An ideal gas closed system reversible process

In summary, the given problem involves an ideal gas at T = 70 C and 1 bar, undergoing three reversible processes: a) adiabatic compression to 150 C, b) cooling from 150 to 70 C at constant pressure, and c) isothermal expansion back to the original state. To solve for the final pressure in part c), you can use the pressures calculated in part b) to go back to the original state.
  • #1
HethensEnd25
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Homework Statement


an Ideal gas at T = 70 C and 1 bar undergoes following reversible
processes:
a: Adiabatically compressed to 150 C
b: then, cooled from 150 to 70 C at constant pressure
c: finally, expanded isothermally to the original state (T=70 C and P = 1 bar)

Homework Equations


1-dU=dQ+dW
2-dU=CvdT
3-(T2/T1)=(P2/P1)(R/Cp) from TP(1-ϒ)/ϒ=constant
4-CpdT=dH

The Attempt at a Solution


So far for part a) I have done the equation dU=dQ+dW, since it is an adiabatic process I have dQ=0, thus leaving dU=dW. I know that dW=-PdV so I can calculate dU=CvdT to get my change in internal energy. From there I calculated P2 using equation 3.

For b) knowing it is an Cp problem I used equation 4 having CpdT=dH where my change in temperature is 150 to 70 C

I am stuck on part c)

I know it is an isotherm that means that there is no change in temperature thus dU and dH are equal to zero since they are temperature dependent. So dQ=dW. Do I make it so that T is a constant and solve with partial derivatives giving me dT=(V/R)dP+(P/R)dV. With dT=0 we get -PdV=VdP. Using the ideal gas equation to give V=RT/P ==> -(RT/P) dP. Integrating gives me -RTln(P2/P1)

My problem is I don't know what pressure to use. Do i just use the pressures calculated in my part b going from there back to the original state?
 
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  • #2
You haven't told us the questions you need to answer!
 
  • #3
HethensEnd25 said:

Homework Statement


an Ideal gas at T = 70 C and 1 bar undergoes following reversible
processes:
a: Adiabatically compressed to 150 C
b: then, cooled from 150 to 70 C at constant pressure
c: finally, expanded isothermally to the original state (T=70 C and P = 1 bar)

Homework Equations


1-dU=dQ+dW
2-dU=CvdT
3-(T2/T1)=(P2/P1)(R/Cp) from TP(1-ϒ)/ϒ=constant
4-CpdT=dH

The Attempt at a Solution


So far for part a) I have done the equation dU=dQ+dW, since it is an adiabatic process I have dQ=0, thus leaving dU=dW. I know that dW=-PdV so I can calculate dU=CvdT to get my change in internal energy. From there I calculated P2 using equation 3.

For b) knowing it is an Cp problem I used equation 4 having CpdT=dH where my change in temperature is 150 to 70 C

I am stuck on part c)

I know it is an isotherm that means that there is no change in temperature thus dU and dH are equal to zero since they are temperature dependent. So dQ=dW. Do I make it so that T is a constant and solve with partial derivatives giving me dT=(V/R)dP+(P/R)dV. With dT=0 we get -PdV=VdP. Using the ideal gas equation to give V=RT/P ==> -(RT/P) dP. Integrating gives me -RTln(P2/P1)

My problem is I don't know what pressure to use. Do i just use the pressures calculated in my part b going from there back to the original state?
Yes.
 

1. What is an ideal gas closed system?

An ideal gas closed system is a theoretical model used in thermodynamics to describe a gas that follows certain assumptions. These assumptions include that the gas particles have no volume and do not interact with each other, and that the gas obeys the ideal gas law (PV = nRT).

2. What is a reversible process in an ideal gas closed system?

A reversible process in an ideal gas closed system is a process that can be undone by reversing the direction of the process without causing any changes in the system or its surroundings. This means that the system and its surroundings return to their original states after the process is reversed.

3. How is work calculated in an ideal gas closed system during a reversible process?

In an ideal gas closed system, work is equal to the change in volume of the gas multiplied by the external pressure. This is represented by the equation W = -PΔV, where W is work, P is pressure, and ΔV is the change in volume.

4. What is the relationship between temperature and volume in an ideal gas closed system during a reversible process?

In an ideal gas closed system, temperature and volume have a direct relationship during a reversible process. This means that as the volume of the gas increases, the temperature also increases, and vice versa. This relationship is described by the ideal gas law (PV = nRT).

5. How does entropy change in an ideal gas closed system during a reversible process?

In an ideal gas closed system, entropy (a measure of disorder) remains constant during a reversible process. This means that the system does not become more or less disordered, and the net change in entropy is zero. This is because a reversible process is a balanced exchange of energy and does not result in any overall changes in the system.

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