Introduction:
Thermodynamics is basically the study of the relationship between the heat (or energy) and work. Enthalpy is the central factor in thermodynamics. This is the heat content of a system. The heat which passes to or out of the system throughout a reaction is the enthalpy change. Whether the enthalpy of the system rises (that is, whenever energy is added) or reduces (as energy is given off) is a crucial factor which finds out whether a reaction can occur.
At times, we state the energy of the molecules undergoing change the 'internal enthalpy'. At times, we state it the 'enthalpy of the system'. Such two phrases refer to the similar thing. Likewise, the energy of the molecules which don't take part in the reaction is termed as the 'external enthalpy' or the 'enthalpy of the surroundings'.
Enthalpy and enthalpy changes:
Enthalpy is the amount of heat content utilized or liberated in a system at constant pressure. Enthalpy is generally deduced as the change in enthalpy. The change in enthalpy is associated to the change in internal energy (E) and a change in the volume (V), which is multiplied via the constant pressure of the system.
Define Enthalpy: Enthalpy (H) is the sum of internal energy (E) and the product of pressure and volume (PV) represented by the equation:
H = E + PV
Whenever a process takes place at constant pressure, the heat evolved (either liberated or absorbed) is equivalent to the change in enthalpy. Enthalpy is the state function that totally depends on the state functions T, P and E. The enthalpy is generally represented as the change in enthalpy (ΔH) for a procedure between the initial and final states:
ΔH = ΔE + ΔPV
If pressure and temperature remain constant via the process and the work is limited to pressure-volume work, then the enthalpy change is represented by the equation:
ΔH = ΔE + PΔV
As well at constant pressure, the heat flow (q) for the process is equivalent to the change in enthalpy stated by the equation:
ΔH = q
Define Enthalpy change:
Enthalpy change (?H) is the amount or quantity of heat energy taken in or given out throughout any change in a system given the pressure is constant.
By examining at whether 'q' is exothermic or endothermic we can find out a relationship between ΔH and q. If the reaction absorbs heat, it is endothermic implying the reaction uses heat from the surroundings therefore, q > 0 (positive). Thus, at constant pressure and temperature, by the equation above, if 'q' is positive then ΔH is as well positive. And the similar goes for if the reaction liberates heat, then this is exothermic, meaning that the system gives off heat to its surroundings, therefore q < 0 (negative). If 'q' is negative, then ΔH will as well be negative.
If H_{1} and H_{2} are the enthalpies of the initial and final states of a system, then the enthalpy change accompanying the procedure is represented by,
ΔH = H_{2} - H_{1}
= (E_{2} + P_{2}V_{2}) - (E_{1} + P_{1}V_{1})
= ΔE + (P_{2}V_{2} - P_{1}V_{1})
In case of a constant pressure process P_{1} = P_{2} = P. Therefore, the equation can be written as,
ΔH = ΔE + P (V_{2} - V_{1})
For a finite change, we obtain:
q_{p }= ΔE + PΔV
For a small change in enthalpy, we can write:
dq_{P }= dH
Supposing that there is no phase change or chemical reaction, we have,
dH = C_{P}dT = n C_{p}dT
If an ideal gas is heated from temperature T_{1} to T_{2} at constant pressure, then the integrated form of equation is to be employed.
dH = _{T1}∫^{T2} C_{P}dT = _{T1}∫^{T2 }n C_{p}dT
Take a change in the state of a system,
Initial state, H_{1} = E_{1} + P_{1}V_{1}
Final state, H_{2} = E_{2} + P_{2}V_{2}
Therefore, H_{2} - H_{1} = dH = dE + P (dV)
At constant pressure:
dH = dE + PdV
For an ideal gas:
dH = dE + dnRT at constant temperature.
The unit of 'R' based on the unit of energy being utilized. We must not forget that SI units as well apply here. The table below illustrates the units of energy and 'R' for any conversion we might need in our computation.
Unit of energy
Unit of R
Cal
1.987 Cal mol^{-1}K^{-1}
Joules
8.314 J mol^{-1} K^{-1}
Lit-atm
0.08205 Liter-atm/mol. K
If the energy needed is more than the energy liberated, the total result will be the absorption of energy and the reaction will be endothermic, that is, (dH = + ve). If energy liberated is greater than the energy needed, the total result is the discharge of energy and the reaction will be exothermic (dH = - ve).
Enthalpy changes associated with some typical methods are given special names. For illustration: enthalpy of vaporization or evaporation is the enthalpy change accompanying the conversion of one mole of a liquid to its vapor. Likewise, enthalpy of fusion or sublimation is the enthalpy change accompanying fusion or sublimation of one mole of the substance. For a chemical reaction, the enthalpy of reaction is the difference in the enthalpies of the products and the reactants as per the stoichiometric provided in the chemical equation.
