91Ó°ÊÓ

Gas laws are fundamental principles that describe the behavior of gases. They explain how pressure, volume, and temperature are interrelated. One important gas law relevant to our exercise is Gay-Lussac's Law, which focuses on the relationship between pressure and temperature of a gas with constant volume. Gay-Lussac's Law states that the pressure of a given mass of gas is directly proportional to its absolute temperature, as long as the volume remains constant. Mathematically, we express this law as: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] where \(P_1\) and \(P_2\) are the initial and final pressures, and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin. Understanding gas laws like this helps us predict how gases will respond to changes in their environment.

When working with gas laws, it's essential to use temperatures in Kelvin. This is because Kelvin is the absolute temperature scale and it ensures direct proportionality in gas law equations. To convert Celsius to Kelvin, we use the formula: \[ T(K) = T(^{\rm o}C) + 273.15 \] In this exercise, we need to convert the given temperatures from Celsius to Kelvin.
For example: \[ T_1 = -15^{\rm o}C + 273.15 = 258.15 \text{ K} \] \[ T_2 = 46^{\rm o}C + 273.15 = 319.15 \text{ K} \] By converting to Kelvin, we can correctly apply Gay-Lussac's Law to find the relationship between pressure and temperature.

Once we have the temperatures in Kelvin, we can calculate the final pressure. Using Gay-Lussac's Law \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \], we rearrange to solve for the final pressure \(P_2\): \[ P_2 = P_1 \times \frac{T_2}{T_1} \] Here, \(P_1 = 2.35 \text{ atm}\), \(T_1 = 258.15 \text{ K}\), and \(T_2 = 319.15 \text{ K}\).
Substituting these values into the equation: \[ P_2 = 2.35 \text{ atm} \times \frac{319.15 \text{ K}}{258.15 \text{ K}} ≈ 2.91 \text{ atm} \] This final step ensures that we’ve correctly accounted for the temperature change and accurately calculated the new pressure of the gas.