# Calculate work and heat in a cyclic process ?

• rash219
In summary, the problem involves considering n moles of ideal gas in a cylinder with two heat reservoirs at different temperatures, and a reversible process involving heat exchange and pressure changes. The goal is to calculate the work done by the system, as well as the amounts of heat exchanged with each reservoir. Some questions arise about the validity of the problem, but using the first and second laws of thermodynamics, equations are derived to calculate the net work done and the amounts of heat exchanged.
rash219

## Homework Statement

Consider n moles of ideal gas kept in a cylinder with a piston. Two heat reservoirs 1 and 2 with the temperatures T1 < T2 are available, and at any given moment of time the heat exchange is established with only one of the reservoirs. In the initial equilibrium state the external pressure is p1, and heat contact is with reservoir 1. At the same time moment the external pressure is quickly changed to p2, whereas heat exchange switches to reservoir 2, and we wait for the system to equilibrate. Then the external pressure is quickly returned to its initial value p1, whereas heat exchange switches back to reservoir 1, and we wait for the system to equilibrate.

Calculate the amount of work (w) the system produce on the environment, the amount of heat (q2) transferred from reservoir 2 (heat source) to the system, and the amount of heat (q1) transferred to reservoir 1 from the system.

dU = 0
U = q - w
w = -∫ Pext dV

## The Attempt at a Solution

The system returns to its initial state and therefore we can call our entire process a cycle.

To calculate work total

Wtotal = w1 + w2

w = -∫ Pext dV

w1 = p2( v2 - v1)
w2 = - p1 (v1 -v2)

∴ Wtotal = p1v2 + p2v2 - p1v1 - p2v1

we do not know if this is correct and we have no idea how to proceed to calculate q1 and q2

rash219 said:

## Homework Statement

w1 = p2( v2 - v1)
w2 = - p1 (v1 -v2)
You're saying w1 and w2 both have the same sign which you know is not right.

Actually, I have problems with this problem - it doesn't seem to be a reversible process since we assume instantaneous state changes. But, assuming reversibility anyway:

w2→1 = -p1(V2-V1)

Then sum these two, substitute V = nRT/p as appropriate, and come up with net work done per cycle as a function of p1, p2, T1, T2, n and R.

Of course, you know that net work must equal Q2 -Q1. What second equation can you produce to enable solving for Q1 and Q2 separately? (Hint: what does the 2nd law require?)

Note: hopefully others will join in here to maybe give you a second opinion ...

## 1. How is work calculated in a cyclic process?

In a cyclic process, work is calculated by taking the integral of the pressure-volume (PV) curve. This means that you must calculate the area under the PV curve to determine the work done by the system.

## 2. What is the formula for calculating work in a cyclic process?

The formula for calculating work in a cyclic process is W = ∫PdV, where W represents work, P is pressure, and dV is the change in volume. This integral can be solved by breaking the curve into smaller sections and calculating the area for each section.

## 3. How can heat be calculated in a cyclic process?

Heat can be calculated in a cyclic process by using the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system. Therefore, Q = ΔU + W.

## 4. What is the difference between work and heat in a cyclic process?

Work and heat are both forms of energy, but they differ in how they are transferred. Work is a transfer of energy due to a force acting over a distance, while heat is a transfer of energy due to a temperature difference.

## 5. Can the work and heat in a cyclic process be negative?

Yes, both work and heat in a cyclic process can be negative. This occurs when the system loses energy through work or heat, rather than gaining it. This can happen, for example, when the system releases heat to the surroundings or when the work done by the system is against an external force.

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