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Thermal Physics
First Law of Thermodynamics
The first law of thermodynamics is a statement of the principle of conservation of energy applied to thermodynamic systems. It establishes that internal energy is a state function and links heat, work, and internal energy changes.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Sign convention for heat and work, cyclic processes, application to isothermal, isobaric, isochoric, and adiabatic processes, and fundamental definitions of internal energy.
PGT extension
Differential formulation dQ = dU + dW, molar heat capacities C_p and C_v relation (C_p - C_v = R), application to non-quasi-static processes, and microscopic link with the kinetic theory of gases.
Prerequisites
Concept of Temperature
Thermal equilibrium and Zeroth Law
Mechanical work and energy in mechanics
Statement of the First Law
Energy can be neither created nor destroyed, only transformed from one form to another.
When a system undergoes a thermodynamic change from an initial state to a final state, the net heat supplied to the system minus the work done by the system depends only on the initial and final states.
Expressed mathematically as d$Q = dU + dW$, where dQ is heat added, dU is change in internal energy, and dW is work done by the system.
Internal Energy as a State Function
Internal energy U is a macroscopic property representing the total microscopic kinetic and potential energy of the molecules in the system.
It is a state function, meaning its change depends solely on the initial and final equilibrium states, independent of the path taken.
For a complete cyclic process, the net change in internal energy is zero (delta U = 0).
Sign Conventions
Heat (Q) added to the system is positive (+Q); heat lost by the system is negative (-Q).
Work (W) done by the system (expansion) is positive (+W); work done on the system (compression) is negative (-W).
Increase in internal energy (U) is positive (+dU); decrease is negative (-dU).
Thermodynamic Processes
Isothermal process: Temperature is constant, leading to delta U = 0, hence Q = W.
Adiabatic process: No heat exchange (Q = 0), leading to delta $U = -W (work done at the expense of internal energy)$.
Isochoric process: Volume is constant (dW = 0), leading to Q = delta U.
Derivations
Relation between Molar Heat Capacities (C_p - C_v = R)
Conditions: One mole of an ideal gas
Apply the first law for a constant volume (isochoric) process: d$Q_v = dU = C_v dT$, since dW = P dV = 0.
Apply the first law for a constant pressure (isobaric) process: d$Q_p = dU + P dV = C_p dT$.
Substitute internal energy change d$U = C_v dT into the isobaric equation: C_p dT = C_v dT + P dV$.
For one mole of an ideal gas, the equation of state is PV = RT, differentiating gives P dV = R dT.
Substitute P d$V = R dT into the energy balance equation: C_p dT = C_v dT + R dT$.
Divide throughout by dT to yield the final relation.
Result: C_p - $C_v = R$
Work Done in an Isothermal Reversible Expansion of an Ideal Gas
Conditions: Quasi-static isothermal expansion of n moles of an ideal gas
Start with the definition of work done: $W = \int_{V_1}^{V_2} P \$, dV.
Use the ideal gas equation of state to express pressure: P = \frac{nRT}{V}.
Substitute P into the work integral: $W = \int_{V_1}^{V_2} \frac{nRT}{V} \$, dV.
Since temperature T is constant during an isothermal process, factor out nRT: $W = nRT \int_{V_1}^{V_2} \frac{1}{V} \$, dV.
Integrate the logarithmic term: $W = nRT \ln\left(\frac{V_2}{V_1}\right)$.
dQ = heat supplied, dU = change in internal energy, dW = work done by system Joule (J)
General thermodynamic system
Work Done in a Thermodynamic Process
$$W = \int P \, dV$$
P = pressure, V = volume Joule (J)
Quasi-static process
Molar Heat Capacity Relation
$$C_p - C_v = R$$
$C_p = molar heat capacity at constant pressure$, $C_v = molar heat capacity at constant volume$, R = universal gas constant J \, mol^{-1} \, K^{-1}
Ideal gas
Adiabatic Process Equation
$$PV^{\gamma} = \text{constant}$$
P = pressure, V = volume, $\gamma = ratio of specific heats (C_p / C_v)$ Dimensionless for \gamma
Reversible adiabatic process for an ideal gas
Internal Energy of an Ideal Gas
$$U = n C_v T$$
n = number of moles, $C_v = molar heat capacity at constant volume$, T = absolute temperature Joule (J)
Ideal gas
First Law for Cyclic Process
$$\oint dQ = \oint dW$$
Closed loop integrals for heat and work over a cycle Joule (J)
System returns to initial state
Adiabatic Work Done
$$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$$
P_1, $V_1 = initial state variables$, P_2, $V_2 = final state variables$ Joule (J)
Adiabatic process for ideal gas
Poisson's Relation in Terms of T and V
$$T V^{\gamma - 1} = \text{constant}$$
T = temperature, V = volume, $\gamma = C_p / C_v$ Standard SI units
Adiabatic process
Graphs and experiments
P-V Diagram for Isothermal and Adiabatic Processes
Draw: Two curves starting from the same initial state (P_1, V_1) on a P-V graph, showing the adiabatic curve steeper than the isothermal curve.
