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Thermal Physics
Adiabatic Process in Thermodynamics
An adiabatic process is a thermodynamic process in which no heat is exchanged between a system and its surroundings (dQ = 0). During this process, changes in internal energy result entirely from work done by or on the system.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of adiabatic process, first law of thermodynamics applied to dQ = 0, Poisson's relations between P, V, and T, work done during adiabatic expansion and compression, and comparison of adiabatic and isothermal slopes.
PGT extension
Microscopic derivation of Poisson's law using the equipartition of energy and degrees of freedom, adiabatic elasticity of gases, free expansion of ideal and real gases (Joule-Thomson effect), and velocity of sound in air (Newton-Laplace formula).
Prerequisites
Concept of Temperature and First Law of Thermodynamics
Degrees of Freedom and Law of Equipartition of Energy
Equation of State for Ideal Gases (PV = nRT)
Fundamental Definition and First Law Form
By definition, an adiabatic process satisfies the condition dQ = 0 (no heat enters or leaves the system).
Applying the First Law of Thermodynamics (d$U = dQ - dW)$, for an adiabatic process, d$U = -dW$.
If a gas expands adiabatically (dW > 0), its internal energy decreases (dU < 0), causing a drop in temperature.
Poisson's Relations for Ideal Gases
For an ideal gas undergoing a reversible adiabatic process, the variables are related by PVγ = constant, where $\gamma = C_{p} / C_{v}$ is the adiabatic index.
Alternative forms include TVγ-1 = constant and P1-γTγ = constant.
The value of γ depends strictly on the atomicity (degrees of freedom) of the gas: $\gamma = 1.67$ for monatomic, $\gamma = 1.40$ for diatomic, and $\gamma = 1.33$ for polyatomic gases.
Comparison with Isothermal Processes
In an isothermal process, temperature remains constant (dT = 0), and PV = constant.
In an adiabatic process, pressure changes more rapidly with volume because the adiabatic curve is steeper than the isothermal curve by a factor of γ.
The slope of the adiabatic curve on a P-V diagram is given by (dP/dV)ad = -γ (P/V), whereas the isothermal slope is (dP/dV)iso = -(P/V).
Adiabatic Elasticity of Gases
The bulk modulus of elasticity for a gas undergoing an isothermal process is equal to its pressure P.
For an adiabatic process, differentiating PVγ = constant yields the adiabatic elasticity Ead = γP.
This higher rigidity modulus is directly utilized in Laplace's correction for the velocity of sound in gases.
Derivations
Derivation of Poisson's Law (PV^gamma = constant)
Conditions: Reversible adiabatic process for an ideal gas
Start with the First Law of Thermodynamics: d$Q = dU + dW$. For an adiabatic process, dQ = 0, so dU + dW = 0.
Substitute d$U = n C_v dT$ and d$W = P dV into the equation: n C_v dT + P dV = 0$.
Use the ideal gas equation, PV = nRT, and differentiate it: P dV + V dP = n R dT, which gives d$T = (P dV + V dP) / (nR)$.
Substitute dT into the energy equation: n C_v [(P dV + V dP) / (nR)] + P dV = 0. Since $R = C_p - C_v$, let C_p / $C_v = \gamma$.
Divide by C_v and simplify to obtain (P dV + V dP) / (gamma - 1) + P dV = 0, which reduces to V dP + gamma P dV = 0.
Divide the entire equation by PV to get (dP / P) + gamma (dV / V) = 0.
Integrating both sides yields ln P + gamma ln V = constant, leading directly to P$V^\gamma = constant$.
Result: P$V^\gamma = constant$
Work Done in an Adiabatic Process
Conditions: Quasistatic adiabatic expansion or compression of an ideal gas from (P_1, V_1, T_1) to (P_2, V_2, T_2)
The work done in any thermodynamic process is given by $W = \int_{V_1}^{V_2} P dV$.
For an adiabatic process, P$V^\gamma = K (constant)$, so $P = K / V^\gamma$.
