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At 27°C and 1 ATP, a volume of air abruptly shrank by half. Temperature and pressure must be determined.

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Step 1 (a) The ideal-gas equation describes each condition of the air in the tire. We will set up ratios in order to describe these changes. Dividing the equations PV = nRT in the initial and final states, we have PfVf Tf P¡Vi Ti Solving for the final pressure in the tire, gives Pf = = = which gives I Tf P(V)() P₁ Vf T₁ = (1.013 x 105 Pa) x 105 Pa. Note that the volume V¡ cancels out of the equation. = Vf Po = Pr (V ( ) ( ) Pd Pf Vi V₁ Step 2 (b) After the car is driven, the temperature and volume of air in the tire have changed. Let Td be the temperature and Vd be the volume of air in the tire. We have PdVd Td PfVf Tf' x 105 Pa 273 + 273 + Vf Vf Again we see that the volume Vf cancels out of the equation. °C K °℃ K = x 105 Pa.
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