Physics · Thermodynamics · NEET
Because pressure keeps changing during the process. In an isothermal process PV = nRT = constant, so P = nRT / V. As the gas expands, V goes up and P goes down at every instant. Work is W = integral of P dV, so you must integrate nRT/V dV from V1 to V2. The integral of 1/V is ln(V), which is why you get W = nRT ln(V2/V1). The natural log is the mathematical fingerprint of pressure varying as 1/V.
From the ideal gas law PV = nRT. For an isothermal process T is fixed, and n and R are constants, so the whole product nRT is a single constant that can be pulled out of the integral. That is why W = integral (nRT/V) dV = nRT integral (1/V) dV = nRT ln(V2/V1). You can also write nRT as P1V1 or P2V2 since PV is constant along the path.
They are the same value. Because PV = constant in an isothermal process, P1V1 = P2V2, which means V2/V1 = P1/P2. So W = nRT ln(V2/V1) = nRT ln(P1/P2). Notice the pressures are flipped: volume ratio is final over initial, but pressure ratio is initial over final. Mixing them up (using P2/P1) gives the wrong sign, a common mistake.
Look at the volume. Expansion means V2 > V1, so V2/V1 > 1, so ln is positive and W is positive (gas does work on the surroundings). Compression means V2 < V1, so V2/V1 < 1, so ln is negative and W is negative (work is done on the gas). You never memorise a separate sign; the log carries the sign automatically once you plug in V2 and V1 correctly.
W = P delta V only works when pressure is constant (isobaric process). In an isothermal process pressure changes continuously as the gas expands, so a single P does not exist for the whole path. You must integrate: W = integral P dV with P = nRT/V. Using P delta V with one pressure value would be wrong. The only time a NEET question uses P delta V for an expansion is when it clearly states a constant external pressure, which is an irreversible expansion, not the reversible isothermal formula.
If V2 = 2 V1, then V2/V1 = 2 and ln(2) = 0.693. So W = nRT ln(2) = 0.693 nRT. If the volume is halved (compression), V2/V1 = 0.5 and ln(0.5) = -0.693, giving W = -0.693 nRT. Keep ln(2) = 0.693 ready; it appears in most isothermal NEET numericals.
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
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
W = nRT ln(V2/V1) for work done by n moles of an ideal gas expanding from volume V1 to V2 at constant temperature T. It can also be written W = nRT ln(P1/P2) = 2.303 nRT log10(V2/V1).
Positive. In expansion V2 > V1, so ln(V2/V1) is positive, meaning the gas does positive work on the surroundings. In compression it is negative.
Zero. Internal energy of an ideal gas depends only on temperature, and temperature is constant in an isothermal process, so delta U = 0. From the first law Q = delta U + W, this gives Q = W: all the heat absorbed is converted to work.
In an isobaric process pressure is constant so W = P delta V. In an isothermal process pressure changes as 1/V, so you must integrate P dV, which gives the natural log form W = nRT ln(V2/V1).
It is the same isothermal work written with base-10 log for easy calculation: W = 2.303 nRT log10(V2/V1), because ln(x) = 2.303 log10(x). It gives an identical numerical answer.
