For an irreversible isothermal expansion of an ideal gas, the gas's temperature remains constant. This constancy is key to understanding the change in internal energy (\( \Delta U \)).
Step 1: First Law of Thermodynamics.The First Law of Thermodynamics is expressed as: \[\Delta U = q + w,\] where \( \Delta U \) represents the change in internal energy, \( q \) is the heat exchanged, and \( w \) is the work done on or by the system.
Step 2: Isothermal Process for an Ideal Gas.In an isothermal process, temperature is constant. For an ideal gas, internal energy is solely dependent on temperature. Therefore, for an isothermal process, the change in internal energy (\( \Delta U \)) is zero: \[\Delta U = 0.\]Substituting this into the First Law: \[0 = q + w.\]This simplifies to: \[q = -w.\]This means the heat absorbed by the system is equal in magnitude but opposite in sign to the work done by the system.
Step 3: Work and Heat Direction.During gas expansion:- Work (\( w \)) is positive as the gas expands against external pressure.- To maintain constant temperature during expansion, the gas must absorb heat (\( q \)) equal to the work done but with opposite sign, hence \( q = -w \).
Conclusion:For an irreversible isothermal expansion of an ideal gas: \[\Delta U = 0, \quad q = -w.\]Therefore, the correct option is \( \mathbf{(2)} \).