Thermodynamic Processes Thermodynamic Processes Imagine a gas trapped in a smooth cylinder with a movable piston, like a giant syringe. The molecules rush around like hyperactive ping‑pong balls, bouncing off the walls. A thermodynamic process is simply the rule by which we change the state of this gas from one equilibrium state to another. If we heat slowly while allowing the piston to move, the gas may expand. If we clamp the piston, pressure rises instead. If we keep the cylinder in contact with a heat reservoir, the temperature can be held fixed even as the gas does work while expanding. And if we wrap the cylinder in perfect insulation and compress it quickly, the temperature shoots up because no heat can escape. These different “rules” lead to the classic named processes: isobaric (constant pressure), isochoric (constant volume), isothermal (constant temperature), and adiabatic (no heat exchange). Each path has a definite equation connecting P , V , and T , a characteristic curve on a P – V diagram, and characteristic values of heat Q , work W , and change in internal energy U . Preparing for NEET means becoming fluent with the equations, but also with the physical picture: what is being controlled, how quickly the change occurs, and which walls allow heat to pass. Throughout, we follow the Physics sign convention: Q is heat supplied to the system (positive when added), W is work done by the system on the surroundings (positive for expansion), and U is the change in internal energy of the system. The first law, Q = U + W , ties these together and acts like a budget for energy in any process. remember Crowded dance floor analogy: To keep the “pressure” of bumping constant as the music speeds up (temperature rises), you must expand the floor (isobaric). If the door is locked (isochoric) and music speeds up, bumping gets harder (pressure increases). If the room is perfectly insulated and you suddenly squeeze the crowd (adiabatic compression), it heats up strictly because of the squeeze, not because the DJ changed the music. Thermodynamic process A path connecting two equilibrium states of a system, specified by how P , V , T change and how heat and work flow. Quantities like P , V , T , and U that depend only on the current state, not on the path taken. State variables Path functions Heat Q and work W ; values depend on the path followed during a process. A boundary that allows heat exchange with surroundings; needed for isothermal processes. Diathermic wall Adiabatic wall A thermally insulating boundary that blocks heat transfer; used for adiabatic processes. Quasi-static (reversible) process A change that is carried out infinitely slowly so the system passes through a sequence of equilibrium states; equations like P ,dV apply cleanly. Work sign convention W>0 when the system does work on the surroundings (expansion); W<0 for compression. Internal energy of an ideal gas Depends only on temperature: U = n C v T . Adiabatic index = C p/C v , typically 1.67 (monoatomic), 1.4 (diatomic), 1.33 (nonlinear triatomic). First law framework with NEET sign convention: Q = U + W . It is useful to read this as “heat in equals rise in internal energy plus work done by the gas.” For an isochoric change dV=0 , so W= P ,dV=0 and heat goes entirely into U . For isothermal change of an ideal gas, U=0 because U depends only on T , so any heat absorbed becomes work. For an adiabatic change Q=0 , so W=- U : when the gas expands without heat input, it cools while doing work; when it is compressed adiabatically, it warms as work is done on it. Keep pressures absolute (Pa), temperatures in Kelvin, and volumes in m 3 . Use R=8.314 , J ,mol -1 K -1 unless the question gives a rounded value. We will assume processes are quasi-static unless stated otherwise, so that W= P ,dV with the process equation P(V) well defined. Heat supplied = increase in internal energy + work done by the system. First Law It describes how heat transfer, work done, and internal energy change relate during processes like adiabatic or isothermal changes. We will use the Cartesian sign convention common in Physics: W positive for expansion work by the gas. Many Chemistry texts use the opposite sign; always check the convention in the question. tip Isothermal process (T constant) In an isothermal process the temperature stays fixed: T= constant . For an ideal gas, that means the average molecular kinetic energy is unchanged and U=0 . To maintain T constant while the gas expands or compresses, the cylinder must be in contact