Question

The volume of a weather balloon is \(200.0 \mathrm{L}\) and its internal pressure is 1.17 atm when it is launched at \(20^{\circ} \mathrm{C}\). The balloon rises to an altitude in the stratosphere where its internal pressure is \(63 \mathrm{mmHg}\) and the temperature is \(210 \mathrm{K} .\) What is the volume of the balloon at this altitude?

Short answer

Answer: The volume of the balloon at this altitude is approximately 2018 L.

Step-by-step solution

Write down the known and unknown values

We are given: - Initial volume \(V_1 = 200.0 \ \text{L}\) - Initial pressure \(P_1 = 1.17 \ \text{atm}\) - Initial temperature \(T_1 = 20^{\circ} \mathrm{C}\) - Final pressure \(P_2 = 63 \ \mathrm{mmHg}\) - Final temperature \(T_2 = 210 \ \mathrm{kK}\) We need to find the final volume \(V_2\).

Convert the initial and final pressure units to a common unit

Since we have pressure in two different units, we'll convert these values to a common unit, in this case, \(atm\). We'll use the following conversion factor: \(1 \ \text{atm} = 760 \ \mathrm{mmHg}\) Initial pressure is already in \(atm\). Convert the final pressure: \(P_2 = \dfrac{63 \ \mathrm{mmHg}}{760 \ \mathrm{mmHg/atm}} = 0.0829 \ \text{atm}\)

Convert the initial temperature to Kelvin

Convert the initial temperature \(T_1\) from Celsius to Kelvin using the following formula: \(T_1 = T_1^\circ \mathrm{C} + 273.15 = 20^\circ \mathrm{C}+ 273.15 = 293.15 \ \mathrm{K}\)

Apply the Ideal Gas Law

Use the Ideal Gas Law with initial and final conditions: \(P_1V_1/T_1=P_2V_2/T_2\) Rearrange the equation to solve for \(V_2\): \(V_2 = \dfrac{P_1V_1T_2}{P_2T_1}\)

Substitute the known values into the equation and solve for \(V_2\)

Substitute the known values of initial and final conditions into the equation and solve for \(V_2\): \(V_2 = \dfrac{(1.17 \ \text{atm})(200.0 \ \text{L})(210 \ \mathrm{K})}{(0.0829 \ \text{atm})(293.15 \ \mathrm{K})}\) \(V_2 = \dfrac{49042}{24.301}\) \(V_2 ≈ 2018 \ \text{L}\) The volume of the balloon at this altitude is approximately \(2018 \ \text{L}\).

Key concepts

Pressure Conversion

In the context of the ideal gas law, pressure needs to be consistent in its units for accurate calculations. It's common to encounter different pressure units, such as atmospheres (atm) and millimeters of mercury (mmHg), in problems. To convert from one unit to another, a conversion factor is employed.

For example, in this exercise, the initial pressure was already given in atmospheres (1.17 atm). The final pressure, however, was provided in mmHg (63 mmHg). To convert it to atm, we use the fact that 1 atm equals 760 mmHg. The conversion is performed by dividing the given pressure in mmHg by the conversion factor:

• \(P_2 = \frac{63 \ \mathrm{mmHg}}{760 \ \mathrm{mmHg/atm}} = 0.0829\ \text{atm}\)

Doing this ensures both pressures are expressed in the same unit, allowing the ideal gas law to be applied correctly.

Temperature Conversion

Temperature conversions are essential in gas law problems since the temperature in Kelvin is a key part of the formula. The ideal gas law requires absolute temperature, which is why we convert Celsius to Kelvin.

In this exercise, the initial temperature was given as \(20^{\circ} \mathrm{C}\). The formula to convert from Celsius to Kelvin is:

• \(T_1 = T_1^{\circ} \mathrm{C} + 273.15\)

Applying the conversion, we find:

• \(T_1 = 20 + 273.15 = 293.15\ \mathrm{K}\)

This step ensures that all temperatures are expressed in Kelvin, which is necessary as negative values or Celsius do not suit the equations derived from the ideal gas law.

Initial and Final Conditions

Understanding the significance of initial and final conditions is crucial in solving gas law problems. These conditions dictate the parameters which need to be compared and adjusted in the equation.

In this exercise, we had initial parameters:

• Volume \(V_1 = 200.0 \ \text{L}\)
• Pressure \(P_1 = 1.17\ \text{atm}\)
• Temperature \(T_1 = 293.15\ \mathrm{K}\)

And final parameters:

• Pressure \(P_2 = 0.0829\ \text{atm}\)
• Temperature \(T_2 = 210\ \mathrm{K}\)

By plugging these into the ideal gas law

• \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\) and rearranging for \(V_2\).

We can solve for the desired final volume, which is a representation of all these physical conditions at a new state for the gas, demonstrating its behavior under varied parameters.