Question

A gas sample has a volume of \(4250 \mathrm{~mL}\) at \(15^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg}\). What is the new temperature \(\left({ }^{\circ} \mathrm{C}\right)\) after the sample is transferred to a new container with a volume of \(2.50 \mathrm{~L}\) and a pressure of \(1.20 \mathrm{~atm} ?\)

Short answer

The new temperature is approximately \(-59.37^{\circ}C\).

Step-by-step solution

Convert volume units

Convert the initial volume from milliliters to liters. \ \ Initial volume: \( 4250 \, \text{mL} \) = \( 4.250 \, \text{L} \)

Convert pressure units

Convert the initial pressure from millimeters of mercury to atmospheres. Use the conversion factor 1 atm = 760 mmHg. \ \ Initial pressure: \( 745 \, \text{mmHg} \) = \( \frac{745}{760} \, \text{atm} \approx 0.980 \, \text{atm} \)

Convert initial temperature to Kelvin

Convert the initial temperature from Celsius to Kelvin using the formula: \( K = ^{\circ}C + 273.15 \). \ \ Initial temperature: \( 15^{\circ} \text{C} \) = \( 15 + 273.15 = 288.15 \, \text{K} \)

Apply the Combined Gas Law

Set up the Combined Gas Law equation: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). Substitute the known values into the equation: \ \ \[ \frac{0.980 \, \text{atm} \times 4.250 \, \text{L}}{288.15 \, \text{K}} = \frac{1.20 \, \text{atm} \times 2.50 \, \text{L}}{T_2} \]

Solve for the new temperature

Rearrange the equation to solve for \( T_2 \): \ \ \[ T_2 = \frac{1.20 \, \text{atm} \times 2.50 \, \text{L} \times 288.15 \, \text{K}}{0.980 \, \text{atm} \times 4.250 \, \text{L}} \] \ \ Calculate the value: \ \ \[ T_2 \approx 213.78 \, \text{K} \]

Convert the final temperature back to Celsius

Convert the final temperature from Kelvin to Celsius using the formula: \( ^{\circ}C = K - 273.15 \). \ \ \( T_2 \approx 213.78 \text{ K} = 213.78 - 273.15 \approx -59.37^{\circ} \text{C} \)

Key concepts

Gas Laws

Gas laws describe how gases behave under different conditions of pressure, volume, and temperature. The most common gas laws include Boyle’s Law, Charles’s Law, and Avogadro’s Law. These laws come together in the Combined Gas Law, represented by the equation:

• Boyle’s Law (\(P_1V_1 = P_2V_2\)) explains that the pressure and volume of a gas are inversely proportional when temperature remains constant.
• Charles’s Law (\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)) states that the volume and temperature of a gas are directly proportional when pressure is held constant.
• Avogadro’s Law (\(\frac{V_1}{n_1} = \frac{V_2}{n_2}\)) explains that the volume and the number of gas molecules are directly proportional at constant temperature and pressure.
These individual gas laws can be combined into the Combined Gas Law formula: \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\] This formula helps us calculate the unknown variable when the gas undergoes changes in pressure, volume, and temperature but the number of molecules (amount of gas) remains constant.

Pressure Conversion

Pressure conversion is essential when dealing with gas laws, as pressure values can be given in different units, such as atmospheres (atm), millimeters of mercury (mmHg), or pascals (Pa). To solve gas law problems accurately, all pressure measurements need to be in the same units. For example, in this exercise, the initial pressure of the gas is given as 745 mmHg. Using the conversion factor of 1 atm = 760 mmHg, we can convert this to atmospheres:\[ P_{initial}= \frac{745 \text{mmHg}}{760 \text{mmHg/atm}} \approx 0.980 \text{atm}\]Always ensure you use the proper conversion factors and be meticulous when converting units. Consistency in units is key to accurate results.

Temperature Conversion

When working with gas laws, temperature must always be in Kelvin (K), not Celsius (\({}^{\text{°}}\text{C}\)). This is because Kelvin is the absolute temperature scale and starts from absolute zero, making the gas law equations work correctly. The conversion between Celsius and Kelvin is straightforward:\[K = {}^{\text{°}}\text{C} + 273.15\]For instance, if the initial temperature of the gas is 15°C, it needs to be converted to Kelvin as follows:\[T_{initial} = 15 + 273.15 = 288.15 \text{K}\]Similarly, after solving for the temperature in Kelvin using the Combined Gas Law, if you need the temperature in Celsius, you would convert it back by subtracting 273.15:\[{}^{\text{°}}\text{C} = K - 273.15\]Thus, proper conversion ensures your calculations are accurate and meaningful in terms of physical reality.

Volume Conversion

Just like temperature and pressure, converting volume units is crucial in gas law calculations. Volumes can be given in milliliters (mL), liters (L), or cubic meters (m³). Generally, liters are used in gas law problems because they are the standard unit for gas volume in the metric system. For example, in the exercise, the initial volume of the gas is 4250 mL, which we need to convert to liters for consistency:\[V_{initial} = 4250 \text{mL} = 4.250 \text{L}\]This conversion is simple because there are 1000 mL in 1 L. Ensuring that all volumes are in the same unit before applying the Combined Gas Law makes the mathematical process smoother and more reliable.