with a large thermal reservoir (diathermic wall), and the process must be slow enough for heat to flow in or out continuously. Using the ideal gas law PV=nRT with constant T , the product PV stays constant. The P – V curve is a rectangular hyperbola (Boyle’s law). When the gas expands isothermally, it draws heat from the reservoir equal to the work it does: Q= W>0 . During isothermal compression, the surroundings do work on the gas and an equal amount of heat flows out: W<0 and Q<0 , but U=0 in both cases. Work done by an ideal gas in a reversible isothermal change. Isothermal Work Isothermal Alternative Same work written with pressures using P 1V 1=P 2V 2=nRT . Calculate the work done by an ideal gas undergoing a reversible process while maintaining a constant temperature. Slope and limits on an isotherm: differentiating PV= constant gives dP dV = - P V . At a given point (P,V) , the adiabatic curve passing through the point is steeper by a factor (discussed later). Boundary cases: as V 0 , the ideal gas model breaks down and the predicted pressure would diverge; as V , P 0 . For finite reversible changes, ensure V 2/V 1>0 . Using base‑10 logs is fine if you multiply by 2.303; for natural logs use directly. easy 1.0 mol of an ideal gas at T=300 , K expands isothermally and reversibly to double its initial volume. Find the work done by the gas. n = 1.0 mol T = 300 K V 2 / V 1 = 2.0 R = 8.314 J mol -1 K -1 Use W = nRT (V 2/V 1) . Transparent cylinder and piston over a flame with blue gas molecules and a temperature display. Diathermic cylinder with movable piston heated by a Bunsen burner; blue molecules show increased kinetic energy as the piston rises. The attached display reads 75 °C. Ideal for illustrating isobaric expansion or slow isothermal control. Isobaric process (P constant) For an isobaric process, pressure is held constant by allowing the piston to move freely against a constant external pressure. The ideal‑gas equation gives V T (Charles’ law). Work is the area under the horizontal line on the P – V diagram, so W = P(V 2 - V 1) = nR(T 2 - T 1) . Heat added increases both internal energy and the work done: Q = nC p T , U = nC v T , with C p - C v=R (Mayer’s relation). Because expansion also requires pushing the piston, C p exceeds C v by exactly R for an ideal gas. In isobaric cooling, the gas contracts, W<0 , and if T<0 then heat flows out ( Q<0 ) while internal energy falls. Isobaric Work & Heat Relations that connect work, heat, and energy at constant pressure. This set of relations applies when the external pressure remains constant throughout the thermodynamic process (isobaric process). Practical reading on a P – V plot: a horizontal line means isobaric. The sign of W is set by (V 2-V 1) ; to the right is expansion ( W>0 ), to the left compression ( W<0 ). In NEET numericals, if the gas type is given (monoatomic/diatomic), choose C v and C p accordingly. If not specified, avoid assuming a type unless a value of is provided. Keep T in Kelvin when using T ; the difference in °C equals the difference in K, but formulas with ratios require absolute K. medium 2.0 mol of an ideal monoatomic gas is heated isobarically so that its temperature rises by 50 K. Find W, ΔU, and Q. Use W = nRΔT, ΔU = nC vΔT, Q = nC pΔT. n = 2.0 mol Monoatomic gas: C v = (3/2)R, C p = (5/2)R ΔT = +50 K R = 8.314 J mol -1 K -1 W, ΔU, Q Split diagram showing isochoric (locked piston) vs isobaric (rising piston) processes with flames. Left: Isochoric heating with a locked piston and rising pressure. Right: Isobaric heating with a rising piston at constant pressure. Clean side‑by‑side comparison. Isochoric process (V constant) In an isochoric process the piston is clamped so the volume stays fixed: V= constant . Because dV=0 , the work W= P ,dV=0 regardless of how pressure varies. Any heat supplied changes only the internal energy: Q = U = nC v T . Pressure scales with temperature (Gay‑Lussac’s law): P T = constant . On a P – V diagram the path is a vertical line. On a P – T diagram it is a straight line through the origin. Isochoric heating is an efficient way to raise temperature when work output is not desired; conversely, cooling at fixed volume is a direct way to lower U without extracting work. At constant volume, heat only changes internal energy. Isochoric Relations This set of relations applies specifically to thermodynamic processes occurring at constant volume (isochoric process). neet-alert Trap: In any isochoric process, W=0 even if pressure changes a lot. Students sometimes plug