Substitute P into the integral: $W = \int_{V_1}^{V_2} (K / V^\gamma) dV = K \int_{V_1}^{V_2} V^{-\gamma} dV$.
Evaluate the integral: $W = K [ (V_2^{1-\gamma} - V_1^{1-\gamma}) / (1 - \gamma) ]$.
Since $K = P_1 V_1^\gamma = P_2 V_2^\gamma$, substitute K at the respective limits: $W = [ (P_2 V_2^\gamma V_2^{1-\gamma} - P_1 V_1^\gamma V_1^{1-\gamma}) / (1 - \gamma) ]$.
Simplify the expression to obtain $W = (P_1 V_1 - P_2 V_2) / (\gamma - 1)$.
P = Pressure, V = Volume, gamma = Adiabatic exponent Pa/m^3
Slope of adiabatic curve
Graphs and experiments
P-V Diagram Comparing Isothermal and Adiabatic Processes
Draw: A set of curves on a P-V plane starting from the same initial state (P_1, V_1). The adiabatic curve drops faster and lies below the isothermal curve during expansion.
Interpret: The adiabatic curve has a slope γ times steeper than the isothermal curve because heat cannot enter to sustain the temperature during expansion.
Exam use: Frequently tested in multiple-choice questions regarding work done, final temperatures, and curve steepness.
Clement and Desormes' Experiment
Draw: Schematic setup of a large glass flask connected to a manometer and a pump, demonstrating adiabatic expansion of air.
Interpret: Used to determine the ratio of specific heats (γ) of a gas by noting pressure changes during sudden opening and closing of a stopcock.
Exam use: Core experimental setup for PGT physics practical theory questions.
Adiabatic Free Expansion (Joule Effect)
Draw: Two insulated bulbs connected by a valve; one contains gas at pressure P, the other is evacuated.
Interpret: When the valve opens, the gas expands into vacuum without doing external work (W = 0) and without heat exchange (Q = 0), resulting in dU = 0.
Exam use: Used to test conceptual understanding of internal energy dependence on temperature for ideal versus real gases.
Solved numericals
TGT core
A certain mass of an ideal gas (? = 1.4) is compressed adiabatically from an initial volume of 8 liters to 1 liter. If the initial pressure is 1 atm, find the final pressure.
Two moles of an ideal monatomic gas (? = 5/3) at 300 K are compressed adiabatically and reversibly to half of their initial volume. Calculate the final temperature and the work done on the gas.
Identify given data: n = 2, $T_1 = 300 K$, V_2 / $V_1 = 1/2$, gamm$a = 5/3$.
Use the temperature-volume relation: T_1 $V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
The process must be perfectly insulated (dQ = 0), which is practically difficult to achieve as complete thermal isolation does not exist.
For the relations P$V^\gamma = constant$ to hold, the process must be quasistatic (infinitely slow) so that the system remains in internal thermal and mechanical equilibrium at all intermediate stages.
Common misconceptions
Common error: An adiabatic process must always be sudden.
Correct: An adiabatic process can be carried out either rapidly or slowly, provided the system is thermally insulated from its surroundings.
Rapidity ensures no heat exchange due to time constraints, but even a slow process is adiabatic if the boundaries are perfectly lagging (insulated).
Common error: Temperature remains constant during an adiabatic expansion because no heat is added.
Correct: Temperature decreases during adiabatic expansion.
Since dQ = 0 and the gas does positive work (dW > 0), the internal energy must decrease (dU < 0), directly leading to a drop in temperature.
Common error: The adiabatic exponent gamma is a universal constant for all gases.
Correct: The value of gamma depends entirely on the atomicity and degrees of freedom of the specific gas.
gamm$a = 1 + (2/f)$ where f is the degrees of freedom, which varies between monatomic, diatomic, and polyatomic molecules.
Quick revision
Adiabatic condition: dQ = 0 and d$U = -dW$.
Fundamental governing law: P$V^\gamma = constant$.
Adiabatic curves are steeper than isothermal curves by a factor of gamma.
