Work, Heat and Energy 

1.0Introduction to Work, Heat and Energy

In thermodynamics and physical chemistry, work, heat, and energy are fundamental concepts that describe energy transfer and transformation.

• Work refers to energy transfer due to a force acting over a distance.
• Heat is energy transfer resulting from a temperature difference.
• Energy is the capacity to do work, including both kinetic and potential forms.

These concepts form the foundation of thermodynamics, chemical reactions, and physical processes studied in JEE Chemistry.

2.0Heat and Work done in Thermodynamic Process

$$\begin{gathered} \text{Work and Heat:} \\ W\propto Q(\text{Heat}) \\ W=JQ \\ Here\text{ J = conversion factor} \\ \text{= Joule’s constant (mechanical equivalent of heat)} \\ J=4.18\frac{\text{joule}}{\text{calorie}} \end{gathered}$$

Special Point : 

(a) Work and heat both are the form of energy. Work is ordered energy where as heat is disordered form of energy. 
(b) Conversion of mechanical work into heat is completely possible but its reverse is not true.

Work done by thermodynamic system
One of the simple example of a thermodynamic system is a gas in a cylinder with a movable piston.

• If the gas expands against the piston Gas exerts a force on the piston and displace it through a distance and does work on the piston.
• If the piston compresses the gas When piston moved inward, work is done on the gas.
• The work associated with volume changes
  If pressure of gas on the piston = P.
  Then the force on the piston due to gas is F = PA
• When the piston is pushed outward an infinitesimal distance
  dx, the work done by the gas is $dW=F\times dx=PAdx$
  The change in volume of the gas is dV = Adx,
  $\therefore dW=PdV$          

For a finite change in volume from Vi to Vf, this equation is then integrated between Vi to Vf to find

the net work done = $W={\int }_{V_i}^{V_f}dW={\int }_{V_i}^{V_f}PdV$

Hence the work done by a gas is equal to the area under P–V graph.

Work Done in Various Processes

(1) Mathematical method $W={\int }_{V_i}^{V_f}PdV$

3.0Concept of Heat

Heat Definition

Heat (q) is the form of energy transfer due to temperature difference between a system and its surroundings.

Modes of Heat Transfer

1. Conduction: Transfer through direct contact (metals).
2. Convection: Transfer through fluid motion (liquids and gases).
3. Radiation: Transfer through electromagnetic waves (no medium required).

Heat Capacity and Specific Heat

• Heat Capacity (C): Amount of heat required to raise the temperature of a system by 1 K.
  $C=\frac{q}{\Delta T}$
• Specific Heat (c): Heat required to raise the temperature of 1 gram of substance by 1 K.
  $q=mc\Delta T$
  Where m = mass of substance, ΔT = temperature change

4.0Concept of Internal Energy

Internal Energy

Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration. The energy due to molecular motion is called internal kinetic energy$(U_k)$ and that due to molecular configuration is called internal potential energy$(U_p)$.

$dU=d{U}_k+d{U}_p$

If there no intermolecular forces, then $d{U}_p=0\text{ and }dU=d{U}_k=m{c}_{v}dT\text{ [ideal gas]}$

$$\begin{gathered} c_v=\text{Specific heat at constant volume}\text{ and}\text{dT = Infinitesimal change in temperature} \\ \text{m = Mass of system}\text{M = Molecular weight} \\ \text{Molar heat capacity }{C}_v=M{c}_{v}For\mu \text{-moles of ideal gas = }dU=\mu C_{v}dT=\frac{m}{M}{C}_{v}dT \\ \text{Internal energy in the absence of inter–molecular forces is simply the function of temperature and state only, it is independent of path followed.} \\ \Delta U=U_f-U_i \\ {U}_i=\text{Internal energies in initial state and}{U}_f=\text{ Internal energies in final state} \end{gathered}$$

5.0Properties of Internal Energy

1. It depends mainly on states of matter and temperature of system.
2. For ideal gas it a function of temperature, identity and moles.
3. For real gas it is a function of T and V or T and P as well as identity and moles.
4. Absolute value of internal energy is impossible to determine, so it is generally expressed as a difference between the two states of system, or by applying zero value to a reference state.
5. Internal energy is an Exact quantity or a state function.
6. Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration.
7. The energy due to molecular motion is called internal kinetic energy $(U_k)$ and that due to molecular configuration is called internal potential energy $(U_p)$.

