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 75 ⁰C. Neglecting the thermal expansion of 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.)

Use the ideal gas equation for initial and final conditions. After that, by comparing those equations, find the final pressure in the air pocket. According to an ideal gas equation,

.$\mathbf{pv}\mathbf{=}\mathbf{nRT}$

Formula is as follow:

$p_i{v}_i=nR{T}_i$

Here, p is pressure, v is volume, T is temperature, R is universal gas constant and n is number of moles.

From the problem, initial volume and final volume is same. Using ideal gas equation,

$p_i{v}_i=nR{T}_i$ ……. (1)

And,

$p_f{v}_f=nR{T}_f$ ……. (2)

By dividing equation (2) by (1),

$$\begin{gathered} \frac{p_f{v}_f}{p_i{v}_i}=\frac{nR{T}_f}{nR{T}_i} \\ \frac{p_f{v}_f}{p_i{v}_i}=\frac{T_f}{T_i} \end{gathered}$$

Calculate the final pressure. So, rearrange the equation for final pressure,

$p_f=\frac{p_i{v}_i{T}_f}{v_f{T}_i}$

As,

$$\begin{gathered} v_i=v_f \\ {p}_f=\frac{p_i{T}_f}{T_i} \\ {p}_f=\frac{1\times 348.15}{278.15} \\ {p}_f=1.25\text{atm} \end{gathered}$$

So, the final pressure in the air pocket trapped in the bottle is $p_f=1.25\text{atm}$.