Q.Derive the equation of state for an adiabatic process.
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Adiabatic Compression Factor: From Intuition to Precision
Imagine you pump air into a bicycle tyre. The pump gets noticeably warm. That warmth isn't coming from outside — it's generated inside the air you're compressing. Why? Because you're doing work on the gas, and since the compression happens too fast for heat to escape, all that work stays inside as internal energy, raising the temperature.
This is the core idea: adiabatic means "no heat exchange with the surroundings." When you compress a gas adiabatically, its temperature rises. The adiabatic compression factor is the ratio that tells you how much the temperature rises for a given compression.
The Intuition First
Think of a gas as a swarm of tiny, fast-moving particles. When you push a piston in, you're moving the wall toward the particles. Each time a particle bounces off the approaching wall, it rebounds with a higher speed than it had — like a tennis ball hit by a moving racket. Faster particles mean higher temperature.
If the compression is slow enough that heat can leak out (isothermal), the temperature stays constant. But if it's fast (adiabatic), the temperature climbs. The adiabatic compression factor captures exactly this: the ratio of final temperature to initial temperature when a gas is compressed without heat loss.
The Precise Statement
For an ideal gas undergoing a reversible adiabatic process, the relationship between temperature (T) and volume (V) is:
TVγ−1=constant
where γ (gamma) is the adiabatic index — the ratio of specific heats: γ=CvCp.
If you compress from volume V1 to V2 (so V2<V1), the temperature changes from T1 to T2 according to:
T2=T1(V2V1)γ−1
The factor (V2V1)γ−1 is the adiabatic compression factor for temperature. Since V1/V2>1 and γ−1>0, this factor is always greater than 1 — confirming that temperature rises.
Adiabatic compression factor (temperature)=(V2V1)γ−1
You can also express it in terms of pressure. Using PVγ=constant, you get:
T2=T1(P1P2)γγ−1
Here (P1P2)γγ−1 is the pressure-based version.
What γ Means
γ depends on the number of degrees of freedom of the gas molecule:
| Gas type | Degrees of freedom | γ | Example |
|---|---|---|---|
| Monatomic | 3 (translation only) | 5/3 ≈ 1.67 | He, Ar |
| Diatomic / linear triatomic (rigid) | 5 (3 translation + 2 rotation) | 7/5 = 1.40 | N₂, O₂; CO₂ (theoretical) |
| Non-linear triatomic | 6 (3 translation + 3 rotation) | 4/3 ≈ 1.33 | H₂O vapour |
A higher γ means the temperature rises more sharply for the same compression. Monatomic gases heat up the most — they have only translational motion to store energy, so all the work of compression goes into raising temperature. …
For an adiabatic process (no heat exchange with surroundings), applying the first law of thermodynamics with dQ=0 to an ideal gas leads to a relation between pressure and volume. …
For an adiabatic process on an ideal gas, combining the first law (with dQ=0) with the ideal gas equation gives P*V^gamma = constant.
In an adiabatic process, no heat enters or leaves the system: dQ = 0. The first law of thermodynamics, dQ = dU + dW, then gives:
dU = -dW
For n moles of an ideal gas, dU = nCvdT, and the work done by the gas during a small expansion is dW = PdV. So: nCvdT = -PdV ... (1)
From the ideal gas equation PV = nRT, we get P = nRT/V. Substituting into (1): nCvdT = -(nRT/V)dV CvdT/T = -RdV/V
Integrating both sides:
Cvln(T) = -Rln(V) + constant
ln(T^Cv) + ln(V^R) = constant
ln(T^Cv * V^R) = constant
T^Cv * V^R = constant ... (2)
Since R = Cp - Cv, we have R/Cv = (Cp-Cv)/Cv = Cp/Cv - 1 = gamma - 1, where gamma = Cp/Cv. Raising (2) to the power 1/Cv:
T * V^(R/Cv) = constant
T * V^(gamma-1) = constant ... (3)
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- CBSE 2026Set ANNUAL1 markMCQQ.In an adiabatic process, which of the following quantities remains constant?(a) Temperature(b) Pressure(c) Volume(d) None of these
›Reveal solutionSolution
"Adiabatic" specifically means no heat transfer into or out of the system (Q = 0) -- it says nothing about P, V, or T individually staying constant; in fact all three typically DO change in an adiabatic process.
By definition, an adiabatic process is one in which no heat enters or leaves the system: Q = 0. This is different from:
- Isothermal process: temperature (T) stays constant.
- Isobaric process: pressure (P) stays constant.
- Isochoric (isovolumetric) process: volume (V) stays constant. …
- CBSE 2026Set ANNUAL1 markMCQQ.The relation between pressure (P) and volume (V) of an ideal gas in an adiabatic process is (where γ = Cp/Cv)(a) (PV)^γ = constant(b) P^γ V = constant(c) PV = constant(d) PV^γ = constant
›Reveal solutionSolution
Adiabatic process: P V^gamma = constant. Answer (D).