6.0Relationship Between Work, Heat and Energy

First Law of Thermodynamics

The first law of thermodynamics states that:
"The change in internal energy (ΔU) of a system is equal to the heat added to the system (q) minus the work done by the system (W)."

If some quantity of heat is supplied to a system capable of doing external work, then the quantity of heat absorbed by the system is equal to the sum of the increase in the internal energy of the system and the external work done by the system.

$\boxed{\Delta Q=\Delta U+\Delta W}\text{or}\boxed{Q=W+\Delta U}$

• This law is applicable to every process in nature.
• Thermodynamic force is non-conservative force.
• The first law of thermodynamics introduces the concept of internal energy.
• The first law of thermodynamics is based on the law of conservation of energy.
• $\Delta Q,\Delta U\text{ and }\Delta W$ must be expressed in the same units (either in units of work or in units of heat).
• This law is applicable to all the three phases of matter, i.e., solid, liquid and gas.
• dU is a characteristic of the state of a system, it may be any type of internal energy–translational kinetic energy, vibrational, rotational kinetic energy, binding energy etc.

Sign Convention used in Physics Thermodynamics

$\text{FLOT }\Delta Q=\Delta U+\Delta W\text{ or }[Q=\Delta U+W]$

Rules:

$(1)Q/\; \Delta Q=\{\begin{array}{l}\text{+ve}\Rightarrow \text{heat given to system}/\text{heat absorbed by system.}\\ \text{–ve}\Rightarrow \text{heat rejected }/\text{ released by system or heat taken out from system}\end{array}$

$(2)\Delta U/dU=\{\begin{array}{l}\text{+ve}\Rightarrow \text{U }\uparrow \text{ (internal energy) T }\uparrow \text{ (increases)}\\ \text{–ve}\Rightarrow \text{U }\downarrow \text{ (internal energy) T }\downarrow \text{ (decreases)}\end{array}$

$(3)W/\Delta W=\{\begin{array}{l}\text{V }\uparrow \text{ (expansion) dV = +ve}\Rightarrow \Delta \text{W = PdV = +ve}\\ \text{(work is done by system)}\\ \\ \text{V }\downarrow \text{ (compression) ; dV = –ve}\Rightarrow \Delta \text{W = PdV = –ve}\\ \text{(work is done on system)}\end{array}$

7.0Work and Heat in Different Thermodynamic Processes

Isothermal Process

In this process pressure and volume of the system change but temperature remains constant. In an isothermal process, the exchange of heat between the system and the surroundings is allowed. Isothermal process is carried out by either supplying heat to the substance or by extracting heat from it. A process has to be extremely slow to be isothermal.

