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HomeACSEE (Form 6)PhysicsHeat (Thermometers and First Law of Thermodynamics) › Air escaping from a suddenly bursting tyre cools…

Air escaping from a suddenly bursting tyre cools. The correct relation to use is:

A$PV=K$
B$PT^{-1}=K$
C$PV=nRT$
D$P^{\gamma-1}T^{-\gamma}=K$
Answer & Solution
Correct answer: D. $P^{\gamma-1}T^{-\gamma}=K$
1. A tyre bursting is sudden, so no appreciable heat is exchanged with the surroundings. 2. A process with no heat transfer is adiabatic, not isothermal and not isobaric. 3. The adiabatic relation between pressure and temperature is $P^{\gamma-1}T^{-\gamma}=K$. 4. Equivalently $T_1^{\gamma}P_1^{1-\gamma}=T_2^{\gamma}P_2^{1-\gamma}$ connects the two states. 5. Boyle's law holds the temperature fixed and the pressure law holds the volume fixed, so neither describes a burst; both were recorded errors. _Source: NECTA ACSEE 2023 Physics 131/1, Question 5:Heat (Thermometers and First law of Thermodynamics)_
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