Work done in adiabatic expansion: $W = (P_1V_1 - P_2V_2) / (\gamma - 1)$.
Laplace's correction uses adiabatic bulk modulus $E_{ad} = \gamma P$ for sound velocity in gases.
Must remember
In an adiabatic process, the system is thermally insulated so that $dQ = 0$.
The first law applied to an adiabatic process yields $dU = -dW$.
Poisson's relations for an ideal gas are $PV^{\gamma} = \text{constant}$, $TV^{\gamma-1} = \text{constant}$, and $P^{1-\gamma}T^{\gamma} = \text{constant}$.
The adiabatic exponent $\gamma = C_p / C_v$ depends strictly on atomicity: $1.67$ for monatomic, $1.40$ for diatomic, and $1.33$ for polyatomic gases.
The slope of an adiabatic curve on a $P-V$ diagram is steeper than an isothermal curve by a factor of $\gamma$: $(dP/dV)_{ad} = -\gamma (P/V)$.
The adiabatic bulk modulus of elasticity is given by $E_{ad} = \gamma P$.
Laplace's correction for the velocity of sound in a gas is $v = \sqrt{\gamma P / \rho}$.
Glossary
Adiabatic ProcessA thermodynamic process in which no heat enters or leaves the system ($dQ = 0$).
Adiabatic Index ($\gamma$)The ratio of molar heat capacity at constant pressure to that at constant volume ($C_p / C_v$).
Adiabatic Elasticity ($E_{ad}$)The bulk modulus of elasticity for a gas under adiabatic conditions, equal to $\gamma P$.
Laplace's FormulaThe expression $v = \sqrt{\gamma P / \rho}$ used to calculate the speed of sound in gases, accounting for rapid adiabatic compression and expansion.
Isothermal vs Adiabatic Process
Parameter
Isothermal Process
Adiabatic Process
Thermal Condition
Constant Temperature ($dT = 0$)
Thermal Insulation ($dQ = 0$)
Governing Equation
$PV = \text{constant}$
$PV^{\gamma} = \text{constant}$
Slope on $P-V$ Diagram
$-\frac{P}{V}$
$-\gamma \frac{P}{V}$
Elasticity
$E_{iso} = P$
$E_{ad} = \gamma P$
Practice Quiz
Choose an option to reveal the explanation.
1. Which of the following conditions correctly defines an adiabatic process?
Explanation: By definition, an adiabatic process is one in which no heat is exchanged between the system and its surroundings, meaning $dQ = 0$.
2. In an adiabatic expansion of an ideal gas, the internal energy of the gas:
Explanation: Since $dQ = 0$ and the gas does positive work during expansion ($dW > 0$), the first law $dU = -dW$ dictates that internal energy decreases ($dU < 0$).
3. Poisson's relation between pressure and volume for an adiabatic process is expressed as:
Explanation: For a reversible adiabatic process of an ideal gas, the relation between pressure and volume is given by $PV^{\gamma} = \text{constant}$.
4. What is the value of the adiabatic index $\gamma$ for a monatomic gas?
Explanation: For a monatomic gas, degrees of freedom $f = 3$, giving $\gamma = 1 + 2/3 = 1.67$.
5. The slope of an adiabatic curve on a $P-V$ diagram compared to an isothermal curve is:
Explanation: The adiabatic slope is $(dP/dV)_{ad} = -\gamma (P/V)$, whereas the isothermal slope is $-(P/V)$. Thus, the adiabatic curve is steeper by a factor of $\gamma$.
6. The adiabatic bulk modulus of elasticity ($E_{ad}$) for an ideal gas is given by:
7. Which of the following relations is correct for an adiabatic process involving temperature and volume?
Explanation: Substituting $P = nRT/V$ into $PV^{\gamma} = \text{constant}$ yields the temperature-volume relation $TV^{\gamma-1} = \text{constant}$.