Interpret: The slope of the adiabatic curve is -\gamma P / V, whereas the slope of the isothermal curve is -P / V. Since \gamma > 1, the adiabatic curve is steeper.
Exam use: Frequently tested in multiple-choice questions comparing work done and final temperatures.
Cyclic Process on a P-V Diagram
Draw: A closed loop (clockwise or counter-clockwise) on a P-V plane.
Interpret: The net work done during a complete cycle equals the enclosed area of the loop. Clockwise represents net positive work (heat engine), counter-clockwise represents net negative work (heat pump/refrigerator).
Exam use: Used to evaluate net heat absorbed and net work done in thermodynamic cycles.
Joule's Paddle-Wheel Experiment
Draw: A schematic of a falling weight connected via a string and pulley to a paddle-wheel submerged in an insulated container of water.
Interpret: Demonstrated the mechanical equivalence of heat. Work done by falling masses converted directly into thermal energy, increasing the temperature of water, establishing that work and heat are interchangeable forms of energy.
Exam use: Conceptual questions on the equivalence of mechanical work and thermal energy.
Solved numericals
TGT core
In a thermodynamic process, a system absorbs 500 J of heat and simultaneously does 200 J of work on its surroundings. What is the change in internal energy of the system?
Identify given values using sign convention: d$Q = +500 J (heat absorbed)$, d$W = +200 J (work done$ by system).
Write the first law of thermodynamics: d$Q = dU + dW$.
Rearrange to solve for change in internal energy: d$U = dQ - dW$.
Substitute values: d$U = 500 J - 200 J = 300 J$.
Answer: 300 J
PGT extension
Two moles of an ideal monoatomic gas ($C_v = 3R/2)$ are heated at constant pressure from 27 °C to 127 °C. Calculate the heat supplied, change in internal energy, and work done by the gas. (Take $R = 8.314 J mol^-1 K^-1)$
Convert temperatures to Kelvin: $T_1 = 27 + 273 = 300 K$, $T_2 = 127 + 273 = 400 K$. Temperature change dT = 100 K.
Calculate change in internal energy: d$U = n C_v dT = 2 \times (3/2 R) \times 100 = 300 R$.
An ideal gas undergoes a cyclic process 1-2-3-1. From 1 to 2, 300 J of heat is added and 100 J of work is done by the gas. From 2 to 3, the process is adiabatic and 150 J of work is done on the gas. From 3 to 1, the gas is compressed back to state 1 with 50 J of heat rejected to the surroundings. Find the internal energy at state 2 if $U_1 = 100 J$, and find the work done from 3 to 1.
The first law assumes macroscopic equilibrium states where pressure and temperature are well-defined throughout the system.
It does not specify the direction of heat transfer or natural processes (which is the domain of the Second Law of Thermodynamics).
Common misconceptions
Common error: Heat and internal energy are the same thing.
Correct: Heat is energy in transit across a boundary due to a temperature difference, whereas internal energy is an inherent property contained within the system.
Heat appears only during a process, whereas internal energy is a state function possessing a definite value at any given equilibrium state.
Common error: In an isothermal process, because temperature is constant, no heat is exchanged.
Correct: Heat is exchanged in an isothermal process (Q = W), but it is entirely converted into or derived from work without changing the internal energy.
Constant temperature means delta U = 0, but d$Q = dU + dW implies dQ = dW when dW$ is non-zero.
Quick revision
The first law of thermodynamics is simply the law of conservation of energy: d$Q = dU + dW$.
First Law of Thermodynamics (Differential): d$Q = dU + dW$
Work Done in a Thermodynamic Process: W = \int P \, dV
Must remember
The First Law of Thermodynamics is a statement of the principle of conservation of energy applied to thermodynamic systems: $dQ = dU + dW$.
Internal energy ($U$) is a state function; its change in a cyclic process is always zero: $\oint dU = 0$.
Sign conventions: Heat added to the system is $+Q$, heat lost is $-Q$. Work done by the system is $+W$, work done on the system is $-W$.
For an isothermal process of an ideal gas, temperature is constant, so $\Delta U = 0$ and $Q = W$.
For an adiabatic process, there is no heat exchange ($Q = 0$), leading to $dU = -dW$.