• Equation of state: $PV=\text{constant }(\mu RT)\text{ (T is constant)}$
• Work Done$$\begin{gathered} \text{Consider }\mu \text{ moles of an ideal gas, enclosed in a cylinder, at absolute temperature T, fitted with a frictionless piston. Suppose that gas undergoes an isothermal expansion from the initial state }(P_1,V_1)\text{ to the final state}(P_2,V_2). \\ \text{Work done: }W={\int }_{V_1}^{V_2}PdV\therefore PV=\mu RT\text{ then }P=\frac{\mu RT}{V} \\ \therefore W={\int }_{V_1}^{V_2}\frac{\mu RT}{V}dV=\mu RT{\int }_{V_1}^{V_2}\frac{dV}{V}=\mu RT[{\log }_{e}V{]}_{V_1}^{V_2}=\mu RT[{\log }_e{V}_2-{\log }_e{V}_1]=\mu RT{\log }_e\left[\frac{V_2}{V_1}\right] \\ \Rightarrow W=2.303\mu RT{\log }_{10}\left[\frac{P_1}{P_2}\right][\because P_1{V}_1=P_2{V}_2] \end{gathered}$$
• Form of First Law
  $$\begin{gathered} \text{There is no change in temperature and internal energy of the system depends on temperature only} \\ \text{So }\Delta U=0,Q=2.303\mu RT{\log }_{10}\left[\frac{V_2}{V_1}\right] \\ \text{It is clear that whole of the heat energy supplied to the system is utilized by the system in doing external work. There is no change in the internal energy of the system.} \end{gathered}$$
• Slope of the isothermal curve
  $$\begin{gathered} \text{For an isothermal process, }PV=\text{constant} \\ \text{Differentiating, }PdV+VdP=0 \\ \Rightarrow VdP=-PdV\Rightarrow \frac{dP}{dV}=-\frac{P}{V} \\ \text{Slope of isothermal curve, }{\left[\frac{dP}{dV}\right]}_{isothermal}=-\frac{P}{V} \end{gathered}$$

Adiabatic Process

$$\begin{gathered} \text{It is that thermodynamic process in which pressure, volume and temperature of the system do change but there is no exchange of heat between the system and the surroundings.} \\ \text{A sudden and quick process will be adiabatic since there is no sufficient time available for exchange of heat so process adiabatic.} \\ \text{Equation of state :}PV=\mu RT \\ \text{Equation for adiabatic process }P{V}^{\gamma }=\text{constant},T{V}^{\gamma -1}=\text{constant},T^{\gamma }{P}^{1-\gamma }=\text{constant} \\ \text{Work done} \\ \text{Let initial state of system is }(P_1,V_1,T_1)\text{ and after adiabatic change final state of system is }(P_2,V_2,T_2)\text{ then we can write }{P}_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }=K\text{ (here K is const.)} \\ SoW={\int }_{V_1}^{V_2}PdV=K{\int }_{V_1}^{V_2}{V}^{-\gamma }dV=K{\left(\frac{V^{-\gamma +1}}{-\gamma +1}\right)}_{V_1}^{V_2}=\frac{K}{(-\gamma +1)}[V_2^{-\gamma +1}-V_1^{-\gamma +1}](\because K=P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }) \\ \Rightarrow W=\frac{1}{(\gamma -1)}[P_1{V}_1^{\gamma }.V_1^{-\gamma }.V_1-P_2{V}_2^{\gamma }.V_2^{-\gamma }.V_2]=\frac{1}{(\gamma -1)}[P_1{V}_1-P_2{V}_2]\Rightarrow W=\frac{\mu R}{(\gamma -1)}(T_1-T_2)(\because PV=\mu RT) \\ \text{Form of first law : }\Delta U=-\Delta W\text{It means the work done by an ideal gas during adiabatic expansion (or compression) is on the cost of change in internal energy proportional to the fall (or rise) in the temperature of the gas.} \\ \text{If the gas expands adiabatically, work is done by the gas. So, }{W}_{adia}\text{ is positive.}\text{The gas cools during adiabatic expansion and }{T}_1>T_2.\text{If the gas is compressed adiabatically, work is done on the gas. So, }12W_{adia}\text{is negative} \\ \text{The gas heats up during adiabatic compression and }{T}_1<T_2 \end{gathered}$$

Slope of the adiabatic curve

$$\begin{gathered} \text{For an adiabatic process, }P{V}^{\gamma }=\text{constant} \\ \text{Differentiating,}\gamma P{V}^{\gamma -1}dV+V^{\gamma }dP=0 \\ \Rightarrow V^{\gamma }dP=-\gamma P{V}^{\gamma -1}dV\Rightarrow \frac{dP}{dV}=-\frac{\gamma P{V}^{\gamma -1}}{V^{\gamma }}=-\gamma \frac{P}{V}=\gamma \left(-\frac{P}{V}\right) \\ \text{Slope of adiabatic curve,}{\left[\frac{dP}{dV}\right]}_{\text{adiabatic}}=-\frac{\gamma P}{V} \end{gathered}$$

