Question

Water bottle in a hot car. In the American Southwest, the temperature in a closed car parked in sunlight during the summer can be high enough to burn flesh. Suppose a bottle of water at a refrigerator temperature of $5.00^\circ C$ is opened, then closed and then left in a closed car with an internal temperature of $75.0^\circ C$. Neglecting the thermal expansion of the water and the bottle, find the pressure in the air pocket trapped in the bottle. (The pressure can be enough to push the bottle cap past the threads that are intended to keep the bottle closed)

Answer 1

To understand the behaviour of gases there was an equation that was discovered by Benoit Paul Emile. The equation is known as the ideal gas equation. This ideal gas equation is appropriate for many gases under many circumstances. Use the ideal gas equation to find the unknown pressure.

Complete step by step solution:
Find the pressure in the air pocket trapped in the bottle:
Here we would be using the ideal gas equation which is given by:
$PV = nRT$;
Here:
P = Pressure;
n = Number of moles;
V = Volume;
R = Universal gas constant;
T = Temperature;
Now, there would be two equations, one for the bottle at $5.00^\circ C$and another at $75.0^\circ C$.
${P_1}{V_1} = nR{T_1}$;
Put in the given values:
$\Rightarrow 1 \times {V_1} = nR \times 278$; …($5.00^\circ C$= 278K)
$\Rightarrow {V_1} = nR \times 278$;
Now, the second equation:
${P_2}{V_2} = nR{T_2}$;
$\Rightarrow {P_2}{V_2} = nR \times 348.15$; …($75.0^\circ C$= 348.15K)
Now, divide the ideal gas equation for the first part ${V_1} = nR \times 278$by the second part${P_2}{V_2} = nR \times 348.15$;
$\dfrac{{{V_1}}}{{{P_2}{V_2}}} = \dfrac{{nR \times 278}}{{nR \times 348.15}}$;.
Here, the number of moles and the constant will remain the same:
$\Rightarrow \dfrac{{{V_1}}}{{{P_2}{V_2}}} = \dfrac{{278}}{{348.15}}$;
We apply the ideal gas law $\left( {{V_1} = {V_2}} \right)$:
$\Rightarrow \dfrac{1}{{{P_2}}} = \dfrac{{278}}{{348.15}}$;
Solve for the unknown pressure:
$\Rightarrow {P_2} = \dfrac{{348.15}}{{278}}$;
$\Rightarrow {P_2} = 1.252 atm$;

The pressure in the air pocket trapped in the bottle is 1.252 atm.

Note: Here, we need to keep in mind that the bottle was opened at $5.00^\circ C$ and closed. Also, here the pressure at $5.00^\circ C$ will be the atmospheric pressure 1 atm. The volume would be the same for both the cases. Equate the two equations together and find the unknown.