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For an isothermal expansion of an ideal gas, $\vartriangle \text{U}$ is zero. If true enter 1 else 0.
A. 1
B. 0

Last updated date: 23rd Apr 2024
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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 $\text{(}\vartriangle \text{T=0)}$.

Complete answer:
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.
Suppose we consider an Ideal gas. Let it undergo an Isothermal process.
-If Average kinetic energy is a constant, then total kinetic energy will also be constant. Internal energy for an ideal gas is total kinetic energy which makes internal energy constant.
-Internal energy is a state function which depends on temperature. Thus, the internal energy change is zero.
-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 (\[\vartriangle \text{U}=0\]).
So, it is true that the internal energy of the system remains constant during isothermal expansion of gases.

So, the correct answer is “Option A”.

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.
-Real gases have intermolecular attractions between molecules at low pressure and feel repulsion at high pressure. Its internal energy changes with change in pressure when temperature is constant. Consider boiling water. Water is a liquid; we cannot neglect the attraction between particles so potential energy cannot be neglected. While boiling, all the heat energy supplied increases the potential energy of the system, but keeps the total kinetic energy constant. So there is a constant temperature, by definition this is an isothermal process, but potential energy is increasing the internal energy of the system is also increasing.
-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, \[\vartriangle \text{U}=0=\text{Q}-\text{W thus Q}=\text{W}.\]

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