Question

In a cyclic \(pV\) process forming a square loop from \((p = 1 \text{ atm}, V = 2\text{L})\) to \((p = 3 \text{ atm}, V = 4\text{L})\), then, the net heat absorbed by the gas is:

• A. \(2\text{L} \cdot \text{atm}\)
• B. \(4\text{L} \cdot \text{atm}\)
• C. \(8\text{L} \cdot \text{atm}\)
• D. \(6\text{L} \cdot \text{atm}\)

Hint:

Whenever a problem asks for the "net heat absorbed" or "total work done" in a cyclic loop, skip calculating separate step-by-step pathways entirely! Simply find the enclosed area of the shape. Just be careful to double-check the direction: Clockwise = Positive Work, Counter-clockwise = Negative Work.

Answer 1

The Correct Option is B

Concept: According to the First Law of Thermodynamics, the net heat change (\(\Delta Q_{\text{net}}\)) in a thermodynamic process is equal to the sum of the change in internal energy (\(\Delta U\)) and the net work done (\(W_{\text{net}}\)) by the system: \[ \Delta Q_{\text{net}} = \Delta U + W_{\text{net}} \] For any complete cyclic process, the system returns precisely to its initial state. Because internal energy is a state function depending solely on the current state parameters, the net change in internal energy over a complete cycle is identically zero: \[ \Delta U_{\text{cyclic}} = 0 \quad \Rightarrow \quad \Delta Q_{\text{net}} = W_{\text{net}} \] The net work done during a cyclic process plotted on a pressure-volume (\(p\)-\(V\)) diagram is equal to the geometric area enclosed by the loop. If the cycle runs in a clockwise direction, the work done is positive; if it runs counter-clockwise, the work done is negative. Step 1: Determine the sign of the net work done from the cycle direction.
Looking closely at the arrow directions marked on the square loop inside the given diagram:
• The top horizontal line shows expansion to the right (clockwise movement).
• The right vertical line tracks a downward path, completing a standard clockwise loop orientation. Therefore, the net work done (\(W_{\text{net}}\)) and the net heat absorbed (\(\Delta Q_{\text{net}}\)) are both positive values.

Step 2: Calculate the geometric dimensions of the enclosed square area.
The loop forms a geometric square on the graph. Let us determine its side lengths along both axes:
• Change in Volume along the horizontal axis (\(\Delta V\)): \[ \text{Width} = V_{\text{max}} - V_{\text{min}} = 4\text{L} - 2\text{L} = 2\text{ L} \]
• Change in Pressure along the vertical axis (\(\Delta p\)): \[ \text{Height} = p_{\text{max}} - p_{\text{min}} = 3\text{ atm} - 1\text{ atm} = 2\text{ atm} \]

Step 3: Compute the enclosed area to find the net heat absorbed.
Since it is a square loop, the area is simply the product of its dimensions: \[ W_{\text{net}} = \text{Enclosed Area} = \text{Width} \times \text{Height} \] \[ W_{\text{net}} = 2\text{ L} \times 2\text{ atm} = 4\text{ L} \cdot \text{atm} \] By substituting this value back into our derived First Law relation: \[ \Delta Q_{\text{net}} = W_{\text{net}} = 4\text{ L} \cdot \text{atm} \]