Magnitude of slope of adiabatic curve is greater than the slope of isotherm

${\left|\frac{dP}{dV}\right|}_{\text{adiabatic}}=\gamma \frac{P}{V}=\gamma {\left|\frac{dP}{dV}\right|}_{\text{isothermal}}$

$\Rightarrow \frac{\text{Slope of adiabatic changes}}{\text{Slope of isothermal changes}}=\gamma$

$\text{Since }\gamma \text{ is always greater than one so an adiabatic curve is steeper than an isothermal curve}$

Isobaric Process

Isobaric process is a thermodynamic process that takes place at constant pressure, but volume and temperature varies for change in state of the system.

• $\text{Equation of state}V/T=\text{Constant or }V\propto T$
• $$\begin{gathered} \text{Work done}\text{ In this process pressure remains constant}\therefore dP=0 \\ \text{Work done }W={\int }_{V_i}^{V_f}PdV=P(V_f-V_i) \end{gathered}$$

$\Rightarrow \text{Fraction of Heat given which –}$

$(i)\text{is converted into internal energy }\boxed{\frac{\Delta U}{\Delta Q}=\frac{1}{\gamma }}$

$(ii)\text{does work against external pressure }\boxed{\frac{\Delta W}{\Delta Q}=1-\frac{1}{\gamma }}$

• Form of first Law
  $$\begin{gathered} Q=\Delta U+P(V_f-V_i) \\ {C}_p(T_f-T_i)=\mu C_v(T_f-T_i)+P(V_f-V_i) \\ \text{It is clear that heat supplied to the system is utilized for:} \\ \text{(i) Increasing internal energy and } \\ \text{(ii) Work done against the surrounding atmosphere.} \end{gathered}$$
• Slope of the PV curve:
  ${\left(\frac{dP}{dV}\right)}_{isobaric}=0$

Isochoric Process

Isometric or Isochoric Process

Isochoric process is a thermodynamic process that takes place at constant volume of the system, but pressure and temperature varies for change in state of the system.

• Equation of state P/T =constant (P and T are variable, V is constant)
• Work done In this process volume remains constant so $dV=0\Rightarrow W={\int }_{V_i}^{V_f}PdV=0$

Form of First Law

Q=U

$\text{Slope of the P-V curve }\frac{dP}{dV}=\infty$

8.0Solved Examples

Problem 1: In a thermodynamic process, work done on the system is 100 J and heat given to system is 500 cal. Then calculate change in internal energy of system.

Solution:

$$\begin{gathered} \text{As from FLOT : }\Delta Q=\Delta U+\Delta W \\ \Rightarrow (500\times 4.2)=\Delta U-100 \\ \Delta U=2100+100=2200\text{ J}=524\text{ cal }\Delta U\text{ is positive so temperature will increase} \end{gathered}$$

Problem 2: If Q amount of heat is given to a diatomic ideal gas in process in which the gas perform a work $\frac{2Q}{3}$ its surrounding. Find the molar heat capacity (in terms of R) for the process.

Solution:

$$\begin{gathered} \text{According to FLOT} \\ Q=W+\Delta U \\ Q=\frac{2}{3}Q+\mu C_v\Delta T \\ \frac{Q}{3}=\mu \times \frac{5}{2}R\Delta T \\ \mu C\Delta T=\frac{15}{2}\mu R\Delta T \\ C=\frac{15}{2}R \end{gathered}$$

Problem 3: The pressure of one mole monoatomic gas increases linearly from $4\times 10^5{\text{ Nm}}^{-2}to8\times 10^{+5}{\text{ Nm}}^{-2}$ when its volume increases from ${\text{0.2 m}}^3{\text{ to 0.5 m}}^3$. Calculate.