For an isochoric process, volume is constant ($dW = 0$), so $Q = \Delta U$.
Meyer's relation connects molar heat capacities for an ideal gas: $C_p - C_v = R$.
Poisson's ratio relation for adiabatic processes is given by $PV^{\gamma} = \text{constant}$ and $TV^{\gamma-1} = \text{constant}$.
Glossary
First Law of ThermodynamicsFundamental law stating that $dQ = dU + dW$, representing conservation of thermal and mechanical energy.
Internal Energy (U)Macroscopic state function representing the total microscopic kinetic and potential energy of molecules in a system.
State FunctionA property whose value depends only on the current state of the system and not on the path taken to reach it.
Isothermal ProcessA thermodynamic process occurring at constant temperature ($\Delta T = 0$ and $\Delta U = 0$ for ideal gases).
Adiabatic ProcessA process taking place without any heat transfer between the system and its surroundings ($Q = 0$).
Isochoric ProcessA constant-volume thermodynamic process where no work is done by or on the system ($dW = 0$).
Molar Heat Capacity Ratio ($\gamma$)The ratio of molar heat capacity at constant pressure to that at constant volume ($\gamma = C_p / C_v$).
Comparison of Thermodynamic Processes
Process
Constant Variable
First Law Implication
Isothermal
Temperature ($T$)
$\Delta U = 0 \implies Q = W$
Adiabatic
Heat ($Q = 0$)
$dU = -dW$
Isochoric
Volume ($V$)
$dW = 0 \implies Q = \Delta U$
Isobaric
Pressure ($P$)
$Q = \Delta U + P\Delta V$
Practice Quiz
Choose an option to reveal the explanation.
1. Which of the following represents the differential form of the First Law of Thermodynamics?
Explanation: The First Law of Thermodynamics is expressed mathematically as $dQ = dU + dW$, where $dQ$ is heat added, $dU$ is change in internal energy, and $dW$ is work done by the system.
2. Internal energy of a system is a state function because:
Explanation: Internal energy $U$ is a macroscopic property and a state function, meaning its change depends exclusively on the initial and final states, independent of the thermodynamic path.
3. For a complete cyclic process, the net change in internal energy ($\Delta U$) is:
Explanation: Since internal energy is a state function and the system returns to its exact initial state in a complete cyclic process, the net change in internal energy is zero ($\oint dU = 0$).
4. According to standard sign conventions, work done *by* the system during expansion is taken as:
Explanation: Work done by the system (expansion) is considered positive ($+W$), whereas work done on the system (compression) is negative ($-W$).
5. In an isothermal process for an ideal gas, the change in internal energy ($\Delta U$) is:
Explanation: An isothermal process occurs at constant temperature. Since internal energy of an ideal gas depends only on temperature, $\Delta U = 0$, leading to $Q = W$.
6. An adiabatic process is characterized by which of the following conditions?
Explanation: By definition, an adiabatic process is one in which no heat enters or leaves the system, hence $Q = 0$.
7. In an isochoric process, the heat supplied to the system is completely converted into:
Explanation: In an isochoric process, volume is constant ($dW = P \cdot dV = 0$). From $dQ = dU + dW$, it follows that $dQ = dU$.
8. What is the relation between molar heat capacities $C_p$ and $C_v$ for an ideal gas?
Explanation: Meyer's relation states that for an ideal gas, the difference between molar heat capacity at constant pressure and at constant volume is the universal gas constant: $C_p - C_v = R$.
9. The equation governing a reversible adiabatic process for an ideal gas is:
Explanation: For a reversible adiabatic process involving an ideal gas, the pressure-volume relation is $PV^{\gamma} = \text{constant}$, where $\gamma = C_p / C_v$.
10. What is the internal energy of $n$ moles of an ideal gas at absolute temperature $T$?
Explanation: The internal energy of an ideal gas is given by $U = n C_v T$, derived from the equipartition of energy.
11. Which of the following is a misconception regarding heat and internal energy?
Explanation: Heat is energy in transit across a boundary due to temperature difference, whereas internal energy is an inherent property contained within the system. They are fundamentally different.
12. For an adiabatic expansion of an ideal gas, the internal energy of the gas:
Explanation: In an adiabatic expansion, $Q = 0$ and $W > 0$. From $dU = -dW$, internal energy must decrease, resulting in cooling of the gas.
13. Which variable combination represents Poisson's relation in terms of temperature and volume for an adiabatic process?
Explanation: Substituting ideal gas law into $PV^{\gamma} = \text{constant}$ yields Poisson's relation in terms of temperature and volume: $TV^{\gamma-1} = \text{constant}$.
