# Reversable process with adiabatic expansion

• osheari1
In summary, the problem involves finding the energy added to and leaving a monatomic ideal gas in a reversible cycle, as well as the net work done by the gas and the efficiency of the cycle. Using the ideal gas law and adiabatic condition, the temperatures at points a, b, and c are determined. The heat flow into the system is from the isochoric process, while the heat flow out of the system is from the isobaric process. The work done by the gas is equal to the net heat flow, which is 909.9 J.
osheari1

## Homework Statement

One mole of a monatomic ideal gas is taken through the reversible cycle shown below.
http://img508.imageshack.us/img508/2595/newbitmapimageqm.png
process bc is an adiabatic expansion, with P_b = 10.00 atm and V_b = 10^-3 m^3.
Find (a) the energy added to the gas as heat. (b) find the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.

## Homework Equations

PV=nRT (ideal gas)
PV^(gamma) = constant
gamma = C_p/C_v
V^(gamma-1)T = constant
U = (3/2)nRT (for a monatomic gas)

## The Attempt at a Solution

I really don't know how to start this problem
I know for the bc section of the cycle Q = 0 because its an adiabatic expansion.
I also solved for the temperature at point b T = PV/(nR) = (1.0132*10^6Pa)(10^-3m^3)/(1mole)(8.315) = 121.85 K

from here I am not sure what to do because I am not given gamma or the pressure at a or c

Last edited by a moderator:
osheari1 said:

## Homework Statement

One mole of a monatomic ideal gas is taken through the reversible cycle shown below.
http://img508.imageshack.us/img508/2595/newbitmapimageqm.png
process bc is an adiabatic expansion, with P_b = 10.00 atm and V_b = 10^-3 m^3.
Find (a) the energy added to the gas as heat. (b) find the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.

## Homework Equations

PV=nRT (ideal gas)
PV^(gamma) = constant
gamma = C_p/C_v
V^(gamma-1)T = constant
U = (3/2)nRT (for a monatomic gas)

## The Attempt at a Solution

I really don't know how to start this problem
I know for the bc section of the cycle Q = 0 because its an adiabatic expansion.
I also solved for the temperature at point b T = PV/(nR) = (1.0132*10^6Pa)(10^-3m^3)/(1mole)(8.315) = 121.85 K

from here I am not sure what to do because I am not given gamma or the pressure at a or c
[Note: A reversible cycle has to be in equilibrium with its surroundings at all time. This means heat flow can only occur when there is an infinitessimal temperature difference between the system and the hot or cold reservoirs. So heatflow has to occur isothermally. This cannot occur on the isochoric or isobaric parts of this cycle so I fail to see how this can be a reversible cycle. But that is just an aside.]

You have to determine the temperatures at a, b and c.

First, determine the temperature at b using the ideal gas law.

Second, determine the temperature and pressure at c using the adiabatic condition.

Then determine the temperature at a from that.

Since a-b is isochoric you can work out the heatflow into the gas from the change in temperature (using what specific heat capacity and what value?)

Since c-a is isobaric, you can work out the heatflow out of the gas from the change in temperature (using what specific heat capacity and what value?).

Since b-c is adiabatic, what can you say about the heat flow from b-c?

AM

Last edited by a moderator:
after working it out I ended up with temperature values of T_a = 3.86K, T_b = 121.85, T_c = 30.88. however, these values seems extremely low, but I just assumed it was because of the very small volume.
I determined these values by calculated gamma. sense the gas is monatomic and ideal, 5/2R = C_p and C_p = C_v +R so Gamma = 5/2/(5/2-1) = 1.66

so, the heat flowing into the system is from the Isochoric process, Q = U + 0 = 3/2nRT = 1471.6J
the heat flowing out of the system is from the isobaric process, Q = 5/2nRT = -561.68J

