INTERNAL ENERGY (E or U)

Internal energy: The internal energy of a system is the total energy stored within the system due to the microscopic motion and interactions of its molecules. It includes:

Kinetic energy of molecules (translation, rotation, vibration)

Potential energy due to intermolecular forces and electronic energy

It is a state function, and it has the SI unit J. It is included in extensive properties and depends on pressure, temperature, volume, and the nature of the matter. The change in internal energy is written as ΔE. mathematically, it is represented as: ΔE = ±q ± w, where ’ΔE’ is the change in internal energy, ‘q’ is the heat, and ‘w’ is the work done. In various thermodynamic processes, the values of internal energy change. For instance, in an isothermal process, ΔE = 0, an adiabatic process, ΔE = w, an isochoric process, ΔE = q, and in an isobaric process, ΔE = q-ΔPV. The change in internal energy can be measured indirectly by calorimetry or using equations that relate internal energy to other thermodynamic properties, such as temperature and volume. The change in internal energy (ΔE) can be derived using the first law of thermodynamics or the law of equipartition of energy.

First law of thermodynamics: According to the first law of thermodynamics, the energy of a system cannot be created or destroyed, only converted from one form to another. Mathematically, it can be expressed as: ΔE = q–w. Where ΔE is the change in total energy, q is the heat added to the system, and w is the work done by the system.

Suppose a gas having internal energy E₁ is enclosed in a cylinder having a frictionless piston. When a quantity of heat ‘q’ is added to the system, the internal energy changes from E₁ to E₂, and the piston moves upward. The change in internal energy can be written as: E₂-E₁=ΔE =q-w. If the work is the pressure-volume work (in case of gases, we always consider the pressure-volume work. Then the change in internal energy can be written as: ΔE =q-PΔV. Or when the work done is on the system, then the change in internal energy can be written as ΔE =q+PΔV. Now, two possibilities arise: 1) when q=0 (adiabatic process), ΔE =PΔV, and ii) when volume remains constant (isochoric process), ΔE =q_v. In addition to this, the values of internal energy at some other process, like an isothermal process (ΔT=0), ΔE=0 (because ΔE depends upon T). Similarly, in an isobaric process (ΔP=0), ΔE is converted into enthalpy (H) by the following relation, ΔE =ΔH-Δ(PV).

Law of equipartition of energy: According to the law of equipartition of energy, the energy associated with each degree of freedom is (½)K_BT (or (½) RT per mole). Thus, for all types of degrees of freedom, the total energy can be calculated as:

Translational degrees of freedom. In translational motions, three degrees of freedom exist. Therefore, the total energy is 3 x (½)K_BT = 1.5 K_BT.

 Rotational degrees of freedom. In rotational motions, three degrees of freedom exist for polyatomic non-linear Therefore, the total energy is 3 x (½)K_BT = 1.5 K_BT. Whereas two degrees of freedom exist for diatomic or polyatomic linear Therefore, the total energy is 2 x (½)K_BT = 1K_BT.

Vibrational degrees of freedom. In vibrational degrees of freedom, 1k_BT per mode exists (½k_BT kinetic + ½k_BT potential), but only at high temperature. At room temperature, most molecules only exhibit translational and rotational DOF, while vibrational DOF are frozen out” at room temperature and become important only at ~10³–10⁴ K.

            This law is followed by the ideal gas, and the derivation of the change in internal energy is carried out at constant volume. Mathematically, it can be written as: 

ΔE= (f/2) RΔT→ for n moles of gas, ΔE= n(f/2) RΔT or ΔE= nC_vΔT (because, f/2)R=C_v).

Summary: Thus, at constant volume, the internal energy can be calculated in the following ways:

From the heat: ΔE =q_v.

From degrees of freedom: ΔE= n(f/2)RΔT (for ideal gases)

From the heat capacity: ΔE= nC_vΔT    

IMPORTANT NUMERICALS RELATED TO E/U

Q1. A gas absorbs 150 J of heat and performs 50 J of work. What is the change in internal energy?

Given: Heat absorbed by the gas = q=+150J, work done by the gas = W = −50 J, the change in internal energy ΔE =?

Solution: from the formula of internal energy, it is obtained that, ΔE = q+w, by putting the values of q and w. ΔE = 150+(-50)= 100J. 

Q: In an isothermal process, 200 J of work is done by the gas. Find the change in internal energy.

Solution: The internal energy is a function of temperature by the following relations ΔE = (f/2)nRΔT. So, when ΔT is 0, the ΔE = 0  

Q: 2 moles of an ideal monoatomic gas are heated from 300 K to 500 K at constant volume. Calculate the change in internal energy.

