Q.Why is the p-V curve for an adiabatic process steeper than that for an isothermal process?

A process with $Q = 0$ throughout — no heat enters or leaves the system — is called adiabatic.

By definition, an adiabatic process is one that occurs without any exchange of heat between the system and its surroundings, i.e. $dQ = 0$ at every stage. This is achieved either by thermally insulating the system (perfectly non-conducting walls) or by carrying out the process so quickly that there is no time for heat to flow.

From the first law of thermodynamics, $dQ = dU + dW$, so for an adiabatic process $dU = -dW$: any work done by the system comes entirely at the expense of its internal energy (and temperature changes as a result), even though no heat is absorbed or released.

(Contrast: isobaric = constant pressure, isochoric = constant volume, isothermal = constant temperature, all of which generally do involve heat exchange.)

A sudden tyre burst happens so fast that there is no time for heat to be exchanged with the surroundings, which is the defining condition of an adiabatic process.

An adiabatic process is one in which no heat enters or leaves the system (Q = 0); any change in internal energy occurs purely due to work done by or on the gas.

When a cycle tyre bursts, the compressed air inside expands almost instantaneously. This expansion happens on a timescale far too short for heat to flow between the escaping air and its surroundings.

Since Q ≈ 0 during this rapid expansion, the process is adiabatic (not isothermal, which would require the temperature to stay constant via heat exchange; not isobaric, since pressure changes rapidly; not isochoric, since volume clearly changes).