A balloon will burst at a volume of \(2.00 \mathrm{~L}\). If it is partially filled at \(20.0^{\circ} \mathrm{C}\) and \(65 \mathrm{~cm}\) Hg to occupy \(1.75 \mathrm{~L}\), at what temperature will it burst if the pressure is exactly \(1 \mathrm{~atm}\) at the time that it bursts?

Understanding how to convert units of pressure is crucial when solving problems in physics and chemistry.

• Pressure is the force exerted per unit area and is measured in various units, such as atmospheres (atm), millimeters of mercury (mm Hg), and pascals (Pa).
• In this exercise, the initial pressure was given in centimeters of mercury (cm Hg). We need to convert this to atmospheres to use it in the ideal gas law equation.

We used the conversion factor: 1 atm = 76 cm Hg.
Therefore, 65 cm Hg is converted by: \[ \text{Initial pressure} = \frac{65 \text{ cm Hg}}{76 \text{ cm Hg/atm}} = 0.855 \text{ atm} \]
This pressure conversion is important because it allows us to use the pressure value in equations that require atmospheres, making calculations accurate.

Temperature conversion is another key step in this problem because the ideal gas law requires temperature in Kelvin (K).

• Celsius (°C) is commonly used in everyday contexts, but Kelvin is the SI unit for thermodynamic temperature.
• To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.

In this exercise, the initial temperature was given as 20.0°C. The conversion to Kelvin is calculated as follows:\[ T_1 (K) = T (°C) + 273.15 = 20 + 273.15 = 293.15 \text{ K} \]
Knowing this conversion allows us to use the temperature in the ideal gas law equation, thus ensuring we have a consistent set of units for our calculations.

The ideal gas law ( \[ PV=nRT \] ) can be used to solve for various unknowns. In this case, we need to determine the final temperature at which the balloon bursts.

• Given the initial volume, pressure, and temperature and the final volume and pressure, we use the combined gas law:
• \[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \]

First, we convert all measurements to appropriate units:

• Initial pressure ( \[ P_1 \] ) = 0.855 atm.
• Initial volume ( \[ V_1 \] ) = 1.75 L.
• Initial temperature ( \[ T_1 \] ) = 293.15 K (converted from Celsius).
• Final pressure ( \[ P_2 \] ) = 1 atm.
• Final volume ( \[ V_2 \] ) = 2.00 L.

Plugging these into the combined gas law equation and solving for the final temperature ( \[ T_2 \] ):\[ \frac{0.855 \text{ atm} \times 1.75 \text{ L}}{293.15 \text{ K}} = \frac{1 \text{ atm} \times 2.00 \text{ L}}{T_2} \]Rearranging to solve for \[ T_2 \] :\[ T_2 = \frac{1 \text{ atm} \times 2.00 \text{ L} \times 293.15 \text{ K}}{0.855 \text{ atm} \times 1.75 \text{ L}} = 393.52 \text{ K} \]Finally, converting \[ T_2 \] back to Celsius:\[ T (°C) = T (K) - 273.15 = 393.52 - 273.15 = 120.37 \text{ °C} \]This step-by-step process demonstrates how to isolate and solve for an unknown temperature using the ideal gas law and the combined gas law.