8. If a diatomic gas undergoes an adiabatic compression, its adiabatic index $\gamma$ is taken to be:
Explanation: For a typical diatomic gas (like $O_2$ or $N_2$ at room temperature), the degrees of freedom are $5$, resulting in $\gamma = 7/5 = 1.40$.
9. Laplace corrected Newton's formula for the velocity of sound in gases by assuming that sound propagation is:
Explanation: Laplace pointed out that compression and rarefaction in a sound wave happen too rapidly for heat conduction, making the process adiabatic.
10. The work done during an adiabatic expansion of $n$ moles of an ideal gas from temperature $T_1$ to $T_2$ is:
Explanation: Integrating $P dV$ using $PV^{\gamma} = K$ for an adiabatic process yields $W = \frac{n R (T_1 - T_2)}{\gamma - 1}$.
11. What happens to the temperature of a gas during adiabatic compression?
Explanation: During adiabatic compression ($dW < 0$), work is done on the gas, increasing its internal energy ($dU > 0$), which manifests as a rise in temperature.
12. For a polyatomic gas with non-linear molecules, the value of $\gamma$ is approximately:
Explanation: A non-linear polyatomic gas has $6$ degrees of freedom, giving $\gamma = 1 + 2/6 = 1 + 1/3 = 1.33$.
13. Which thermodynamic variable remains constant in both isothermal and adiabatic processes if no expansion/compression occurs?
Explanation: A reversible adiabatic process is an isentropic process, meaning entropy remains constant ($dS = 0$).
14. In the relation $P^{1-\gamma} T^{\gamma} = \text{constant}$, which variables are being related?
Explanation: This relation connects Pressure ($P$) and Temperature ($T$) by eliminating Volume using the ideal gas law.
15. Why is an adiabatic curve steeper than an isothermal curve?
Explanation: Because $\gamma > 1$, the slope factor $-\gamma(P/V)$ has a greater magnitude than $-(P/V)$, meaning pressure drops faster for the same volume increase due to falling temperature.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. An ideal gas at pressure $P$ and volume $V$ is expanded adiabatically to $8$ times its initial volume. If $\gamma = 1.5$, the final pressure is:
Explanation: Using $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$, we get $P_2 = P_1 (V_1/V_2)^{\gamma} = P (1/8)^{1.5} = P / (8^{3/2}) = P / (2^3)^{3/2} = P / 2^4 = P / 16$.
2. The velocity of sound in a gas at pressure $P$ and density $\rho$ according to Laplace's correction is:
Explanation: Newton assumed isothermal conditions giving $v = \sqrt{P/\rho}$, but Laplace applied adiabatic correction, giving $v = \sqrt{\gamma P / \rho}$ where $E_{ad} = \gamma P$.
3. If the volume of an ideal gas ($\gamma = 1.5$) is compressed to $1/4$th of its initial volume, find the ratio of final pressure to initial pressure.
4. During an adiabatic process, the pressure $P$ of a gas is found to be proportional to the cube of its absolute temperature $T$ ($P \propto T^3$). The adiabatic index $\gamma$ for the gas is:
Explanation: The adiabatic $P-T$ relation is $T^{\gamma} P^{1-\gamma} = \text{constant}$, which means $P^{(1-\gamma)/} T^{\gamma} = \text{constant}$ or $P \propto T^{-\gamma / (1-\gamma)}$. Given $P \propto T^3$, we have $-\gamma/(1-\gamma) = 3 \implies -\gamma = 3 - 3\gamma \implies 2\gamma = 3 \implies \gamma = 1.5$.
5. Two cylinders $A$ and $B$ containing equal amounts of an ideal gas at the same initial temperature are expanded adiabatically to the same final volume. Cylinder $A$ contains helium (${\gamma} = 1.67$) and cylinder $B$ contains hydrogen (${\gamma} = 1.40$). Which gas will have a higher final temperature?
