here?
What is the expression for W?
How is W related to Q and U?

AM
 
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  • #4
Please HELP!

i am stuck with the internal energy for part a) isothermal process.

here is what I've done so far. since its isothermal:

P1*V1 = P2*V2​
and thus,
V2= (P1/P2)*V1 = 0.8333333 m^3

For isothermal, Ti = Tf = T = constant = 293.15 K

W = -nRT*(Integral: Pi to Pf) (dP/P) = -nRT*ln(P2/P1) = nRT*ln(P1/P2)
thus W = -2.134 * 10^6 J, work done on the system

Now, for U i am stuck. i keep reading in the book, that for an ideal gas U only depends on T, and they give me this formula.

U - U_0 = dU = Cv*dT, where Cv is the specific heat at constant volume. but volume is not constant in this problem, and also, this leads to an integral with dT and for isothermal dT = 0
If this is true, then the integral yields a constant U

and that means U_final = U_initial, but that still doesn't tell me what U is.

Anything further? where am i going wrong?
 
  • #5
please help, need to turn this in tomorrow and I am stuck
 
  • #6
ChronicQuantumAddict said:
i am stuck with the internal energy for part a) isothermal process.
here is what I've done so far. since its isothermal:
P1*V1 = P2*V2​
and thus,
V2= (P1/P2)*V1 = 0.8333333 m^3
For isothermal, Ti = Tf = T = constant = 293.15 K
W = -nRT*(Integral: Pi to Pf) (dP/P) = -nRT*ln(P2/P1) = nRT*ln(P1/P2)
thus W = -2.134 * 10^6 J, work done on the system
Now, for U i am stuck. i keep reading in the book, that for an ideal gas U only depends on T, and they give me this formula.
U - U_0 = dU = Cv*dT, where Cv is the specific heat at constant volume. but volume is not constant in this problem, and also, this leads to an integral with dT and for isothermal dT = 0
If this is true, then the integral yields a constant U
and that means U_final = U_initial, but that still doesn't tell me what U is.
Anything further? where am i going wrong?
U is the internal energy due to kinetic energy of the molecules. If you keep the volume constant and increase the temperature, the gas does no work. So the increase in internal energy is equal to the heat input. Since the heat flow is just the heat capacity of the gas at constant volume multiplied by the change in temperature, [itex]dU = nC_vdT => U = \int dU = \int nC_vdt = nC_vT[/itex].

So [itex]U = nC_vT[/itex] (T in K).

AM
 
Last edited:
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  • #7
understood now, thanks very much AM
 

What is an ideal diatomic gas?

An ideal diatomic gas is a theoretical model of a gas that consists of two particles, typically atoms, that are separated by a fixed distance and interact only through collisions. This model is used in thermodynamics to simplify calculations and understand the behavior of real gases.

What is the ideal gas law?

The ideal gas law is a mathematical equation that relates the pressure, volume, temperature, and number of moles of an ideal gas. It is expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

What is the difference between an ideal gas and a real gas?

An ideal gas is a theoretical model that assumes the gas particles have no volume and do not interact with each other. In reality, real gases have volume and do interact with each other, especially at high pressures and low temperatures.

What is the significance of the ideal diatomic gas in thermodynamics?

The ideal diatomic gas is used as a simplified model in thermodynamics to understand the behavior of real gases. It helps in the development of thermodynamic laws and equations, which can then be applied to real gases with some adjustments.

What are some examples of diatomic gases?

Oxygen (O2), nitrogen (N2), hydrogen (H2), and chlorine (Cl2) are some examples of diatomic gases. These gases consist of two atoms of the same element bonded together, and they behave similarly to an ideal diatomic gas at low pressures and high temperatures.

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