Solution: As in this problem, the information about temperature and a monoatomic gas is given, therefore, the formula used for internal energy will be ΔE= nC_vΔT. The value of C_v is obtained by the degrees of freedom. i.e. Cv​=(f/2)R. thus, ΔE= n(f/2)RΔT. For a monoatomic gas, the value of f = 3/2R. Putting the values, ΔE = 2×3/2×8.314×(500−300) = ΔE=3×8.314×200=4988.4 J.

Q:5 moles of a diatomic ideal gas undergo a temperature rise of 100 K at constant volume. Find the internal energy change.

Solution: As in this problem, the information about temperature and a diatomic gas is given, therefore, the formula used for internal energy will be ΔE= nC_vΔT. The value of C_v is obtained by the degrees of freedom. i.e. Cv​=(f/2)R. thus, ΔE= n(f/2)RΔT. For a diatomic gas, the value of f = 5/2R. Putting the values, ΔE = 5×5/2×8.314×100=10392.5 J.

Q: A gas expands from 4 L to 10 L against an external pressure of 2 atm and absorbs 600 J of heat. Calculate the change in internal energy. (1 atm·L = 101.3 J)

Solution: As in this problem, the information about heat, pressure, and volume is provided, so the formula used to calculate the internal energy will be, ΔE=q-w or ΔE=q-PΔV.

Given: ΔV=10-4=6L, P = 2atm, q =600J. ΔE=600J-(2atmx6L) = 600J-12atmL. As in this case, heat has the unit of joule and PV has the unit of atmL, so first of all, it is needed to convert the atmL into joules by the following relationship. 1atmL=101.3J. 12atmL=12×101.3= -1215.6J. ΔE= 600 – (1215.6) =−615.6J​     

Q: Calculate ΔU for 1 mole of ideal gas undergoing adiabatic expansion from 500 K to 300 K.

Solution: As in this problem, the information about temperature and an ideal gas is given, therefore, the formula used for internal energy will be ΔE= nC_vΔT. The value of C_v is obtained by the degrees of freedom. i.e. Cv​=(f/2)R. thus, ΔE= n(f/2)RΔT. For one mole of a gas, the value of f = 3/2R. Putting the values, ΔE = 1×3/2×8.314×-200=-2494.2 J.

Q: An ideal gas (diatomic) undergoes a process where the temperature increases from 300 K to 600 K, and the volume doubles adiabatically. Calculate ΔU and W.

Solution: In this problem, first of all, the value of ΔU or ΔE is calculated. As in this question, the information about temperature and an ideal gas is given, therefore, the formula used for the calculation of internal energy will be ΔE= nC_vΔT. The value of C_v is obtained by the degrees of freedom. i.e. Cv​=(f/2)R. thus, ΔE= n(f/2)RΔT. For a diatomic gas, the value of f = 5/2R. suppose the number of moles =1.

 Putting the values, ΔE = 1×5/2×8.314×300 = 6235.5 J. for the calculation of w, the following relation will be used. ΔE = q-w (the work is done by the gas, as the volume of gas increases). The system is adiabatic; therefore, the value of q=0. Thus, ΔE = 0-w or w = – ΔE, w = -6235.5J.

Q: Calculate the change in internal energy of an ideal gas that undergoes a thermodynamic cycle consisting of two isothermal processes and two adiabatic processes. Given: n = 2 mol, R = 8.314 J/mol·K, T1 = 300 K, T2 = 400 K.

Solution: ΔE = nCvΔT = 2 mol × (3/2) × 8.314 J/mol·K × (400 K – 300 K) = 2494.2 J

Q: A system undergoes a thermodynamic process in which its internal energy changes by 1500 J. If the system does 800 J of work, calculate the heat transfer.

Solution: ΔE= Q – W, 1500 J = Q – 800 J, Q = 2300 J

Q: A gas is compressed adiabatically, and its temperature increases from 300 K to 400 K. If the specific heat capacity at constant volume is 20 J/mol·K, calculate the change in internal energy per mole.

 Solution: ΔE= nCvΔT = 1 mol × 20 J/mol·K × (400 K – 300 K) = 2000 J

ENTHALPY (H)

Enthalpy measures the force or total energy present in any material. This force contains the internal energy (E or U) of material + the pressure present in the product or the multiplication of volume. Thus, we conclude that the total power of any substance can be increased or decreased by changing the pressure. The power or force of any material is directly proportional to pressure.

