Exchange of heat is an important characteristic to note down in the fields of chemistry and thermal engineering. Considering the significance of this process, we are going to learn about a chemical procedure called isothermal expansion. After reading through this context, we will be able to know how the work done in an isothermal process is different in a reversible and irreversible state, and how heat gets transferred to another medium, along with important formulas and other pointers of study.
But before everything, let us first quickly grasp the isothermal process definition along with 2 important gas forms namely ‘Real’ and ‘Ideal’ in brief.
Isothermal Expansion
To gain basic knowledge about the isothermal expansion of an ideal gas and real, it is essential to know what both these gases mean.
An ideal gas possesses atoms and molecules that are highly elastic. Since the molecules of an ideal gas move faster than any other source, there is an absence of any intermolecular force of attraction between the elements. Moreover, the atoms and molecules in an ideal gas are present quite far away (distantly) and hence interaction is not possible at all.
Also, ideal gases have their heat stored in the form of kinetic energy within each particle. This change in the internal energy leads to the change in the temperature, thus resulting in what exchange. Helium is a classic example to state as an ideal gas.
Note that an ideal gas, under a certain reasonable tolerance condition, can change its medium into a real gas.
On the other hand, when a gaseous element has a minimal level of intermolecular attractive forces between their molecules and atoms, then it can be termed as real gas. In the case of an ideal gas, it cannot exist and thrive naturally in the ecosystem. But, real gases can ideally act in both high-temperature conditions as well as in low-pressure situations. The common examples for a real gas include nitrogen (N), Helium (He), Oxygen (O), and more.
Now, let’s move onto the topic of the isothermal expansion process. An isothermal process is defined by the change in a particular system where the temperature will remain constant. To be more precise, isothermal expansion gives ∆T = 0 (no change in the temperature).
When the vacuum gets expanded, it leads to the free expansion of a gas. In the case of an ideal gas, the rate of free expansion is NIL, that is, the work done is 0. The value of 0 is the result regardless of whether the process is irreversible or reversible.
Some of the reversible cases of isothermal expansion include converting ice from its solid-state to the liquid state as water, dehydrogenation and hydrogenation in milling a chemical and more. Examples for an irreversible condition include work that is done against the friction, Joule’s heating effect, magnetic hysteresis and so on.
P-V Diagram to Represent Isothermal Process
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When a system has its internal energy changed, then it is given by the following condition:
∆U = q + w --(1)
Check out the keys to the formula below.
‘∆U’ to represent a change in internal energy.
‘q’ is labelled to denote the heat given by that respective system.
‘w’ is the specific amount of work done over the system.
The following cases are given as isothermal process examples and types….
∆U = q + pex ∆V
w = pex ∆V is the method to denote a work done in the condition of vacuum, Hence, the equation denoted as 1 (from the top) can also be represented as:
∆U = q + pex ∆V
Moving on, if the volume in the condition is 0, therefore…
∆V = 0 → ∆U = q + pex ∆V
∆U is the work done in this NIL vacuum condition. This also results in the fact that ∆U = Q. ‘qv’ is used to symbolize that the volume i.e. getting heat supplied at a constant rate.
Let us consider one more instance where an ideal gas like Helium (He) is subjected to isothermal expansion in the presence of vacuum (∆T = 0). Here, the work done for this vacuum will be NIL, that is w = 0 since the pex=0. According to the experiments of Joule, q =0 and therefore it is concluded that work done is NIL that is ∆U = 0.
Lastly, take this formula: ∆U = q + w. Now, we can express this statement for both reversible and irreversible isothermal expansions processes with the pointers given below:
Isothermal reaction for a reversible change is mentioned as q = -w = pex (Vf-Vi)
Isothermal reaction for an irreversible change is given by q = -w = nRTln (Vf/Vi) = 2.303 nRT log (Vf/Vi)
For in the case of an Adiabatic change, it is written as Q =0, ∆U = wad
Remember that the work done in an isothermal process of expansion for any given gas and vacuum condition is to be denoted as T = Constant, ∆T = 0 and dT = 0.
FAQs on Isothermal Expansion of an Ideal Gas
1. What is meant by the isothermal expansion of an ideal gas?
The isothermal expansion of an ideal gas is a thermodynamic process in which the gas expands and its volume increases, while its temperature remains constant. To achieve this, the system must be in perfect thermal contact with a surrounding heat reservoir, allowing it to absorb heat to compensate for the energy used in doing work on its surroundings.
2. What is the formula for calculating the work done during a reversible isothermal expansion of an ideal gas?
The work done (W) by an ideal gas during a reversible isothermal expansion from an initial volume V₁ to a final volume V₂ is calculated using the formula: W = -nRT ln(V₂ / V₁). Because pressure and volume are inversely related for an ideal gas at constant temperature (Boyle's Law), the formula can also be written in terms of pressure: W = -nRT ln(P₁ / P₂). Here, 'n' is the number of moles of the gas, 'R' is the ideal gas constant, and 'T' is the absolute temperature.
3. Why are the changes in internal energy (ΔU) and enthalpy (ΔH) zero for the isothermal expansion of an ideal gas?
For an ideal gas, its internal energy (U) is solely a function of its temperature. Since an isothermal process occurs at a constant temperature (meaning ΔT = 0), the change in internal energy (ΔU) is necessarily zero. Enthalpy (H) is defined as H = U + PV. For an ideal gas, this becomes H = U + nRT. As both U and T are constant during this process, the change in enthalpy (ΔH) must also be zero.
4. How does the First Law of Thermodynamics apply to an isothermal expansion process?
The First Law of Thermodynamics states that ΔU = q + w, where ΔU is the change in internal energy, q is heat, and w is work. For the isothermal expansion of an ideal gas, we know that ΔU = 0. Therefore, the equation simplifies to 0 = q + w, which can be rearranged to q = -w. This shows that all the heat absorbed by the system from the surroundings (q) is converted into work done by the system on the surroundings (-w).
5. What is the key difference between reversible isothermal expansion and free expansion?
The primary difference lies in the work done and the conditions of the expansion.
- In a reversible isothermal expansion, the gas expands slowly against an external pressure. This process performs the maximum possible work, and heat (q) must be supplied from the surroundings to keep the temperature constant.
- In a free expansion, the gas expands into a vacuum where the external pressure is zero (P_ext = 0). Consequently, no work is done (w = 0). For an ideal gas, this also means no heat is exchanged (q = 0), and the process is therefore both isothermal and adiabatic.
6. What is a real-world example of a process that approximates isothermal expansion?
A good example is the slow expansion of a gas inside a thin, conductive metal cylinder fitted with a piston. If this cylinder is submerged in a large container of water (a heat reservoir), the water will supply heat to the gas as it expands. If the expansion is done slowly enough, the temperature of the gas will remain nearly constant, closely mimicking an isothermal process.
7. How is the formula for work done in a reversible isothermal expansion derived?
The derivation begins with the fundamental equation for pressure-volume work, dW = -P_ext dV. For a reversible process, the external pressure (P_ext) continuously matches the internal gas pressure (P). Using the ideal gas law, P = nRT/V, we can substitute for P:
- Start with the differential work: dW = - (nRT/V) dV
- Integrate this expression from the initial volume (V₁) to the final volume (V₂): W = ∫ - (nRT/V) dV
- Since n, R, and T are constants in this process, they are moved outside the integral: W = -nRT ∫ (1/V) dV
- Solving the integral gives the final expression: W = -nRT ln(V₂/V₁).
