Zeroth Law of Thermodynamics:
- If two bodies (say A and B) are in thermal equilibrium of the third body (say C) then body A and B will also be in thermal equilibrium with each other
First Law of Thermodynamics:
Different forms of the first law of thermodynamics are as follows
- Energy can’t be created nor destroyed but it can be converted from one form into the other (or forms) or into work.
- When a quantity of energy of one kind disappears, then an equivalent amount of energy of another kind makes its appearance.
- The total amount of energy of an isolated system remains constant, it may change from one form to another.
Mathematical Statement of the First Law:
- If ΔU is the change in the internal energy of the system, q i the amount of heat supplied to system, W is the work done by the system on the surroundings. Hence mathematically the first law of thermodynamics can be expressed as ΔU = q + W
- In the pressure-volume type of work W = PΔV, then the first law of thermodynamics can be expressed as ΔU = q + PΔV
Sign Conventions for q, W, and ΔU:
- When heat is absorbed by the system q is positive.
- When heat is rejected or given out by the system q is negative.
- When the work is done on the system by surroundings (Work of compression) then W is positive.
- When the work is done by the system on the surroundings (Work of expansion) then, W is negative.
- When there is an increase in internal energy of the system ( increase in temperature) ΔU is positive.
- When there is a decrease in internal energy of the system ( decrease in temperature) ΔU is negative.
Application of First Law of Thermodynamics to Different Processes:
- Internal energy is a function of temperature. As the temperature is constant, the internal energy is also constant. Hence there is no change in internal energy, then we have ΔU = 0, q = -W or W = – q
- In adiabatic process there is no exchange of heat q = 0, then W = q
- In isochoric process there is no change in volume ΔV = 0, Then W = 0 and ΔU = q
- In isobaric process, there is no change in pressure ΔP = 0 Thus work done W = – Pext ΔV and ΔU = q – Pext ΔV
- Most of the chemical reactions take place at constant pressure.
The concept of Enthalpy of System:
- The enthalpy of a system is defined as the sum of the internal energy of the system and energy that arises due to pressure and volume. It is denoted by letter H
- Mathematically, H = U + PV, where, H = Enthalpy U = Internal Energy P = Pressure V = Volume
- As the internal energy E, Pressure P and Volume V are state functions, the enthalpy of the system is a state function.
Change in Enthalpy of System:
- The change in enthalpy is given by ΔH = H1 – H2 ………………. (1)
By definition of enthalpy H = U + PV
For initial state, H1 = U1 + P 1V1
For final state, H2 = U2 + P2 V2
Substituting these values in equation (1)
Δ H = (U2 + P2 V2 ) – (U1 + P1V1)
∴ Δ H = (U2 – U1) + (P2 V2 – P1 V1)
At constant pressure P1 = P2 = P
∴ Δ H = (U2 – U1) + (P V 2 – P V1)
∴ Δ H = (U2 – U1) + P( V2 – V1)
∴ Δ H = ΔU + PΔ V
This is mathematical expression for change in enthalpy of system.
- If Δ H is positive then the reaction is endothermic and when ΔH is negative the reaction is exothermic.
Change in Enthalpy of System at Constant Pressure (Isobaric Process ):
- At constant pressure we have Δ H = Δ U + PΔ V ………… (1)
- By the first law of thermodynamics at constant pressure qp = ΔU + PΔV ………. (2)
From equations (1) and (2) we have
ΔH = qp
Change in enthalpy of System at Constant Volume (Isochoric Process ):
- We have Δ H = Δ U + PΔ V
- In isochoric process ΔV = 0, ∴ Δ H = Δ U + P(0) , hence Δ H = Δ U ………….. (1)
- The mathematical statement of the first law of thermodynamics is Δ U = q + PΔV ………….. (2)
Substituting ΔV = 0 in equation (2)
q = ΔU
Hence ΔH = ΔU = qv
The Relation between ΔH and ΔU:
- Change in the enthalpy of a system is given by
Δ H = ΔU + PΔV
∴ Δ H = ΔU + P( V2 – V1)
∴ Δ H = ΔU + P V2 – PV1
∴ Δ H = ΔU +n2RT – n1RT
∴ Δ H = ΔU + (n2 – n1 )RT
∴ Δ H = ΔU + ΔnRT
Work Done in Chemical Reaction:
- The work done by a system at constant pressure and temperature is given by
W = – PextΔV
Assuming Pext = P
W = – P( V2 – V1)
∴ W = – (PV2 + PV1)
∴ W = – (n2RT – n1RT)
∴ W = – (n2 – n1 )RT
∴ W = – ΔnRT
Note: These relations are only to be used for gaseous products and gaseous reactants
- For a particular reaction, the system absorbs 6 kJ of heat and does 1.5 kJ of work on its surroundings. What are the change in internal energy and enthalpy change of the system? (Ans: 6 kJ)
- An ideal gas expands from a volume of 6 dm³ to 16 dm³ against constant external pressure of 2.026 x 105 Nm-2. find Enthalpy change if ΔU is 418 J. (Ans: 2444 J)
- A sample of gas absorbs 4000 kJ of heat. a) If volume remains constant, What is ΔU? b) Suppose that in addition to absorption of heat by the sample, the surrounding does 2000 kJ of work on the sample. What is ΔU? c) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. What is ΔU? (Ans: 4000 kJ, 6000 kJ, 3400 kJ)
- Calculate the work done in the following reaction when 2 moles of HCl are used at constant pressure and 423 K. State whether work is on the system or by the system. (Ans: 1758 J)
- Calculate the work done in the following reaction when 1 mol of SO2 is oxidised at constant pressure at 5o °C. State whether work is on the system or by the system. (Ans: 1343 J)
- Calculate the work done in the following reaction when 2 mol of NH4NO3 decomposes at constant pressure at 10o °C. State whether work is on the system or by the system. (Ans: -18.61 kJ)
- CO reacts with O2 according to the following reaction. How much pressure volume work is done and what is the value of ΔU for the reaction of 7.0 g of CO at 1 atm pressure, if the volume change is – 2.8 L. (Ans: -70.47 kJ)
- The enthalpy change for the following reaction is – 620 J, when 100 mL of ethylene and 100 mL of H2 react at 1 atm pressure. Calculate pressure volume type work and ΔU. (Ans: W = 10.13 J and Δ U = – 609.9 J)