In an adiabatic process no heat is exchanged (Q = 0). Combining the first law with the ideal-gas law leads to the relation:
P V^gamma = constant,
…
- CBSE 2025Set ANNUAL1 markMCQQ.For adiabatic process, the relation between pressure and volume is (A) PV^γ = constant (B) P^(1-γ)V^γ = constant (C) PV^(γ-1) = constant (D) P^γV^γ = constant
›Reveal solutionSolution
For an adiabatic process on an ideal gas, PVγ=constant.
In an adiabatic process, dQ=0. Applying the first law of thermodynamics (dQ=dU+PdV) with dU=nCVdT, combined with the ideal gas law PV=nRT, and eliminating T using Mayer's relation CP−CV=R, leads (after integration) to the relation:
PVγ=constant
…
- CBSE 2025Set ANNUAL1 markMCQQ.Assertion (A): Air quickly leaking out of a balloon becomes cooler. Reason (R): The leaking air undergoes adiabatic expansion. Select the correct answer from the codes below.(a) Both (A) and (R) are true and (R) is the correct explanation of (A).(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).(c) (A) is true, but (R) is false.(d) (A) and (R) both are false.
›Reveal solutionSolution
The leaking air's rapid, essentially heat-free expansion is a textbook case of adiabatic cooling — the reason given is exactly why the observation is true.
Assertion (A): Air leaking quickly out of a balloon becomes cooler — this is true and can be felt directly (a deflating balloon's nozzle feels cold).
Reason (R): The leaking air undergoes adiabatic expansion — also true. The process happens so fast that essentially no heat is exchanged with the surroundings (Q≈0).
…
- CBSE 2025Set ANNUAL1 markMCQQ.The thermodynamic process in which no heat exchange takes place is called -(a) Adiabatic(b) Isochoric(c) Isobaric(d) Isothermal
›Reveal solutionSolution
A process with no heat exchange with the surroundings is called an adiabatic process.
In thermodynamics, processes are classified by what is held constant or restricted:
- Isochoric: volume is held constant.
- Isobaric: pressure is held constant.
- Isothermal: temperature is held constant (system in thermal contact with a reservoir, heat CAN flow). …
- CBSE 2024Set ANNUAL1 markMCQQ.An adiabatic process occurs at constant :(a) Temperature(b) Pressure(c) Heat(d) Temperature and Pressure
›Reveal solutionSolution
"Adiabatic" literally means no heat transfer occurs — the system is thermally insulated from its surroundings during the process.
An adiabatic process is one carried out in a thermally insulated system, or fast enough that there is no time for heat exchange with the surroundings. This means:
Q=0
…
- CBSE 2024Set ANNUAL1 markMCQQ.Two statements are given below: one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer using the codes given. Assertion (A) : Air quickly leaking out of a balloon becomes cooler. Reason (R) : The leaking air undergoes adiabatic expansion.(a) Both (A) and (R) are true and (R) is the correct explanation of (A).(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).(c) (A) is true but (R) is false.(d) (A) is false and (R) is also false.
›Reveal solutionSolution
Rapidly leaking air expands adiabatically (no time to absorb heat), so by the first law its internal energy — and hence temperature — falls, which is exactly why it feels cooler.
Assertion (A): When air rushes quickly out of a balloon, it expands very fast. Because this happens too quickly for any significant heat to be exchanged with the surroundings, the expansion is essentially adiabatic, and the escaping air does indeed feel cooler.
Reason (R): In an adiabatic expansion, Q = 0. By the first law of thermodynamics:
ΔU=Q−W=−W
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- CBSE 2022Set ANNUAL1 markMCQQ.The ratio gamma = Cp/Cv for a gas mixture consisting of 8 g of helium and 16 g of oxygen is :(a) 27/17(b) 23/15(c) 17/27(d) 15/23
›Reveal solutionSolution
Find the number of moles of each gas, use the correct degrees of freedom for each (monatomic He: f=3; diatomic O2: f=5) to get each gas's Cv and Cp, mole-average these for the mixture, and take the ratio.
Step 1 — moles of each gas:
Helium: mass = 8 g, molar mass M(He) = 4 g/mol ⇒ n(He) = 8/4 = 2 mol.
Oxygen: mass = 16 g, molar mass M(O2) = 32 g/mol ⇒ n(O2) = 16/32 = 0.5 mol.
Step 2 — molar specific heats of each gas:
Helium is monatomic (f = 3): Cv(He) = (3/2)R, Cp(He) = (5/2)R.
Oxygen is diatomic (f = 5, rigid, room temperature): Cv(O2) = (5/2)R, Cp(O2) = (7/2)R.
Step 3 — mixture's Cv and Cp (mole-weighted average):
Cv(mix) = [n(He)·Cv(He) + n(O2)·Cv(O2)] / [n(He)+n(O2)] …
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