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A cubic metal box that has 20 cm long edges contains air at a pressure of 1.0 atm and a temperature of 300 K.
The box is sealed so that the enclosed volume remains constant, and it is heated to a temperature of 400 K. Find
the force due to the internal air pressure on each wall of the box.

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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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