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# For an isothermal expansion of an ideal gas, $\vartriangle \text{U}$ is zero. If true enter 1 else 0.A. 1B. 0  Verified
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Hint: The internal energy change of the ideal gas cannot be determined without knowing the actual meaning of these terms and what they represent. Isothermal is the process where the temperature during the process remains constant or unchanged $\text{(}\vartriangle \text{T=0)}$.

To understand this question, we need to understand two terms; Isothermal process and Internal Energy. Isothermal process is a process in which properties of a substance change when temperature constant$\text{(}\vartriangle \text{T=0)}$. Internal Energy is the sum of kinetic energy and potential energy of all particles in the system. Temperature is directly proportional to the average kinetic energy of particles in a system. Ideal gas is a gas in which we neglect attraction between particles then, there is no potential energy and thus here internal energy is total kinetic energy.
-When a system is in contact with an outside heat bath and the change in that system will take place slowly that does not allow the system to adjust to the temperature of the reservoir by heat exchange. So, in an isothermal process, $\text{(}\vartriangle \text{T=0)}$. Thus, the change in internal energy is also zero ($\vartriangle \text{U}=0$).
Note: There are differences in conditions in real gases and ideal gases. It is not always true that $\vartriangle \text{U}=0$in an isothermal process.
-An ideal gas has no interactions between the particles with no intermolecular forces, so the change in pressure at constant temperature does not change its internal energy. For an ideal gas in an isothermal process, $\vartriangle \text{U}=0=\text{Q}-\text{W thus Q}=\text{W}.$