Question

A sample of helium gas has a volume of \(6.50 \mathrm{~L}\) at a pressure of \(845 \mathrm{mmHg}\) and a temperature of \(25^{\circ} \mathrm{C}\). What is the pressure of the gas, in atmospheres, when the volume and temperature of the gas sample are changed to the following, if the amount of gas is constant? a. \(1850 \mathrm{~mL}\) and \(325 \mathrm{~K}\) b. \(2.25 \mathrm{~L}\) and \(12^{\circ} \mathrm{C}\) c. \(12.8 \mathrm{~L}\) and \(47^{\circ} \mathrm{C}\)

Short answer

a) 4.28 atm, b) 3.62 atm, c) 0.50 atm

Step-by-step solution

Understand the given data and formulas

The problem involves changes in volume, pressure, and temperature of a gas sample with a constant amount of gas. Use the Combined Gas Law: \[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]where: \( P_1 \) is the initial pressure, \( V_1 \) is the initial volume, \( T_1 \) is the initial temperature, \( P_2 \) is the final pressure, \( V_2 \) is the final volume, and \( T_2 \) is the final temperature.

Convert given units

1. Convert volumes into liters (if needed). 2. Convert all temperatures to Kelvin by using: \[ T(K) = T(^{\circ}C) + 273.15 \]3. Convert pressure from \( mmHg \) to \( atm \) using: \[ 1 \text{ atm} = 760 \text{ mmHg} \]

Convert initial values

Given initial values: \[ P_1 = 845 \text{ mmHg} \to \frac{845}{760} \text{ atm} \approx 1.112 \text{ atm} \]\[ V_1 = 6.50 \text{ L} \]\[ T_1 = 25 ^{\circ}C \to 298.15 \text{ K} \]

Solve for case (a)

Convert given values: \[ V_2 = 1850 \text{ mL} = 1.850 \text{ L} \]\[ T_2 = 325 \text{ K} \]Apply combined gas law: \[ P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{1.112 \times 6.50 \times 325}{298.15 \times 1.850} \approx 4.28 \text{ atm} \]

Solve for case (b)

Convert given values: \[ V_2 = 2.25 \text{ L} \]\[ T_2 = 12 ^{\circ}C \to 285.15 \text{ K} \]Apply combined gas law: \[ P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{1.112 \times 6.50 \times 285.15}{298.15 \times 2.25} \approx 3.62 \text{ atm} \]

Solve for case (c)

Convert given values: \[ V_2 = 12.8 \text{ L} \]\[ T_2 = 47 ^{\circ}C \to 320.15 \text{ K} \]Apply combined gas law: \[ P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{1.112 \times 6.50 \times 320.15}{298.15 \times 12.8} \approx 0.50 \text{ atm} \]

Key concepts

Gas Laws

Gas laws describe how gases behave and interact with temperature, pressure, and volume.
These laws are fundamental to understanding gas dynamics.
Some major gas laws include Boyle's Law (pressure and volume), Charles's Law (volume and temperature), and Gay-Lussac's Law (pressure and temperature).
The Combined Gas Law merges these individual laws into a single equation, which is useful because gases often experience changes in more than one condition at a time.
It states that \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\), where \(P\) is pressure, \(V\) is volume, and \(T\) is temperature in Kelvin.
This law helps us predict how a gas will respond to changes in its environment, provided the amount of gas remains constant.

Pressure Conversion

Pressure is an important factor in gas laws.
It is essential to convert between different pressure units to apply these laws correctly.
Common pressure units include atmospheres (atm), millimeters of mercury (mmHg), and Pascals (Pa).
To convert \(mmHg\) to \(atm\), you can use the conversion factor: \(\text{1 atm} = 760 \text{ mmHg}\).
For example, a pressure of \(845 \text{ mmHg}\) can be converted to atmospheres by dividing by \(\text{760 mmHg}\):
\(\frac{845}{760} \text{ atm} \approx 1.112 \text{ atm}\).
This step is crucial when inputting values into the Combined Gas Law equation.

Volume Conversion

Volume changes are often involved in gas law problems.
The volume of a gas is typically measured in liters (L), but it can also be found in milliliters (mL).
To convert milliliters to liters, remember that \(\text{1000 mL} = 1 \text{ L}\).
For instance, if you have a volume of \(1850 \text{ mL}\), you can convert this to liters by dividing by \(1000\):
\( \frac{1850}{1000} \text{ L} = 1.850 \text{ L} \).
Accurate volume conversion ensures correct calculations in equations.

Temperature Conversion

Temperature must always be in Kelvin when using gas laws.
This is because gas law equations are derived based on the absolute temperature scale.
The conversion from Celsius (\(^\text{ \circ \ }C\)) to Kelvin (K) is straightforward: \(\text{T(K)} = \text{T}^{\text{ \circ \ }C} + 273.15\).
For example, a temperature of \(25^\text{ \circ \ }C\)) converts to Kelvin as follows:
\( 25^{\text{ \circ \ }C} + 273.15 = 298.15 \text{ K} \).
This conversion ensures that the temperature unit is consistent when applying the Combined Gas Law.