Physics · Thermodynamics · NEET
Work needs a change in volume, because W = integral of P dV. In an isochoric process the volume stays fixed, so dV = 0 at every step and the integral is zero. On a P-V graph this process is a vertical line, and a vertical line has zero area under it. So all the heat given goes into internal energy: Q = delta U.
Adiabatic compression needs MORE work. On a P-V diagram the adiabatic curve (PV^gamma = constant) is steeper than the isothermal curve (PV = constant) because gamma is greater than 1. During compression the adiabatic curve rises above the isothermal, so the area under it (the work) is larger. Physical reason: in adiabatic compression no heat escapes, the gas heats up, pressure climbs faster, so you push harder. (NEET 2016 tested this exact idea.)
For work done BY the gas: expansion (volume increases) gives positive work, compression (volume decreases) gives negative work. This is the physics sign convention used in NCERT first law Q = delta U + W. Just watch the wording: 'work done ON the gas' is the opposite sign of 'work done BY the gas'.
Work done by the gas equals the area between the curve and the volume axis. For a straight horizontal line (isobaric) it is a rectangle P times delta V. For a curve, it is the area under the curve. For a full closed loop (cyclic), work equals the enclosed area: clockwise loop means positive work, anticlockwise means negative work.
Use W = nR(T1 - T2)/(gamma - 1) when you know the two temperatures, or W = (P1V1 - P2V2)/(gamma - 1) when you know pressures and volumes. Both give work done BY the gas. In adiabatic expansion T2 is less than T1, so W is positive and the gas cools. This works because Q = 0, so W = -delta U by the first law.
For the given cycle, the work done during the isobaric process is (pressure P = 200 kPa, volume changes from 1 to 4 litre).
In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (gamma = 5/3) decreases from 60 K to 50 K. The work done by the gas is (R = 8.3 J/mol/K).
An ideal gas expands isothermally from 10^-3 m^3 to 10^-2 m^3 at 300 K against a constant external pressure of 10^5 N/m^2. The work done ON the gas is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Work done by a gas is W = integral of P dV, which equals the area under the P-V curve. Every specific formula (isobaric, isothermal, adiabatic) comes from solving this one integral for that process.
W = nRT ln(V2/V1), which can also be written W = 2.303 nRT log10(V2/V1). Temperature is constant, so for an ideal gas delta U = 0 and Q = W.
W = nR(T1 - T2)/(gamma - 1) = (P1V1 - P2V2)/(gamma - 1). Since no heat is exchanged (Q = 0), the first law gives W = -delta U.
Because volume does not change, so dV = 0 and W = integral of P dV = 0. All heat supplied raises internal energy: Q = delta U = nCv delta T.
Work done equals the area enclosed by the loop on the P-V diagram. A clockwise loop gives positive net work, an anticlockwise loop gives negative net work. Since the gas returns to its start, delta U = 0, so Q = W.
For work done by the gas: expansion (volume up) is positive, compression (volume down) is negative. Watch for the phrase 'work done ON the gas', which flips the sign.
