Question

For each of the following sets of volume/temperature data, calculate the missing quantity. Assume that the pressure and the amount of gas remain constant. a. \(V=22.4 \mathrm{L}\) at \(0^{\circ} \mathrm{C} ; V=44.4 \mathrm{L}\) at? \(\mathrm{K}\) b. \(V=1.0 \times 10^{-3} \mathrm{mL}\) at \(-272^{\circ} \mathrm{C} ; V=?\) at \(25^{\circ} \mathrm{C}\) c. \(V=32.3 \mathrm{L}\) at \(-40^{\circ} \mathrm{C} ; V=1000 . \mathrm{L}\) at? \(^{\circ} \mathrm{C}\)

Short answer

a. \(T_2 \approx 543.9 K\) b. \(V_2 \approx 0.298 \mathrm{mL}\) c. \(T_2 \approx 448.1^{\circ} \mathrm{C}\)

Step-by-step solution

a. Find the missing temperature in Kelvin (K)

Given: \(V_1=22.4\) L, \(T_1=0^{\circ} \mathrm{C}\) \(V_2=44.8\) L, \(T_2=?\) K To calculate the missing temperature, let's first convert the given temperature from Celsius to Kelvin: \(T_1 = 0 + 273 = 273 K\) Now, let's apply Charles's Law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) \(\frac{22.4}{273} = \frac{44.4}{T_2}\) Now, solve for \(T_2\): \(T_2 = \frac{44.4 \times 273}{22.4} \approx 543.9 K\)

b. Find the missing volume

Given: \(V_1=1.0 \times 10^{-3} \) mL, \(T_1=-272^{\circ} \mathrm{C}\) \(V_2=?\), \(T_2=25^{\circ} \mathrm{C}\) First, convert the given temperatures from Celsius to Kelvin: \(T_1 = -272 + 273 = 1 K\) \(T_2 = 25 + 273 = 298 K\) Now, apply Charles's Law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) \(\frac{1.0 \times 10^{-3}}{1} = \frac{V_2}{298}\) Now, solve for \(V_2\): \(V_2 = 1.0 \times 10^{-3} \times 298 \approx 0.298 \mathrm{mL}\)

c. Find the missing temperature in Celsius (°C)

Given: \(V_1=32.3\) L, \(T_1=-40^{\circ} \mathrm{C}\) \(V_2=1000\) L, \(T_2=?\) °C First, convert the given temperatures from Celsius to Kelvin: \(T_1 = -40 + 273 = 233 K\) Now, apply Charles's Law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) \(\frac{32.3}{233} = \frac{1000}{T_2}\) Now, solve for \(T_2\): \(T_2 = \frac{1000 \times 233}{32.3} \approx 721.1 K\) Finally, convert back to Celsius: \(T_2 = 721.1 - 273 = 448.1^{\circ} \mathrm{C}\)

Key concepts

Gas Laws

Gas laws are a fundamental part of understanding how gases behave under various conditions. One of the most important gas laws is Charles's Law. This law explores the relationship between the volume and temperature of a gas, keeping pressure and the amount of gas constant. According to Charles's Law, if you increase the temperature of a gas, its volume will also increase. Likewise, if you decrease the temperature, the volume decreases. This makes sense if you think about how molecules move faster and spread apart when heated, taking up more space. So, when you're solving problems using Charles's Law, you are essentially observing how gases expand or contract with temperature changes.

Temperature Conversion

When dealing with gas laws, especially Charles's, converting temperatures into the proper units is crucial. In these calculations, we always use Kelvin because it is the absolute temperature scale. Unlike Celsius or Fahrenheit, Kelvin does not include negative numbers, which is important for gas law equations.

Here is how you convert Celsius to Kelvin: simply add 273 to the Celsius temperature. For example, converting \(0^{\circ} \mathrm{C}\) gives us 273 K since 0 + 273 = 273. This conversion ensures that the results are accurate in gas law problems. Even though converting temperatures might seem like a small step, it is an important process for ensuring the precision and correctness of the calculations.

Volume-Temperature Relationship

The volume-temperature relationship is at the heart of Charles's Law and it shows a direct proportionality. This means when temperature rises, the volume increases, assuming constant pressure. When you write this law mathematically, using the formula \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), it shows that the ratio of initial volume and temperature is equal to the ratio of the final volume and temperature.

Important steps for solving such problems involve:

• Converting all temperatures into Kelvin
• Plugging in known values into the relationship formula
• Solving for the missing variable, either final volume or temperature

Remember, this law only applies when the amount and pressure of the gas are constant. So, keeping these factors in check is important to accurately use Charles's Law.