values into W=P V using the final pressure; that is wrong because V=0 for the whole path. Adiabatic process (Q = 0) In an adiabatic process the system is thermally insulated so that no heat flows across the boundary: Q=0 . The change must also be fast enough (or the insulation good enough) that heat exchange is negligible. With Q=0 , the first law gives W = - U . Hence adiabatic expansion cools the gas as it does work, while adiabatic compression heats it. For a reversible ideal‑gas adiabatic path, the state variables obey Poisson’s relations: PV = constant , TV -1 = constant , and P 1- T = constant , where =C p/C v . On a P – V diagram an adiabatic curve passing through a given (P,V) is steeper than the corresponding isotherm at that point by a factor : | dP dV | adi = , | dP dV | isoT = , P V . This difference is frequently tested in NEET graph questions. P 1 V 1 = P 2 V 2 Adiabatic equation for a reversible ideal gas: PV γ = constant Ideal gas with constant heat capacities Quasi-static (reversible) adiabatic process No heat exchange: dQ = 0 Adiabatic Work Work by a reversible ideal‑gas adiabatic process (positive for expansion with T 2<T 1 ). Use this formula to calculate the work done by a gas undergoing a reversible adiabatic process, where no heat is exchanged with the surroundings. Interpreting the adiabatic equations: TV -1 = constant links temperature rise directly to compression ratio. For a given volume change, a larger (fewer active degrees of freedom) produces a larger temperature change; hence monoatomic gases heat more on compression than diatomic gases. P 1- T = constant is useful when volumes are not given. Remember applicability: ideal gas, reversible (quasi‑static) and adiabatic. Real processes deviate if the compression is too slow (heat leaks) or too fast and non‑quasi‑static (shock heating). medium 1.0 mol of a diatomic ideal gas ( =1.4 ) is compressed reversibly and adiabatically from V 1=2.0 , L to V 2=1.0 , L at T 1=300 , K . Find the final temperature T 2 and the work done by the gas. T2 and W Use TV -1 = constant to get T 2 , then W= nR(T 1-T 2) -1 . K, J n = 1.0 mol γ = 1.4 V1 = 2.0 L, V2 = 1.0 L T1 = 300 K R = 8.314 J mol -1 K -1 Insulated piston-cylinder with orange hot gas and a pressure gauge indicating high values. Adiabatic compression: thick thermal insulation around the cylinder prevents heat loss; the piston moves rapidly and the gauge shows high pressure while the gas glows hotter inside. Cyclic processes and PV “indicator” diagrams A cyclic process is a closed loop on the P – V diagram where the system returns to its starting state. Because internal energy is a state function, U cycle =0 . The first law reduces to Q net =W net . The signed area enclosed by the loop equals the net work per cycle. Orientation matters: clockwise loops give W net >0 (work done by the gas, typical of heat engines), and anticlockwise loops give W net <0 (work done on the gas, typical of refrigerators/heat pumps). The path segments may be isothermal, isobaric, isochoric, or adiabatic. Many NEET questions give a simple rectangle or triangle; compute the area using basic geometry or PdV if the path is curved. Always identify the direction of traversal before assigning the sign to W net . Comparison of slopes at a common point: |slope| of adiabatic = γ × |slope| of isotherm. Common state (P,V) Pressure P Pa pv Two curves through the same point: the isotherm is a shallow hyperbola; the adiabatic is steeper (falls faster with V). Volume V m 3 Isotherm derived derived Adiabatic Fast PV‑diagram reading: horizontal line → isobaric; vertical line → isochoric; shallow hyperbola → isotherm; steeper curve through the same point → adiabatic. For curved segments, approximate small areas with rectangles or triangles when exact integrals are not required. For a rectangle with pressures P H and P L and volumes V H and V L , W net =(P H-P L)(V H-V L) with the sign decided by orientation. Orientation, W net, and sign of Q net P H = 3.0× 10 5 Pa P L = 1.0× 10 5 Pa V H = 3.0×10 -3 m 3 V L = 1.0×10 -3 m 3 Path A→B (isochoric up), B→C (isobaric expansion at P H ), C→D (isochoric down), D→A (isobaric compression at P L ). hard A rectangular cycle on a PV diagram has corners A (V L,P L) → B (V L,P H) → C (V H,P H) → D (V H,P L) → A. Determine orientation, net work, and the sign of Q net. Take P H=3.0 10 5 , Pa , P L=1.0 10 5 , Pa , V H=3.0 10 -3 , m 3 , V L=1.0 10 -3 , m 3 . Thermodynamic Processes Process Constant Quantity P-V Graph Shape First Law Form ( Q=ΔU+W ) Work Done Formula