Question

A gaseous sample occupies 32.4 mL at ?23°C and 0.75 atm. What volume will it occupy at STP? (Chapter 13)

Short answer

The gaseous sample will occupy a volume of 38.56 mL at standard temperature and pressure (STP).

Step-by-step solution

Write down the given information and convert to appropriate units

The initial volume, pressure, and temperature are given as 32.4 mL, 0.75 atm, and -23°C, respectively. Convert the temperature to Kelvin by adding 273.15: \(-23°C + 273.15 = 250.15 K\) Standard temperature and pressure (STP) are defined as 0°C (273.15 K) and 1 atm.

Write the combined gas law equation

The combined gas law equation is given by: \(P_1V_1/T_1 = P_2V_2/T_2\), where \(P_1\) and \(P_2\) are the initial and final pressures, \(V_1\) and \(V_2\) are the initial and final volumes, and \(T_1\) and \(T_2\) are the initial and final temperatures.

Substitute the given values into the equation and solve for the final volume

We are given the initial volume, pressure, and temperature and want to find the final volume at STP. Substitute the values into the combined gas law equation: \((0.75 \text{ atm})(32.4 \text{ mL})/ 250.15 K = (1 \text{ atm})(V_2)/ 273.15 K\) Now, solve for the final volume, \(V_2\): \(V_2 = (1 \text{ atm})(32.4 \text{ mL})(273.15 K)/ (250.15 K)(0.75 \text{ atm})\)

Calculate the final volume

Perform the calculation: \(V_2 = (1)(32.4)(273.15)/(250.15)(0.75)\) \(V_2 = 38.56 \text{ mL}\)

State the final answer

The gaseous sample will occupy a volume of 38.56 mL at standard temperature and pressure (STP).

Key concepts

Gas Law Calculations

Understanding gas law calculations is pivotal for students tackling chemistry problems related to the behavior of gases. The combined gas law is particularly powerful because it allows us to predict how a gas's volume, pressure, and temperature will change in relation to each other when no moles of gas are added or removed from the system.

The law is an amalgamation of three primary gas laws: Boyle's Law, which shows the inverse relationship between pressure and volume; Charles's Law, which demonstrates the direct relationship between volume and temperature; and Gay-Lussac's Law, highlighting the direct relationship between pressure and temperature.

The combined gas law is expressed mathematically as: \(P_1V_1/T_1 = P_2V_2/T_2\) where subscript 1 denotes initial conditions, and subscript 2 denotes final conditions. To solve for the unknown variable, you need to ensure all other variables are consistent in units and plugged into the equation correctly. It's also essential to use absolute temperature (Kelvin) to avoid negative temperature values, which would invalidate the ratio.

The calculation steps typically involve:

• Identifying the known and unknown variables
• Converting all units to the correct SI units (Kelvin for temperature, atmospheres for pressure, liters or milliliters for volume)
• Rearranging the equation to solve for the unknown
• Performing the arithmetic operation

Standard Temperature and Pressure (STP)

Grasping the concept of standard temperature and pressure (STP) is crucial, as STP serves as a common reference point in many chemistry problems. STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm. These conditions are a baseline that scientists and engineers use to compare different gases under a normalised set of conditions.

When working gas law problems, converting to STP allows us to predict the behavior of a gas as if it were under these standard conditions. This is particularly useful in chemical reactions and when comparing the properties of different gases.

For practice, let's apply it to a hypothetical scenario:

• If a gas occupies 50 mL at 25°C (298 K) and 2 atm, what volume would it occupy at STP?
• Using the combined gas law \(P_1V_1/T_1 = P_2V_2/T_2\), we'd set \(P_2\) to 1 atm and \(T_2\) to 273 K, and solve for \(V_2\).

It's also vital for students to remember that STP is different from NTP (Normal Temperature and Pressure), which is defined as 20°C (293.15 K) and 1 atm. Confusing these two can lead to errors in calculating gas volumes.

Unit Conversion in Chemistry

Unit conversion in chemistry is a fundamental skill that enables one to seamlessly shift between different measurement systems. A thorough understanding is needed because many scientific calculations require units to be consistent to make meaningful conclusions.

In the context of gas law calculations, you often need to convert temperatures to Kelvin, volumes to liters or milliliters, and pressure to atmospheres or another consistent unit. A common mistake is to neglect conversion to absolute temperature (Kelvin), which is crucial since temperature scales like Celsius can have negative values, affecting the ratios in gas law equations.

Here's a quick guide on conversion:

• Temperature: To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature.
• Volume: If you have volume in milliliters and need to convert to liters, divide by 1000 (since 1 L = 1000 mL).
• Pressure: Pressure values might need conversion between units like atmospheres (atm), pascals (Pa), torrs, or millimeters of mercury (mmHg). For example, 1 atm is equivalent to 101.325 kPa or 760 mmHg.

Accurate unit conversion is vital, not only to get the correct answers but also to develop a better conceptual understanding of the quantities involved in chemical equations.