14. Work done in a quasi-static thermodynamic process is represented mathematically by:
Explanation: Work done by a system undergoing a volume change in a quasi-static process is given by the integral $W = \int P \, dV$.
15. For a cyclic process, the net heat supplied $\oint dQ$ is equal to:
Explanation: Since $\oint dU = 0$ for a cyclic process, application of the First Law gives $\oint dQ = \oint dW$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. If an ideal gas undergoes an isothermal expansion, which of the following statements is correct?
Explanation: In an isothermal process, temperature is constant, so $\Delta U = 0$. From $dQ = dU + dW$, it follows that $Q = W$.
2. The molar heat capacities of an ideal gas at constant pressure and constant volume are $C_p$ and $C_v$ respectively. If $R$ is the universal gas constant, then $C_v$ is equal to:
3. During an adiabatic process, $200\text{ J}$ of work is done *on* the system. What is the change in internal energy of the system?
Explanation: In an adiabatic process $Q = 0$. Work done *on* the system is negative, so $W = -200\text{ J}$. From $dU = -dW$, $dU = -(-200\text{ J}) = +200\text{ J}$.
4. A system takes in $300\text{ J}$ of heat and gives out $100\text{ J}$ of work to the surroundings. What is the increase in internal energy?
Explanation: Using $dQ = dU + dW$, where $Q = +300\text{ J}$ and $W = +100\text{ J}$: $300 = dU + 100 \implies dU = 200\text{ J}$.
5. Which of the following thermodynamic quantities is NOT a state function?
Explanation: Heat ($Q$) and work ($W$) depend on the path taken during a thermodynamic process, making them path functions, whereas $U$, $T$, and $H$ are state functions.
6. In a cyclic process, if a system absorbs $500\text{ J}$ of heat, the net work done by the system over the cycle is:
Explanation: For a cyclic process, the net change in internal energy is zero ($\oint dU = 0$). Therefore, net heat absorbed equals net work done: $\oint dQ = \oint dW = 500\text{ J}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. The First Law of Thermodynamics is essentially the principle of conservation of:
Explanation: The First Law of Thermodynamics is the application of the law of conservation of energy to thermal and thermodynamic systems.
2. If a gas is compressed adiabatically and reversibly, its temperature:
Explanation: In an adiabatic compression, work is done on the gas ($W < 0$). Since $Q = 0$, $dU = -W > 0$, so internal energy and temperature increase.
3. For an ideal gas, the adiabatic exponent $\gamma$ is related to degrees of freedom $f$ by:
Explanation: Using $C_v = \frac{f}{2}R$ and $C_p = \left(1 + \frac{f}{2}\right)R$, $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$.
4. Which thermodynamic process involves no change in volume?
Explanation: An isochoric process occurs at constant volume ($dV = 0$), meaning zero work is done ($dW = 0$).
5. In a given process, $700\text{ J}$ of heat is supplied to a system and $300\text{ J}$ of work is done *by* the system. The change in internal energy is:
Explanation: $dQ = dU + dW \implies 700 = dU + 300 \implies dU = 400\text{ J}$.
6. Why is the slope of an adiabatic curve steeper than that of an isothermal curve on a $P-V$ diagram?
Explanation: The adiabatic bulk modulus is $\gamma P$, whereas the isothermal bulk modulus is $P$. Since $\gamma > 1$, the adiabatic curve has a steeper negative slope ($-\gamma P/V$ vs $-P/V$).
7. When a system is taken from state A to state B along path 1, $Q = 50\text{ J}$ and $W = 20\text{ J}$. Along path 2, $Q = 30\text{ J}$. What is the work done along path 2?
Explanation: Internal energy change $\Delta U$ is path-independent. Along path 1, $\Delta U = 50 - 20 = 30\text{ J}$. Thus, along path 2, $\Delta U = 30 = Q_2 - W_2 = 30 - W_2 \implies W_2 = 0\text{ J}$? Wait: $30 = 30 - W_2 \implies W_2 = 0\text{ J}$.
8. The ratio of molar heat capacities $\gamma = C_p/C_v$ for a monoatomic ideal gas is:
Explanation: For a monoatomic gas, $f = 3$, so $\gamma = 1 + 2/3 = 5/3 \approx 1.67$.
9. Which of the following is true for a free expansion of an ideal gas in an insulated container?
Explanation: In free expansion into vacuum, $W = 0$ (no external pressure) and since the container is insulated, $Q = 0$. By First Law, $\Delta U = 0$, and for an ideal gas, temperature remains constant.
10. What happens to the internal energy of an ideal gas when it undergoes isothermal compression?
Explanation: In an isothermal process, temperature is constant. Since the internal energy of an ideal gas depends solely on its temperature, $\Delta U = 0$.