Explanation: Using $TV^{\gamma-1} = \text{constant}$, greater $\gamma$ leads to a steeper drop or rise depending on volume. For expansion, higher $\gamma$ (Helium) results in a lower final temperature because it cools more aggressively, wait, let's re-verify: $T_2 = T_1 (V_1/V_2)^{\gamma-1}$. Since $V_2 > V_1$, $(V_1/V_2) < 1$. A larger exponent $(\gamma-1)$ makes the term smaller, meaning helium drops lower in temperature. Thus, Hydrogen (lower $\gamma$) retains a higher final temperature.
6. The bulk modulus of an ideal gas at pressure $P$ for an isothermal process and an adiabatic process are respectively:
Attempt these questions under a ten-minute limit using a separate timer.
1. An adiabatic process is characterized by which of the following equations representing the First Law of Thermodynamics?
Explanation: Since $dQ = 0$ in an adiabatic process, the first law $dU = dQ - dW$ simplifies to $dU = -dW$.
2. What is the adiabatic elasticity of a gas at atmospheric pressure $P$ where $\gamma = 1.4$?
Explanation: Adiabatic elasticity is given by $E_{ad} = \gamma P = 1.4 P$.
3. If an ideal gas expands adiabatically, its pressure:
Explanation: Because the adiabatic curve has a slope steeper by a factor of $\gamma$, pressure drops more rapidly with volume in an adiabatic expansion.
4. A sample of gas (${\gamma} = 1.5$) is compressed adiabatically to half of its volume. The ratio of final to initial temperature is:
5. The degrees of freedom for a gas having adiabatic index $\gamma = 1.4$ is:
Explanation: Using $\gamma = 1 + 2/f$, $1.4 = 7/5 \implies 1 + 2/5$, so $f = 5$ (diatomic gas).
6. In an adiabatic process, work is done by the gas at the expense of:
Explanation: Since no heat enters ($dQ = 0$), any work done by the gas comes directly from a decrease in its internal energy ($dU = -dW$).
7. Which of the following statements is a common misconception about adiabatic processes?
Explanation: An adiabatic process can be slow as long as the system is perfectly thermally insulated from its surroundings; rapidity is just a common practical way to prevent heat exchange.
8. If the absolute temperature of an ideal gas undergoing an adiabatic change is doubled, what happens to its pressure when $\gamma = 1.5$?
Explanation: Using $T^{\gamma} P^{1-\gamma} = \text{constant}$, $(T_2/T_1)^{\gamma} = (P_1/P_2)^{1-\gamma} \implies (2)^{1.5} = (P_2/P_1)^{0.5} \implies P_2/P_1 = (2^{3/2})^2 = 2^3 = 8$ wait. Let's re-evaluate: $(T_2/T_1)^{\gamma} = (P_2/P_1)^{\gamma-1} \implies 2^{1.5} = (P_2/P_1)^{0.5} \implies P_2/P_1 = (2^{3/2})^2 = 2^3 = 8$. Let's check $T^{\gamma} P^{1-\gamma} = \text{const} \implies T^{\gamma/(1-\gamma)} P = \text{const}$. Here $\gamma/(1-\gamma) = 1.5 / (-0.5) = -3$. So $T^{-3} P = \text{const} \implies P \propto T^3$. If temperature is doubled ($T_2 = 2T_1$), then $P_2 = 2^3 P_1 = 8 P_1$. Wait, let's look at option A ($2\sqrt{2}$) vs $8$. $2^3 = 8$. Let's correct explanation: $P \propto T^3$ when $\gamma = 1.5$, so doubling temperature multiplies pressure by $2^3 = 8$.
9. The ratio of the slope of the adiabatic curve to the slope of the isothermal curve at the intersection point on a $P-V$ diagram is:
Explanation: Slope of adiabatic is $-\gamma (P/V)$ and slope of isothermal is $-(P/V)$. Their ratio is $\gamma$.
10. In a free expansion of an ideal gas into a vacuum ($dW = 0, dQ = 0$), the temperature of the gas:
Explanation: For an ideal gas in free expansion, $dQ = 0$ and $dW = 0$, so $dU = 0$, which means temperature remains completely constant ($dT = 0$).