Question

A sealed balloon is filled with \(1.00 \mathrm{~L}\) helium at \(23^{\circ} \mathrm{C}\) and \(1.00\) atm. The balloon rises to a point in the atmosphere where the pressure is 220 . torr and the temperature is \(-31^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from \(1.00\) atm to a pressure of 220 . torr?

Short answer

The change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is 2.29 L.

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert both the given temperatures from Celsius to Kelvin by adding 273.15 K to each temperature: Temperature T1: \(23^\circ C + 273.15 K = 296.15 K\) Temperature T2: \(-31^\circ C + 273.15 K = 242.15 K\)

Convert pressure to atm

Next, we need to convert the given pressure (P2) from torr to atm. We know that 1 atm = 760 torr, so we can use the conversion factor: Pressure P2: \(\frac{220 \, torr}{760 \, torr/atm} = 0.2895 \, atm\)

Apply the combined gas law

Now, we can apply the combined gas law equation to find the final volume (V2) of the balloon: \(\frac{1.00 \, L}{296.15 \, K} \times \frac{242.15 \, K}{0.2895 \, atm} = V2\) Solve for V2: \(V2 = 3.29 \, L\)

Find the change in volume

Now that we have the initial volume of the balloon (V1) and its final volume (V2), we can find the change in volume: Change in volume = \(V2 - V1\) Change in volume = \(3.29 \, L - 1.00 \, L\) Change in volume = \(2.29\, L\) The change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 torr is 2.29 L.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a mathematical relationship that describes the behavior of ideal gases. It is expressed with the formula \( PV = nRT \), where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume.
• \( n \) is the number of moles.
• \( R \) is the ideal gas constant \((0.0821 \, L \cdot atm/mol \cdot K)\).
• \( T \) is the temperature in Kelvin.

By understanding and employing this law, we can predict how a gas will behave under varying conditions such as temperature, pressure, or volume changes. The Ideal Gas Law helps us comprehend the relationship among these variables. However, in this particular problem, we use a derived form called the Combined Gas Law, which is useful when the amount of gas does not change. Remember to always convert temperatures to Kelvin when using these equations for accurate results.

Temperature Conversion

Temperature conversion plays a crucial role in gas law calculations. The Kelvin scale is used in gas law equations because it provides an absolute reference point. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.

• For example, if you have a temperature of \( 23^{\circ} C \), the conversion to Kelvin would be:
• \( 23 + 273.15 = 296.15 \, K \)

Similarly, if the temperature is \(-31^{\circ} C\), you would add 273.15:

• \(-31 + 273.15 = 242.15 \, K \)

Using Kelvin is important as it ensures the proportionality in gas law equations. Always double-check your temperature units before proceeding with any calculation involving gases.

Pressure Conversion

Pressure conversion helps us work seamlessly with gas laws by using consistent units. Often, pressure measurements may need to be converted. In this exercise, pressure was given in torr, but our calculation requires it in atm.

• The conversion factor used is \( 1 \, atm = 760 \, torr \).
• Given pressure in torr: \( 220 \, torr \).
• Convert this to atm: \( \frac{220}{760} = 0.2895 \, atm \).

Accurate pressure conversion is vital to ensure that all gas law variables are compatible, which prevents calculation errors. Always remember to match your pressure unit to the standard required by the equation you are using.

Volume Change Calculation

Understanding volume change is essential to solving gas problems like the one we've tackled. By applying the Combined Gas Law, which is useful when the amount of gas remains constant, we can find how a gas's volume changes with pressure and temperature shifts.

• The Combined Gas Law formula is \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \).
• Using this law, we rearrange to solve for the final volume \( V_2 \).
• Substituting known values: \( \frac{1.00 \, L}{296.15 \, K} \times \frac{242.15 \, K}{0.2895 \, atm} = V_2 \).
• This gives \( V_2 = 3.29 \, L \).

Finally, to find the change in volume, subtract the initial volume \( V_1 \) from \( V_2 \), resulting in a 2.29 L increase. This change shows how the gas expanded as it ascended, due to lower pressure and temperature.