The Ideal Gas Law: Videos & Practice Problems

The ideal gas law, represented by the equation ${\mathrm{PV}}^{=}\mathrm{nRT}$, relates pressure (P), volume (V), moles (n), and temperature (T) of a gas. The gas constant (R) can be 0.08206 L·atm/(mol·K) for the ideal gas law or 8.314 J/(mol·K) for energy-related calculations. Understanding these constants and their applications is crucial for solving problems involving gas behavior, thermodynamics, and kinetic molecular theory.

The Ideal Gas Law

In the study of gas behavior, the ideal gas law serves as a fundamental equation that describes how gases respond to changes in pressure, volume, moles, and temperature. The ideal gas law is expressed by the formula:

\( PV = nRT \)

In this equation, \( P \) represents pressure, measured in atmospheres (atm), \( V \) denotes volume in liters (L), \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, typically valued at \( 0.08206 \, \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol}) \), and \( T \) signifies temperature in Kelvins (K). Understanding this relationship is crucial, as it underpins various calculations and concepts related to gas behavior throughout the chapter.

To effectively utilize the ideal gas law, one must be familiar with the units associated with each variable, ensuring accurate application in problem-solving scenarios. Mastery of this formula will be essential for tackling a range of questions and practical applications in the study of gases.

The Ideal Gas Law Example 1

To apply the ideal gas law, represented by the equation PV = nRT, it is essential to ensure that all variables are expressed in the correct units. In this scenario, we are working with a 500 milliliter container of nitrogen gas at a pressure of 600 millimeters of mercury and a temperature of 50 degrees Celsius.

First, we need to convert the pressure from millimeters of mercury (mmHg) to atmospheres (atm). The conversion factor is that 1 atmosphere equals 760 mmHg. Therefore, the pressure can be calculated as follows:

P = \frac{600 \, \text{mmHg}}{760 \, \text{mmHg/atm}} \approx 0.789 \, \text{atm}

Next, we convert the volume from milliliters to liters. Since 1 milliliter equals 0.001 liters, we have:

V = 500 \, \text{mL} = 0.500 \, \text{L}

For the number of moles (n) of nitrogen gas (N₂), we start with the mass given, which is 29.3 grams. The molar mass of nitrogen gas is calculated as follows:

M(N_2) = 2 \times 14.01 \, \text{g/mol} = 28.02 \, \text{g/mol}

Now, we can convert grams to moles:

n = \frac{29.3 \, \text{g}}{28.02 \, \text{g/mol}} \approx 1.05 \, \text{mol}

Finally, we need to convert the temperature from Celsius to Kelvin. The conversion is done by adding 273.15:

T = 50 \, \text{°C} + 273.15 = 323.15 \, \text{K}

In summary, the values needed for the ideal gas law are:

• Pressure (P): 0.789 atm
• Volume (V): 0.500 L
• Number of moles (n): 1.05 mol
• Temperature (T): 323.15 K

By ensuring that all variables are in the correct units, we can accurately apply the ideal gas law to solve for any unknowns in gas-related problems.

The Ideal Gas Law

The ideal gas law incorporates the gas constant, denoted as R, which can take on two distinct values depending on the context of the problem. The first value, R = 0.08206 L·atm/(mol·K), is commonly used when applying the ideal gas law. This equation relates the pressure, volume, temperature, and number of moles of an ideal gas.

In contrast, when dealing with scenarios involving speed, velocity, or energy, the gas constant is represented as R = 8.314 J/(mol·K). This value is particularly relevant in calculations involving the kinetic energy of particles or the energy changes in chemical reactions.

To understand the relationship between these two values, it is essential to recognize the conversion factor: 1 L·atm = 101.325 J. This means that when converting from liters and atmospheres to joules, the units of liters times atmospheres can be canceled out, allowing for a straightforward transition to joules. When performing this conversion, the value of R can be expressed as 8.3147295 J/(mol·K), but for simplicity, it is often rounded to 8.314 J/(mol·K).

In summary, the gas constant R is crucial in thermodynamic calculations, with its value varying based on the specific application—either 0.08206 for the ideal gas law or 8.314 for energy-related calculations.

