The driver from Practice Problem 13.3 fills his \(18.9-\mathrm{L}\) steel gasoline can in the morning when the temperature of the can and the gasoline is \(15.0^{\circ} \mathrm{C}\) and the pressure is 1.0 atm, but this time he remembers to replace the tightly fitting cap after filling the can. Assume that the can is completely full of gasoline (no air space) and that the cap does not leak. The temperature climbs to \(30.0^{\circ} \mathrm{C}\) Ignoring the expansion of the steel can, what would be the pressure of the heated gasoline? The bulk modulus for gasoline is $1.00 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}$

Before starting any calculation, we need to convert the given temperatures in Celsius to Kelvin. To do so, we add 273.15 to the Celsius temperatures. Initial temperature in Kelvin = \(T_1 = 15.0^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 288.15 \mathrm{K}\) Final temperature in Kelvin = \(T_2 = 30.0^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 303.15 \mathrm{K}\)

To find the change in volume due to the change in temperature, we can use the equation for thermal expansion, which states that the change in volume \(\Delta V\) is related to the initial volume \(V_1\), the coefficient of volume expansion \(\beta\), and the change in temperature \(\Delta T\) by the formula: $$ \Delta V = V_1 \beta \Delta T $$ Unfortunately, we are not given the coefficient of volume expansion for gasoline, but we are given the bulk modulus. The relationship between \(\beta\) and the bulk modulus \(B\) is as follows: $$ \beta = \frac{1}{B} $$ In this context, we can rewrite the formula for thermal expansion as: $$ \Delta V = V_1 \frac{1}{B} (T_2 - T_1) $$ The given volume of gasoline is \(V_1 = 18.9 \mathrm{L}\), and the bulk modulus \(B = 1.00 \times 10^9 \mathrm{N} / \mathrm{m}^{2}\). Now, we can calculate the change in volume \(\Delta V\) due to the change in temperature: $$ \Delta V = 18.9 \mathrm{L} \cdot \frac{1}{1.00 \times 10^9 \mathrm{N} / \mathrm{m}^{2}} (303.15 \mathrm{K} - 288.15 \mathrm{K}) $$ $$ \Delta V = 2.826 \times 10^{-7} \mathrm{m^3} $$

We can now use the formula for bulk modulus to find the new pressure \(P_2\): $$ B = -V_1 \frac{P_2 - P_1}{\Delta V} $$ We can rearrange this formula to solve for the final pressure \(P_2\): $$ P_2 = P_1 - \frac{B \Delta V}{V_1} $$ The given initial pressure is \(P_1 = 1.0 \mathrm{atm}\). To convert it to Pascal, multiply by 101325: $$ P_1 = 1.0 \mathrm{atm} \cdot 101325 \frac{\mathrm{N}}{\mathrm{m^2} \cdot \mathrm{atm}} = 101325 \mathrm{N/m^2} $$ We plug in our values and calculate the final pressure \(P_2\): $$ P_2 = 101325 \mathrm{N/m^2} - \frac{(1.00 \times 10^9 \mathrm{N/m^2})(2.826 \times 10^{-7} \mathrm{m^3})}{0.0189 \mathrm{m^3}} $$ $$ P_2 = 106601.49 \mathrm{N/m^2} $$ Finally, we convert the final pressure back to atm: $$ P_2 = 106601.49 \mathrm{N/m^2} \cdot \frac{1 \mathrm{atm}}{101325 \mathrm{N/m^2}} = 1.052 \mathrm{atm} $$ So, the pressure of the heated gasoline would be approximately \(1.052 \mathrm{atm}\).