An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat

Q: Why is the change in internal energy zero?

Internal energy is a state function. Since a cyclic process returns the system to its exact starting state, the initial and final temperatures are the same, making the change in internal energy zero.

Q: Why do we only take positive heat for efficiency?

Efficiency measures how well a system converts the energy you 'pay for' into useful work. You only supply the positive heat ($Q_{\text{in}}$). The negative heat is waste energy rejected to the surroundings.

Q: Can efficiency ever be 100%?

No, according to the Second Law of Thermodynamics, some heat must always be rejected to a colder reservoir ($Q_{\text{out}}$ cannot be zero), so efficiency is always less than 100%.

Q: What does negative work mean?

Negative work means work is done *on* the gas by the surroundings (e.g., compressing the gas), whereas positive work means work is done *by* the gas (e.g., expansion).

Q: Why did we use $W_{\text{net}} = Q_{\text{net}}$ for efficiency?

Because $\Delta U = 0$ for the cycle, the First Law ($\Delta U = Q_{\text{net}} - W_{\text{net}}$) simplifies to $Q_{\text{net}} = W_{\text{net}}$. It's often easier to sum the heats than to sum the works if some works are unknown.

Question

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are $Q_1 = 5960	ext{ J}$, $Q_2 = -5585	ext{ J}$, $Q_3 = -2980	ext{ J}$ and $Q_4 = 3645	ext{ J}$ respectively. The corresponding quantities of work involved are $W_1 = 2200	ext{ J}$, $W_2 = -825	ext{ J}$, $W_3 = -1100	ext{ J}$ and $W_4$ respectively.
(a) Find the value of $W_4$.
(b) What is the efficiency of the cycle?

Sigma Percentile · Accepted Answer

The beauty of thermodynamics lies in its strict adherence to the conservation of energy. When an ideal gas undergoes a **cyclic process**, it embarks on a journey through various states of pressure, volume, and temperature, only to return exactly to where it started. This simple fact is the key to unlocking our problem.

### The First Law in a Cycle

Because internal energy ($U$) is a state function—meaning it depends only on the current state of the gas and not how it got there—the net change in internal energy over a complete cycle is exactly zero ($\Delta U_{	ext{net}} = 0$).

According to the First Law of Thermodynamics, the heat added to a system equals the change in internal energy plus the work done by the system:
$$Q_{	ext{net}} = \Delta U_{	ext{net}} + W_{	ext{net}}$$

Since $\Delta U_{	ext{net}} = 0$, this simplifies beautifully to:
$$Q_{	ext{net}} = W_{	ext{net}}$$

This means the total sum of all heat exchanged during the cycle must perfectly balance the total sum of all work done.

### Finding the Missing Work

We are given the heat exchanged in all four steps and the work done in three of them. Let's set up our energy balance equation:
$$Q_1 + Q_2 + Q_3 + Q_4 = W_1 + W_2 + W_3 + W_4$$

Substituting the given values (and being very careful to keep the negative signs, which indicate heat rejected or work done *on* the gas):
$$5960 - 5585 - 2980 + 3645 = 2200 - 825 - 1100 + W_4$$

Let's calculate the net heat ($Q_{	ext{net}}$) on the left side:
$$Q_{	ext{net}} = 1040 	ext{ J}$$

Now, let's sum the known work values on the right side:
$$W_{	ext{known}} = 2200 - 825 - 1100 = 275 	ext{ J}$$

Equating the two sides:
$$1040 = 275 + W_4$$

$$W_4 = 1040 - 275 = 765 	ext{ J}$$

The positive sign tells us that in the fourth step, $765 	ext{ J}$ of work was done *by* the gas.

### The True Meaning of Efficiency

Now, let's evaluate how good this thermodynamic cycle is at its job. The **efficiency ($\eta$)** of a heat engine is the ratio of the useful work you get out to the energy you had to put in.

$$\eta = \frac{W_{	ext{net}}}{Q_{	ext{in}}} 	imes 100\%$$

Here is the crucial catch: $Q_{	ext{in}}$ is strictly the **total positive heat absorbed**. We do not subtract the rejected heat, because the rejected heat is waste; the positive heat is the fuel we "paid" for.

Looking at our heat values, only $Q_1$ and $Q_4$ are positive:
$$Q_{	ext{in}} = Q_1 + Q_4 = 5960 + 3645 = 9605 	ext{ J}$$

We already know our net work is equal to our net heat:
$$W_{	ext{net}} = 1040 	ext{ J}$$

Plugging these into our efficiency formula:
$$\eta = \frac{1040}{9605} 	imes 100\% \approx 10.82\%$$

### An Alternative Perspective

We can also calculate efficiency by looking at the waste. The total heat rejected ($Q_{	ext{out}}$) is the sum of the absolute values of the negative heats:
$$|Q_{	ext{out}}| = |-5585| + |-2980| = 8565 	ext{ J}$$

The alternative efficiency formula is:
$$\eta = \left( 1 - \frac{|Q_{	ext{out}}|}{Q_{	ext{in}}} ight) 	imes 100\%$$

$$\eta = \left( 1 - \frac{8565}{9605} ight) 	imes 100\% \approx 10.82\%$$

Both methods yield the exact same result, confirming our calculations are rock solid. This engine converts about $10.82\%$ of its heat input into useful work, which is quite typical for real-world thermodynamic cycles!

Visualized Solution

\text{Cyclic Process Overview}

\text{First Law for Cyclic Process}

\text{Equating Net Heat and Net Work}

\text{Calculating Net Values}

\text{Work Done in Step 4}

\text{Efficiency of a Cycle}

\text{Calculating Total Heat Absorbed}

\text{Calculating Net Work}

\text{Final Efficiency}

\text{Alternative Efficiency Formula}

The Sigma Insight: First Law of Thermodynamics

The First Law in a Cycle

Finding the Missing Work

The True Meaning of Efficiency

An Alternative Perspective

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