For a reversible process at \(T=300 \mathrm{~K}\), volume of the ideal gas is increased from \(1 \mathrm{~L}\) to \(10 \mathrm{~L}\). If the process is isothermal, the \(\Delta H\) of the process is (a) \(11.47 \mathrm{~kJ}\) (b) \(4.98 \mathrm{~kJ}\) (c) 0 (d) \(-11.47 \mathrm{~kJ}\)

When learning about thermodynamics, an essential concept is the Ideal Gas Law, which is defined by the equation \(PV = nRT\), where \(P\) represents pressure, \(V\) volume, \(n\) moles of gas, \(R\) the ideal gas constant, and \(T\) temperature. This law correlates the pressure, volume, temperature, and amount of an ideal gas within a system in a simple way.

An isothermal process, such as the one described in our exercise, assumes that the temperature (\(T\)) remains constant. Under this condition, the Ideal Gas Law implies that pressure and volume are inversely related; as one increases, the other must decrease. Thus, if you increase the volume of an ideal gas at constant temperature, the pressure must decrease to maintain the equality \(PV = nRT\). However, when dealing with enthalpy changes in isothermal processes, this information alone is not enough, as we also need to consider the first law of thermodynamics and properties of internal energy.

The First Law of Thermodynamics is a statement of energy conservation, which in the context of thermodynamics means that the total internal energy of an isolated system is constant unless acted upon by external work or heat transfer. Expressed mathematically, it is \(\Delta U = Q - W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added to the system, and \(W\) is the work done by the system.

For an isothermal process involving an ideal gas, where temperature is constant, we can infer that internal energy remains unchanged (\(\Delta U = 0\)) because the internal energy of an ideal gas is a function of temperature. Since any heat added to the system (\(Q\)) will be equal to the work done by the system (\(W\)), the internal energy does not change. Importantly, this means that even when work is done or heat is transferred, this does not result in a change in the internal energy or enthalpy in an isothermal process for an ideal gas.

The concept of Internal Energy of an Ideal Gas is tied closely to its temperature. Internal energy (\(U\)) is the total energy contained by a system's particles, mainly due to kinetic energy because ideal gas particles are assumed not to interact with each other except during elastic collisions.

In our exercise, since the temperature remains constant in an isothermal process, and because internal energy for an ideal gas is only temperature-dependent, no change in internal energy occurs. This fact allows us to simplify the first law of thermodynamics to the relationship \(Q = W\) in the context of the given process. It is important for students to understand that in this scenario, despite the volume change, the lack of temperature change means that the internal energy stays the same, as does the enthalpy.

The Enthalpy of an Ideal Gas, represented as \(H\), is another state function, which includes the internal energy plus the product of the pressure and the volume of the gas (\(H = U + PV\)). Because enthalpy is also a function of temperature for an ideal gas, during an isothermal process where the temperature does not vary, neither does the enthalpy.

In the context of the question we reviewed, since the process is isothermal and the gas is ideal, the change in enthalpy (\(\Delta H\)) is effectively zero. It might be confusing when considering that volume changes, as one might think this would affect the \(PV\) term in the enthalpy formula. However, the pressure adjusts inversely to volume due to the Ideal Gas Law, keeping the product \(PV\) constant and thereby keeping the enthalpy (\(H\)) constant as well. This understanding is crucial when sorting through potential answers to homework problems that involve thermodynamics and physical chemistry.