Physics · Thermodynamics · NEET
Only the P-V (pressure vs volume) graph. Work is W = integral of P dV, so you need pressure on the y-axis and volume on the x-axis. The area under a P-T or V-T graph has no direct meaning for work. If a NEET question gives you a P-T or V-T graph, first convert it to a P-V picture (using PV = nRT) before finding any area.
Work depends on the PATH, not just the endpoints. Two curves joining the same start point A and end point B can enclose different areas, so they give different work. This is why work is not a state function. Internal energy change (delta U) does depend only on endpoints (through temperature), but work does not - that is the key difference NEET tests.
A vertical line means the volume does not change (delta V = 0). Since W = integral of P dV and dV = 0, the area under a vertical line is zero, so W = 0. This is an isochoric process. In NEET 2024 the path bc was a vertical segment, so the work along bc was exactly Zero.
When a gas is compressed, its volume decreases, so V2 is less than V1 and delta V is negative. The area is still there but you move right-to-left on the graph, so W = integral P dV comes out negative. Negative work by the gas means work is done ON the gas by the surroundings. Expansion gives positive work; compression gives negative work.
For a closed loop on a P-V diagram, the net work equals the AREA ENCLOSED by the loop. Clockwise loop = positive work done by the gas; anticlockwise loop = negative work. Because the cycle returns to the start, delta U = 0, so by the first law the heat supplied equals this loop area. For a rectangle, area = (delta P) times (delta V).
For the given cycle, the work done during the isobaric process is (constant pressure P = 200 kPa, volume changing from 1 x 10^-3 m^3 to 4 x 10^-3 m^3).
A thermodynamic system is taken through the cycle abcda. The work done by the gas along the path bc is (bc is a vertical segment on the P-V diagram).
One mole of an ideal monatomic gas undergoes a cyclic process shown as a rectangle a to b to c to d in the P-V plane, with P between 100 and 300 N/m^2 and V between 2 and 5 m^3. The total heat supplied to the gas is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
W = integral of P dV, which is the area between the P-V curve and the volume axis. For constant pressure it simplifies to W = P times (V2 - V1).
Force = pressure times area of the piston, and force times displacement = work. Area times displacement is volume change, so pressure times volume change equals work. Adding up all the thin strips P dV gives the total area, which is the work.
Expansion (volume increases, moving right) gives positive work done by the gas. Compression (volume decreases, moving left) gives negative work. For a loop, clockwise is positive and anticlockwise is negative.
No. Work depends on the path taken between two states, not just the start and end points, so it is a path function. Internal energy is a state function; work and heat are not.
Zero. A constant-volume (isochoric) process is a vertical line on the P-V graph, so delta V = 0 and the area is zero, giving W = 0.
