Physics · Thermodynamics · NEET
No. Work is zero in an ISOCHORIC (constant volume) process, not isobaric. In an isobaric process the volume DOES change, so W = P(V2 - V1) is non-zero. Students mix up the two because the names sound similar. Rule to remember: constant Volume -> ΔV = 0 -> W = 0 (isochoric); constant Pressure -> P comes out of the integral -> W = PΔV (isobaric).
Start with W = P(V2 - V1). For an ideal gas, PV = nRT, so at points 1 and 2: PV1 = nRT1 and PV2 = nRT2. Subtracting: P(V2 - V1) = nR(T2 - T1) = nRΔT. So both forms are the same. Use W = PΔV when you know pressure and volumes; use W = nRΔT when you know moles and temperatures.
Use Q = nCp ΔT, where Cp is the molar specific heat at CONSTANT PRESSURE (not Cv). By the first law, Q = ΔU + W = nCv ΔT + nRΔT = n(Cv + R)ΔT = nCp ΔT, which fits Mayer's relation Cp = Cv + R. Part of the heat raises internal energy (ΔU = nCv ΔT) and part does work (W = nRΔT).
W/Q = (nRΔT)/(nCp ΔT) = R/Cp. For a monatomic gas Cp = (5/2)R, so W/Q = R / (5R/2) = 2/5. For a diatomic gas Cp = (7/2)R, so W/Q = 2/7. This exact ratio was asked in NEET 2018.
It is a horizontal straight line, because pressure (the y-axis) does not change. The work done equals the area of the rectangle under this line, W = P x (V2 - V1). If the gas expands (V2 > V1) work is positive (done BY the gas); if it is compressed (V2 < V1) work is negative (done ON the gas).
In isobaric, PRESSURE is fixed and temperature changes, so W = PΔV = nRΔT (a straight rectangle). In isothermal, TEMPERATURE is fixed and pressure changes, so W = nRT ln(V2/V1) (area under a curve). They are two different curves on the P-V graph; do not use PΔV for an isothermal process.
For a given cycle, the work done during the isobaric process is (the isobaric segment is at constant pressure P = 200 kPa, with the volume changing from 1 x 10^-3 m^3 to 4 x 10^-3 m^3).
The volume V of a monatomic gas varies with its temperature T along a straight line through the origin (from A to B), i.e. an isobaric process. The ratio of the work done by the gas to the heat absorbed by it, when it goes from A to B, is
Two gases A and B are at the same pressure in cylinders with movable pistons of radius rA and rB. On supplying equal heat to both at constant pressure, the pistons of A and B move by 16 cm and 9 cm respectively. If the change in their internal energy is the same, the ratio rA/rB is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
A thermodynamic process in which the pressure of the gas stays constant while its volume and temperature change, so work done is W = P(V2 - V1) = nRΔT.
W = P(V2 - V1), which for an ideal gas also equals nRΔT = nR(T2 - T1). Use whichever data you are given.
Cp, the molar specific heat at constant pressure. Heat supplied Q = nCp ΔT, and by the first law this splits into ΔU = nCv ΔT (internal energy) plus W = nRΔT (work).
Yes. The pan is open to the atmosphere, so the pressure on the liquid/gas stays at atmospheric pressure while heating changes volume and temperature. That is why open-air heating and boiling are common real examples of isobaric processes.
As a horizontal straight line. The work done equals the area of the rectangle between that line and the volume axis, W = P x (V2 - V1).
