Physics · Thermodynamics · NEET
No. For an ideal gas, internal energy depends only on temperature. An ideal gas has no intermolecular forces, so there is zero potential energy between molecules. All the internal energy is molecular kinetic energy, and kinetic energy depends only on temperature. So if you compress a gas but keep T the same, U does not change. Two states at the same temperature have the same U even if pressure and volume are totally different.
In an ideal gas, molecules are treated as point particles with no forces between them, so there is no potential energy that could depend on how far apart the molecules are (that is, on volume). Only kinetic energy remains, and by kinetic theory this is fixed by temperature. Real gases DO have intermolecular forces, so their U depends slightly on volume too. NEET problems almost always assume ideal gas, so use U depends only on T.
Yes. In an isothermal process temperature stays constant, so delta T = 0. Since U depends only on T, delta U = n Cv (delta T) = 0. This is why in an isothermal process all the heat supplied goes into work: from the first law, Q = delta U + W becomes Q = W. Do not confuse delta U = 0 with Q = 0 (that would be adiabatic).
Use delta U = n Cv (delta T). This formula works for ANY process, not just constant volume, because U depends only on T. In an adiabatic process Q = 0, so from the first law delta U = -W. If temperature falls, delta U is negative (internal energy decreases) and the gas does positive work. Example: 1 mole monatomic gas cooling 60 K to 50 K gives delta U = (1)(3R/2)(-10), and work done by gas = -delta U.
Internal energy is a STATE function. It depends only on the current state (fixed by temperature for an ideal gas), not on how the gas reached that state. So delta U between two states is the same for every path connecting them. Heat (Q) and work (W) are path functions, they change with the route. This is the whole reason delta U = 0 for any cyclic process.
Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius rA and rB. On supplying an equal amount of heat to both gases reversibly at constant pressure, the pistons of A and B are displaced by 16 cm and 9 cm respectively. If the change in their internal energy is the same, then the ratio rA/rB is
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 in the process is (Take R = 8.3 J/mol/K)
At a certain temperature, during a process, 500 J is absorbed by the system and 200 J of work is done by the system. The change in internal energy of the system is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
U = n Cv T, where n is number of moles, Cv is molar specific heat at constant volume, and T is absolute temperature in kelvin. The change is delta U = n Cv (delta T), and this holds for every process.
Because U of an ideal gas depends only on T. Cv is just the constant that links U to T. So even in an isobaric or adiabatic process, delta U = n Cv (delta T). Cv here is a property of the gas, not a statement that volume is fixed.
Zero. A cycle returns to its starting state, so the temperature returns to its initial value. Since U is a state function fixed by temperature, delta U = 0 over one complete cycle, which means Q = W for the cycle.
In general internal energy = molecular kinetic energy + molecular potential energy. For an ideal gas the potential energy is zero (no intermolecular forces), so U is purely kinetic and depends only on temperature. For real gases the potential part matters.
Yes, for an ideal gas. Same temperature means same U, regardless of differences in pressure or volume. This is the key idea NEET tests again and again.