Remember 'T-Iso, Q-Adia, P-Bar, V-Chor': Temp stays same, Heat is zero, Pressure bars change, Volume chor(e) stays still. Isothermal Process Temperature ( T ) Rectangular Hyperbola ΔU = 0 Q = W W = 2.303nRT 10 ( V f V i ) Adiabatic Process Heat ( Q=0 ) Steep Hyperbola ( P V = constant ) ΔU = -W or Q=0 W = P i V i - P f V f - 1 = nR(T i - T f) - 1 Isobaric Process Pressure ( P ) Horizontal Line ΔU = Q - P V W = P(V f - V i) = nR(T f - T i) Isochoric Process Volume ( V ) Vertical Line W = 0 Q = U W = 0 Cyclic Process Internal Energy ( ΔU=0 ) Closed Loop ΔU = 0 Q net = W net Area enclosed by the P-V loop thermodynamic processes tip Applicability boundaries: Ideal‑gas formulas assume dilute gas and negligible interactions. Reversible adiabatic formulas require quasi‑static steps and good insulation. Isothermal work formula assumes T truly constant; in finite‑time expansion with poor thermal contact, the actual path lies between isothermal and adiabatic. Common traps: (1) Using Celsius in formulas with ratios (e.g., Carnot efficiency) — always use Kelvin. (2) Confusing natural log with 10 . (3) Plugging gauge pressure instead of absolute pressure in PV=nRT . neet-alert Quick recall of process features and PV look. “PV‑TV‑QW”: Isobaric → horizontal (P fixed), Isochoric → vertical (V fixed), Isothermal → TV talk (T fixed so Q=W), Adiabatic → Quick and insulated (Q=0, slope steeper by γ). Isothermal means no heat exchange. Isothermal keeps temperature constant. Heat must usually flow to balance work so that U=0 for an ideal gas; typically Q 0 . Speed alone does not define the process. Adiabatic needs no heat exchange (often fast or well insulated). Isothermal needs perfect thermal contact and enough time; it may still be done briskly if heat exchange keeps up. Adiabatic means fast; isothermal means slow. Only for an isothermal ideal‑gas process does Q=W hold. In cyclic rectangles, W net equals the area but Q along each leg depends on the path; over a cycle Q net =W net , not along each leg. For any process, Q=W if the graph is a rectangle. Adiabatic index is the ratio of molar heat capacities C p/C v ; greater than 1 for stable gases and key in adiabatic relations. C p - C v = R, = C p C v Mayer’s relation and γ = Cp/Cv Ideal gas with equation PV = nRT Heat capacities independent of T over the range Reversible adiabatic ideal‑gas process satisfies P 1V 1 =P 2V 2 with =C p/C v . Carnot efficiency = 1 - T sink /T source is the upper limit for any engine operating between two reservoirs. = 1 - T sink T source Reversible Carnot cycle with two isotherms and two adiabatics Heat exchanges only during isothermal legs at T source and T sink Carnot engine efficiency: η = 1 − T sink/T source Thermal efficiency for any heat engine defined by heat terms: = 1 - Q sink /Q source . = 1 - Q sink Q source Cyclic engine: ΔU cycle = 0 Define W = Q in − Q out with magnitudes Q source, Q sink Thermal efficiency in terms of heats: η = 1 − Q sink/Q source Refrigerator COP (performance): = Q cold /W in , often greater than 1. = Q cold W in Refrigerator COP: α = Q cold / W in Steady cyclic refrigerator Work input drives heat from cold to hot reservoir Translate words to constraints: fix P, V, T, or Q=0. Write the process relation ( PV= const for isothermal, PV = const for adiabatic, etc.). Use W= P ,dV or the compact formula for that process. Compute U = nC v T (ideal gas) and then Q = U + W with the correct signs. Check units (Pa, m 3 , K) and whether the answer sign matches the physical picture. Strategy to crack any ideal‑gas process problem Key exam takeaways: Know which quantity is fixed in each named process and the immediate consequence for Q , W , and U . Remember the slope rule (adiabatic steeper than isothermal through the same point) for graph questions. For rectangular loops, the area gives |W net | and orientation dictates the sign. Use Kelvin for any formula involving ratios of temperature, and always keep pressures absolute. With these habits and the standard formulas, you can navigate even long multi‑step NEET questions with confidence. Quick glossary Isothermal T-constant Temperature constant; U=0 (ideal gas), Q=W . No heat exchange; PV = constant for reversible ideal gas. Q=0 Adiabatic Pressure constant; W=nR T , Q=nC p T . Isobaric Volume constant; W=0 , Q= U=nC v T . Isochoric Isometric Cyclic process Closed loop on PV diagram; U cycle =0 and Q net =W net . =C p/C v , determines adiabatic steepness and temperature change with compression/expansion. Heat capacity ratio Adiabatic index