The only work done by the process is from the adiabatic expansion and the isobaric process, so Wnet = Wbc + Wca = -(Ubc) + (Qca - Uca) = -(3/2*8.315(30.88-121.85)) + (-561.67 - 3/2nRT) = 1134.62 +(337 - 561.67) = 909.95 J

osheari1 said:
after working it out I ended up with temperature values of T_a = 3.86K, T_b = 121.85, T_c = 30.88. however, these values seems extremely low, but I just assumed it was because of the very small volume.
I determined these values by calculated gamma. sense the gas is monatomic and ideal, 5/2R = C_p and C_p = C_v +R so Gamma = 5/2/(5/2-1) = 1.66

so, the heat flowing into the system is from the Isochoric process, Q = U + 0 = 3/2nRT = 1471.6J
the heat flowing out of the system is from the isobaric process, Q = 5/2nRT = -561.68J

The only work done by the process is from the adiabatic expansion and the isobaric process, so Wnet = Wbc + Wca = -(Ubc) + (Qca - Uca) = -(3/2*8.315(30.88-121.85)) + (-561.67 - 3/2nRT) = 1134.62 +(337 - 561.67) = 909.95 J
That looks right. You don't have to do so much calculation to find the work. By the first law, since there is no change in U in one cycle, so the work done by the gas must equal the net heat flow: $W = \Delta Q = Q_{in} - Q_{out} = 1471.6-561.7 = 909.9 J$

AM

!

I would first like to clarify that the given information is not sufficient to solve the problem accurately. We need to know the values of pressure at points a and c, as well as the value of gamma, in order to calculate the energy added and leaving the gas, as well as the net work done and efficiency of the cycle.

However, I can provide some general insights on the process and equations involved. Firstly, the given process is a reversible cycle, which means that the gas goes through a series of changes and returns to its initial state. In this case, the gas goes from point a to point b, then to point c, and finally back to point a. This is a closed system, meaning that no mass can enter or leave the system.

The adiabatic expansion at point bc means that no heat is added or removed from the system during this process. This can also be seen from the fact that the temperature remains constant during an adiabatic process (T = constant in the equation V^(gamma-1)T = constant). Therefore, the energy added to the gas as heat (Q) during this process is equal to zero.

To calculate the energy leaving the gas as heat (Q) at points a and c, we need to know the values of pressure and temperature at these points. This can be done by using the ideal gas law (PV=nRT) and the equation for adiabatic processes (PV^(gamma) = constant). Once we have the values of pressure and temperature at these points, we can use the equation U = (3/2)nRT to calculate the change in internal energy (U) and then use the formula Q = U + W, where W is the work done by the gas, to calculate the energy leaving the gas as heat.

The net work done by the gas can be calculated by using the equation W = P(V_b - V_a), where P is the pressure at point b and V_b and V_a are the volumes at points b and a, respectively. This is the work done during the adiabatic expansion (process bc) and it is equal to the negative of the work done during the isobaric compression (process ca).

Finally, the efficiency of the cycle can be calculated by using the formula η = W/Q_h, where W is the net work done by the gas and Q_h is the heat added to the gas during the entire

## 1. What is a reversible process with adiabatic expansion?

A reversible process with adiabatic expansion is a thermodynamic process in which the system undergoes expansion without exchanging heat with its surroundings, while still maintaining equilibrium. This means that the system's internal energy remains constant throughout the process.

## 2. How is a reversible process with adiabatic expansion different from other types of expansion processes?

A reversible process with adiabatic expansion is different from other types of expansion processes, such as isothermal or isobaric, because it does not involve any heat transfer. This means that the temperature of the system will change during the process, whereas in isothermal or isobaric processes, the temperature remains constant.

## 3. What is the significance of a reversible process with adiabatic expansion in thermodynamics?

A reversible process with adiabatic expansion is significant because it is an idealized process that is used as a standard for comparison with real-world processes. It allows scientists to study the behavior of a system when there is no heat transfer, which can help in understanding the limitations and efficiencies of real processes.

## 4. What is the equation for calculating work done in a reversible process with adiabatic expansion?

The equation for calculating work done in a reversible process with adiabatic expansion is W = -nRTln(V2/V1), where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, and V2 and V1 are the final and initial volumes of the gas, respectively.

## 5. Can a reversible process with adiabatic expansion occur in real-world systems?

While a reversible process with adiabatic expansion is an idealized process, it is possible for real-world systems to closely approximate it. For example, a gas expanding rapidly in a perfectly insulated container can closely resemble a reversible adiabatic expansion process. However, in most real-world scenarios, some heat transfer or friction will occur, making it difficult to achieve a true reversible process.

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