(i) Work done by the gas, 
(ii) Increase in the internal energy,
(iii) Amount of heat supplied,
(iv) Molar heat capacity of the gas $R=8.31{\text{ J mol}}^{-1}{\text{ K}}^{-1}$

Solution:

$$\begin{gathered} P_1=4\times 10^5{\text{ Nm}}^{-2} \\ {P}_2=8\times 10^{+5}{\text{ Nm}}^{-2},V_1=0.2{\text{ m}}^3,V_2=0.5{\text{ m}}^3 \end{gathered}$$

$$\begin{gathered} \text{(i) Work done by the gas} \\ =\text{Area under }P–V\text{ graph (Area ABCDEA)} \\ (AE+BD)\times AC=\frac{1}{2}(4\times 10^5+8\times 10^5)\times (0.5-0.2) \\ =\frac{1}{2}\times 12\times 10^5\times 0.3=1.8\times 10^5\text{ J} \\ \text{(ii) Increase in internal energy} \\ \Delta U=C_V(T_2-T_1)=\frac{C_V}{R}R(T_2-T_1)=\frac{C_V}{R}(P_2{V}_2-P_1{V}_1) \\ \text{For monoatomic gas }{C}_V=\frac{3}{2}R \\ \therefore \Delta U=\frac{3}{2}[(8\times 10^5\times 0.5)-(4\times 10^5\times 0.2)]=\frac{3}{2}[4\times 10^5-0.8\times 10^5]=4.8\times 10^5\text{ J} \\ \text{(iii) Q = }\Delta U+W=4.8\times 10^5+1.8\times 10^5=6.6\times 10^5\text{ J} \\ \text{(iv) C = }\frac{Q}{\mu \Delta T}=\frac{QR}{\mu R\Delta T}=\frac{QR}{(P_2{V}_2-P_1{V}_1)}=\frac{6.6\times 10^5\times 8.31}{1\times 3.2\times 10^5}17.14\text{ J/mol K} \end{gathered}$$

Problem 4

As shown in figure when a system is taken from state a to state b, along the path $a\to c\to b,60\text{ J}$ of heat flow into the system, and 30 J of work is done:

(i) How much heat flows into the system along the path $a\to d\to b$ if the work is 10 J.

(ii) When the system is returned from b to a along the curved path, the work done by the system is -20 J. Does the system absorb or liberate heat, and how much?

(iii) If, $U_a=0\text{ and }{U}_d=22J,$find the heat absorbed in the process $a\to d$ and $d\to b.$

Solution:

$$\begin{gathered} \text{For the path acb }\Delta U=Q-W=60-30=30\text{ J}or{U}_b-U_a=30\text{ J} \\ \text{(i) Along the path adb Q = }\Delta U+W=30+10=40\text{ J} \\ \text{(ii) Along the curved path ba }Q=(U_a-U_b)+W=(-30)+(-20)=-50\text{ J},\text{ heat liberates from system} \\ \text{(iii) }{Q}_{ad}=U_d-U_a+W_{ad} \end{gathered}$$
$$\begin{gathered} \text{but }{W}_{ad}=W_{adb}-W_{db}=10-0=10\text{ Hence }{Q}_{ad}=22-0+10=32\text{ J} \\ \text{and }{Q}_{db}=U_b-U_d+W_{db}=30-22+0=8\text{ J} \end{gathered}$$

Problem 5:

P-V curve of a diatomic gas is shown in the figure.
Find the total heat given to the gas in the process AB and BC.

Solution:

$$\begin{gathered} \text{From first law of thermodynamics} \\ \Delta Q_{ABC}=\Delta U_{ABC}+W_{ABC} \\ {W}_{ABC}=W_{AB}+W_{BC}=0+nR{T}_B\ln \frac{V_C}{V_B}=nR{T}_B\ln \frac{2V_0}{V_0} \\ =nR{T}_B\ln 2=2P_0{V}_0\ln 2 \\ \Delta U=n{C}_V\Delta T=\frac{5}{2}(2P_0{V}_0-P_0{V}_0) \\ \Rightarrow \Delta Q_{ABC}=\frac{5}{2}{P}_0{V}_0+2P_0{V}_0\ln 2 \end{gathered}$$