To find the pressure in the air pocket trapped in the bottle, we can use the ideal gas law, which relates pressure, volume, temperature, and the number of moles of gas:
PV = nRT
where P is the pressure, V is the volume of the air pocket, n is the number of moles of air, R is the ideal gas constant, and T is the temperature of the air pocket in Kelvin.
We can assume that the volume of the air pocket remains constant, so we can rewrite the ideal gas law as:
P = nRT/V
where P, R, and V are constants, and n and T are the variables.
To find the pressure in the air pocket, we need to find the number of moles of air and the temperature of the air pocket.
Assuming that the bottle is initially at a refrigerator temperature of 9.72 °C, or 283.87 K, and is then left in a closed car with an internal temperature of 62.5 °C, or 335.65 K, the change in temperature of the air pocket is:
ΔT = 335.65 K - 283.87 K = 51.78 K
Since the thermal expansion of the water and the bottle is neglected, we can assume that the volume of the air pocket remains constant. Therefore, the number of moles of air also remains constant.
The ideal gas constant R is approximately 8.31 J/(mol K).
Now we can plug in the values into the formula:
P = nRT/V = (nR/V)ΔT
To find n/V, we can use the density of air, which is approximately 1.2 kg/m^3 at room temperature and pressure. The volume of the air pocket in the bottle can be estimated to be about 10% of the volume of the bottle, or about 0.001 m^3.
n/V = (1.2 kg/m^3) / (28.97 g/mol) = 0.04147 mol/m^3
Therefore, the pressure in the air pocket is:
P = (nR/V)ΔT = (0.04147 mol/m^3)(8.31 J/(mol K))(51.78 K)/(0.001 m^3) = 13618.7 Pa
Converting to units of psi, we get:
P = 13618.7 Pa / 6894.76 Pa/psi = 1.975 psi
Therefore, the pressure in the air pocket trapped in the water bottle is approximately 1.975 psi.