An ideal gas undergoes a four step cycle as shown in the P - V diagram below. During this cycle, hea

Q: Why is work done zero in steps 2 and 4?

Because the volume is constant (isochoric process). Work is the area under the P-V curve, and a vertical line has zero area.

Q: How do we know the temperature is increasing in step 1?

From the ideal gas law, $PV = nRT$. In step 1, pressure is constant but volume increases, so the product $PV$ increases, meaning $T$ must increase.

Q: Can heat be absorbed if work is zero?

Yes! In step 4, work is zero, but the internal energy increases because the pressure (and thus temperature) rises. This requires heat input.

Q: Why is heat rejected in step 3?

The gas is compressed (negative work) and its temperature drops (negative change in internal energy). Both terms in the First Law are negative, so heat must flow out.

Q: Is the total heat absorbed in the cycle positive?

Yes, the cycle is clockwise, meaning the net work done is positive. Since it's a closed cycle ($\Delta U_{net} = 0$), the net heat absorbed $\Delta Q_{net}$ must equal the net work done, which is positive.

Question

An ideal gas undergoes a four step cycle as shown in the P – V diagram below. During this cycle, heat is absorbed by the gas in

Options:

(A) steps 1 and 2

(B) steps 1 and 3

(C) steps 1 and 4

(D) steps 2 and 4

Sigma Percentile · Accepted Answer

The study of thermodynamics often feels like an intricate dance between heat, work, and internal energy. In this problem, we are presented with a classic rectangular cycle on a P-V diagram. Our objective is to determine exactly where the gas is absorbing heat from its surroundings. To unlock the secrets of this cycle, we must rely on the First Law of Thermodynamics. ### The Master Equation The First Law of Thermodynamics is essentially the law of conservation of energy applied to thermal systems. It states: $ \Delta Q = \Delta U + W $ Here, $\Delta Q$ is the heat supplied to the gas, $\Delta U$ is the change in its internal energy, and $W$ is the work done by the gas. For an ideal gas, the internal energy depends exclusively on its temperature. According to the ideal gas equation, $PV = nRT$, the temperature is directly proportional to the product of pressure and volume ($PV$). Therefore, if the $PV$ product increases, the temperature rises, and $\Delta U$ is positive. Work done, on the other hand, is geometrically represented as the area under the P-V curve. If the gas expands (volume increases), the work done is positive. If it is compressed, the work done is negative. ### Analyzing Process 1: Isobaric Expansion Let's look at the first step, labeled as process 1. The arrow points to the right, indicating that the volume is increasing while the pressure remains constant. This is an **isobaric expansion**. Because the volume is increasing, the gas is doing work on its surroundings. $ W_1 > 0 $ Simultaneously, since the pressure is constant and the volume is increasing, their product $PV$ is increasing. This means the temperature of the gas is rising, leading to an increase in internal energy. $ \Delta U_1 > 0 $ Plugging these into our master equation, a positive work and a positive change in internal energy guarantee that $\Delta Q_1$ is positive. The gas is **absorbing heat**. ### Analyzing Process 2: Isochoric Pressure Drop Moving to process 2, the path goes straight down. The volume is perfectly constant, meaning the gas is neither expanding nor compressing. This is an **isochoric process**. Since there is no change in volume, the area under the curve is zero. $ W_2 = 0 $ However, the pressure is dropping. A decrease in pressure at a constant volume means the $PV$ product is decreasing. The gas is cooling down, so its internal energy decreases. $ \Delta U_2 < 0 $ With zero work and a negative change in internal energy, $\Delta Q_2$ must be negative. The gas is **rejecting heat**. ### Analyzing Process 3: Isobaric Compression In process 3, the arrow points to the left. The gas is being compressed while the pressure is held constant. Because the volume is decreasing, the work done by the gas is negative. $ W_3 < 0 $ The $PV$ product is also decreasing because the volume is shrinking at a constant pressure. This causes a drop in temperature and a decrease in internal energy. $ \Delta U_3 < 0 $ Adding two negative quantities yields a negative $\Delta Q_3$. Once again, the gas is **rejecting heat**. ### Analyzing Process 4: Isochoric Pressure Rise Finally, we arrive at process 4. The path goes straight up. Just like in process 2, the volume is constant, so no work is done. $ W_4 = 0 $ But look at the pressure—it is rising! An increase in pressure at a constant volume means the $PV$ product is increasing. The temperature goes up, and so does the internal energy. $ \Delta U_4 > 0 $ With a positive change in internal energy and zero work, $\Delta Q_4$ is positive. The gas is **absorbing heat** to increase its pressure. ### The Final Verdict By systematically applying the First Law of Thermodynamics to each step, we have discovered that the gas absorbs heat only during the expansion in step 1 and the pressure rise in step 4. Therefore, the correct answer is **steps 1 and 4**.

Animated Solution for Physics - Thermodynamics: An ideal gas undergoes a four step cycle as shown in the P – V diagram below. During this cycle, heat is absorbed by the gas in

Select Answer:

Visualized Solution

The P-V Cycle

First Law of Thermodynamics

Process 1: Isobaric Expansion

Process 2: Isochoric Pressure Drop

Process 3: Isobaric Compression

Process 4: Isochoric Pressure Rise

Conclusion

The Sigma Insight: First Law of Thermodynamics

The Master Equation

Analyzing Process 1

Analyzing Process 2

Analyzing Process 3

Analyzing Process 4

The Final Verdict

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