Q.For adiabatic process, obtain the relation between pressure and volume for an adiabatic process.

"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. …

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,

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For an adiabatic process on an ideal gas, $PV^\gamma = \text{constant}$.

In an adiabatic process, $dQ = 0$. Applying the first law of thermodynamics ($dQ = dU + PdV$) with $dU = nC_VdT$, combined with the ideal gas law $PV = nRT$, and eliminating $T$ using Mayer's relation $C_P - C_V = R$, leads (after integration) to the relation:

$PV^\gamma = \text{constant}$

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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 ($Q \approx 0$).

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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). …

"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:

$Q = 0$

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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:

$\Delta U = Q - W = -W$

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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)] …