Definition: The total force or heat content of a system at constant pressure is called enthalpy. Mathematically, it is written as, H =E+PV or change in enthalpy is ΔH =ΔE+Δ(PV) or ΔH =ΔE+(P₂V₂-P₁V₁). The unit of H is J or kJ/mol.  H is a state function, and it is an extensive property. The ΔH in a cyclic process is equal to zero. 

At constant pressure: At constant pressure, ΔH can be written as ΔH =ΔE+PΔV. The value of ΔE can be written as, ΔE = ΔQ – W or ΔE = ΔQ – PΔV. On putting the values of ΔE in the equation of enthalpy, we obtain the following equation. ΔH = ΔQ – PΔV+PΔV =   ΔH =ΔQ_p → Q_p=nC_pΔT (from the heat capacity). Thus, ΔH =  nC_pΔT. in any chemical reaction (particularly in gases), the ΔH and ΔE can be related as, ΔH =ΔE+ΔPV → ΔH =ΔE+Δn_gRT (as PV=nRT).    Where, Δn_g is the number of moles of product minus the number of moles of reactants. For all types of chemical reactions, the enthalpy of reaction is calculated by using the formula. Enthalpy of reaction = enthalpy of products-enthalpy of reactants.

Difference between ΔH and ΔE: The difference between ΔH and ΔE can be calculated as, ΔH =ΔE+ΔPV or ΔH-ΔE=ΔPV

In case of a gaseous system, the ΔH and ΔE can be related as:ΔH =ΔE+Δn_gRT or ΔH-ΔE=Δn_gRT, as the ΔH = Q_P and ΔE = Q_{v .} So, Q_P -Q_v = Δn_gRT

Types of ΔH: There are various types of enthalpies, like enthalpy of reaction (ΔH_r), enthalpy of formation (ΔH_f), enthalpy of combustion (ΔH_c), enthalpy of atomization (ΔH_at), enthalpy of neutralization (ΔH_n), enthalpy of solution (ΔH_s), and enthalpy of first electron affinity.            

Summary: The values of ΔH can be correlated with other parameters by the following relations.

From the heat: ΔH =Q_P.

From degrees of freedom: ΔH= (f/2+1) nRΔT (for ideal gases)

From the heat capacity: ΔH= nC_pΔT

In a chemical reaction:  ΔH= ΔE + Δn_gRT 

IMPORTANT NUMERICALS RELATED TO H

Q: Calculate the enthalpy change for a reaction where the internal energy change is 100 kJ/mol and the volume change is 0.01 m³/mol at a constant pressure of 10⁵ Pa.

Solution: By using the formula of enthalpy, ΔH = ΔU + PΔV = 100 kJ/mol + 10⁵ Pa × 0.01 m³/mol. Now, to add the values of PΔV in E, it is necessary to convert the values of PΔV into joules. By using the relation that 1atmL =101.3joule. thus, first of all. We have to convert pascal into atm and m³ into litre. So,1 atm = 101,325 pascals and 10⁵ pascals = 0.986 atm, while 1 litre = 0.001 cubic meters (m³) or conversely,1 cubic meter = 1,000 litres. So,0.01 m³ = 10 L. Thus, the value of PΔV in atmL= 10X0.986 = 9.86atmL. This atmL is converted into joules by following the relations. 1 atmL =101.325 J. so, 9.86 atmL = 999.06J≈ 1000J =1 KJ. Thus, 100 kJ/mol + 1 kJ/mol = 101 kJ/mol.

Q: Calculate the enthalpy change for the reaction: 2H₂(g) + O₂(g) → 2H₂O(l) given that the enthalpy of formation of H₂O(l) is -285.8 kJ/mol.

Solution: ΔH = 2 × ΔHf(H₂O) = 2 × -285.8 kJ/mol = -571.6 kJ/mol

Q: Calculate the enthalpy change for the reaction: C(s) + O₂(g) → CO₂(g) given that the enthalpy of formation of CO₂ is -393.5 kJ/mol.

Solution: ΔH = ΔHf(CO₂) = -393.5 kJ/mol

Q: Calculate the enthalpy change for the reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) given that the enthalpy of formation of CH₄ is -74.8 kJ/mol, CO₂ is -393.5 kJ/mol, and H₂O(l) is -285.8 kJ/mol.

Solution: The enthalpy of reaction can be calculated by using the formula. Enthalpy of reaction = sum of enthalpies of products- sum of enthalpies of reactants. Thus,  ΔHreaction = ΔH_f(CO₂) + 2 × ΔH_f(H₂O) – ΔHf(CH₄) = -393.5 kJ/mol + 2 × -285.8 kJ/mol – (-74.8 kJ/mol) = -890.3 kJ/mol.