Relationship between C_{P} and C_{V} of an ideal gas:
The internal energy of an ideal gas depends just on its temperature and is independent of the volume and pressure. This is quite comprehensible as in an ideal gas, there are no intermolecular interactions, and so forth, attractive or repulsive forces have to be overcome throughout expansion. Though, the enthalpy of the gas changes considerably whenever it expands or contracts.
For one mole of an ideal gas:
H = E + PV = E + RT
On differentiating, we obtain:
dH = dE + RdT (Since R is a constant)
C_{P}dT = C_{V}dT + RdT
C_{P} = C_{V} + R
And therefore, C_{P} - C_{V} = R
As well, for n mole, C_{P} - C_{V} = nR
This signifies that C_{P} is always greater than C_{V} for the Ideal gas. This is because if the temperature of a gas is increased at a constant pressure, there will be expansion of the gas. This will need some extra amount of heat (as compared to heating an ideal gas beneath a constant volume condition). Therefore, more heat will be needed to increase the term nature of the gas via 1K under constant pressure conditions than beneath the constant volume.
Adiabatic expansion:
How does pressure differ with volume change throughout the adiabatic expansion and compression in the ideal gas? Air at atmospheric pressure is almost ideal gas and sound transmission in air is close to the adiabatic for normal frequencies and transmission distances.
1 mole gas (V_{1}, T_{1}) = 1 mole gas (V_{2}, T_{2})
Adiabatic ⇒ dq = 0 Reversible ⇒ dw = - pdV
Ideal gas ⇒ dU = C_{v}dT
∴ From first Law dU = - pdV ⇒ C_{v}dT = - pdV all along the path
C_{V}dT = - pdV (As P = RT/V)
C_{V} (dT/T) = - R (dV/V)
C_{V T1}∫^{T2} (dT/T) = - R _{V1}∫^{V2} (dV/V) => (T_{2}/T_{1}) = (V_{1}/V_{2})^{R/CV} (As C_{P} - C_{V} = R)
(T_{2}/T_{1}) = (V_{1}/V_{2})^{CP/CV-1}
γ = C_{P}/C_{V} => (T_{2}/T_{1}) = (V_{1}/V_{2})^{γ-1}
For monatomic ideal gas: CV = 3/2 R and CP = 5/2 R => γ = 5/3 (>1 generally)
In an adiabatic expansion (V_{2} > V_{1}), the gas cools (T_{2} > T_{1}).
And, in an adiabatic compression (V_{2} < V_{1}), the gas heats up.
For ideal gas (one mole) T = PV/R => (P_{2}/P_{1}) = (V_{1}/V_{2})^{γ}
=> P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ}
PV^{γ} is constant all along the reversible adiabatic
Irreversible adiabatic expansion:
1 mol gas (P_{1}, T_{1}) = 1 mol gas (P_{2}, T_{2}) (P_{ext }= P_{2})
Adiabatic => dq = 0
Constant P_{ext} = P_{2 }=> dw = - P_{2}dV
Ideal gas => dU = C_{V}dT
First Law => dU = - P_{2}dV
Thus, C_{V}dT = - P_{2}dV
On integrating, C_{V} (T_{2} - T_{1}) = - P_{2} (V_{2} - V_{1})
By using PV = RT,
T_{2} (C_{V} + R) = T_{1} [C_{V} + (P_{2}/P_{1}) R]
Note, P_{2} < P_{1 }⇒ T_{2} < T_{1} Again, expansion cools
Note as well, (- w_{rev}) > (- w_{irrev}) Less work is recovered via an irreversible process
The Joule-Thomson effect:
Joule-Thomson effect is the change in temperature which accompanies expansion of the gas without production of work or transfer of heat. At ordinary pressures and temperatures, all real gases apart from helium and hydrogen cool on such expansion; this phenomenon frequently is used in liquefying gases. The phenomenon was introduced in the year 1852 by the British physicists James Prescott Joule and William Thomson (Lord Kelvin). The cooling takes place as work should be done to overcome the long-range attraction between the gas molecules as they move farther apart. Hydrogen and helium will cool on expansion only if their initial temperatures are extremely low due to the long-range forces in these gases are oddly weak.
Joule-Thomson coefficient:
The change of temperature 'T' by a decrease of pressure 'P' at constant enthalpy 'H' in a Joule-Thomson procedure is the Joule-Thomson coefficient represented as μ_{JT} and might be deduced as:
μ_{JT} = (∂T/∂P) H
The value of μ_{JT} is generally deduced in K/Pa or °C/bar (SI units: K/Pa) and based on the specific gas, and also the pressure and temperature of the gas before expansion.
For all the real gases, it will equivalent zero at certain point termed as the inversion point and, as illustrated above, the Joule-Thomson inversion temperature is the temperature where the coefficient changes sign (that is, where the coefficient equivalents zero). The Joule-Thomson inversion temperature based only on the pressure of the gas prior to expansion.
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