The Ideal Gas Law Example 2

To determine the number of moles of ammonia in a 25-liter tank at 190 degrees Celsius and 5.20 atmospheres, we can utilize the ideal gas law, represented by the equation:

PV = nRT

In this equation, P stands for pressure, V for volume, n for the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Our goal is to isolate the variable for moles (n), which can be expressed as:

n = \frac{PV}{RT}

Given the values:

• P = 5.20 atm
• V = 25 L
• R = 0.08206 L·atm/(mol·K)
• T = 190 °C = 190 + 273.15 = 463.15 K

Now, substituting these values into the equation:

n = \frac{(5.20 \, \text{atm})(25 \, \text{L})}{(0.08206 \, \text{L·atm/(mol·K)})(463.15 \, \text{K})}

Calculating the numerator:

5.20 \times 25 = 130 \, \text{atm·L}

Calculating the denominator:

0.08206 \times 463.15 \approx 38.058 \, \text{L·atm/(mol·K)}

Now, dividing the two results:

n \approx \frac{130}{38.058} \approx 3.4205 \, \text{moles}

Considering significant figures, the final answer should reflect the least number of significant figures from the given data, which is 2. Therefore, the number of moles of ammonia is:

3.4 moles

This process illustrates the importance of identifying the variables provided, isolating the desired variable using algebra, and ensuring unit consistency throughout the calculations.

How many grams of carbon dioxide, CO₂, are present in a 0.150 L flask recorded at 525 mmHg and 32 ºC?
a) 1.77 g
b) 0.93 g
c) 0.66 g
d) 0.18 g
e) 0.052 g

• A. 1.77 g
• B. 0.93 g
• C. 0.66 g
• D. 0.18 g
• E. 0.052 g

Answer: (D)

How many liters of HNO₃ gas, measured at 28.0 ºC and 780 torr, are required to prepare 2.30 L of 4.15 M solution of nitric acid?

• A. 56 L
• B. 62 L
• C. 189 L
• D. 230 L
• E. 262 L

Answer: (D)

When 0.670 g argon is added to a 500 cm³ container with a sample of oxygen gas, the total pressure of the gases is found to be 1.52 atm at a temperature of 340 K. What is the mass of the oxygen gas in the bulb?

• A. 0.266 g
• B. 0.335 g
• C. 0.621 g
• D. 0.715 g
• E. 1.72 g

Answer: (B)

Here’s what students ask on this topic:

What is the ideal gas law and its formula?

The ideal gas law is a fundamental equation in chemistry that describes the behavior of an ideal gas. The formula is given by:

$PV=nRT$

where $P$ is the pressure in atmospheres, $V$ is the volume in liters, $n$ is the number of moles of gas, $R$ is the gas constant (0.08206 L·atm/(mol·K)), and $T$ is the temperature in Kelvins. This law helps predict how gases will respond to changes in pressure, volume, and temperature.

What are the units for the gas constant R in the ideal gas law?

In the ideal gas law, the gas constant $R$ is typically given as 0.08206 L·atm/(mol·K). This means that the units for $R$ are liters times atmospheres per mole per Kelvin. These units ensure that when you use the ideal gas law formula $PV=nRT$, all the units cancel out appropriately, giving you a consistent and accurate result.

When should you use the gas constant 8.314 instead of 0.08206?

The gas constant $R$ can take on different values depending on the context of the problem. Use 0.08206 L·atm/(mol·K) when dealing with the ideal gas law, which involves pressure, volume, and temperature of gases. However, use 8.314 J/(mol·K) when dealing with problems related to energy, speed, or velocity, such as in thermodynamics or kinetic molecular theory. The conversion factor between these two values is that 1 L·atm equals 101.325 J.

How do you convert temperature to Kelvins for the ideal gas law?

To convert temperature from Celsius to Kelvins, you use the formula:

$T=T{C}_{}+273.15$

where $T$ is the temperature in Kelvins and $T{C}_{}$ is the temperature in Celsius. For example, if the temperature is 25°C, the temperature in Kelvins would be:

$25+273.15=298.15$

Kelvins. This conversion is essential for using the ideal gas law correctly.

What is the significance of the ideal gas law in chemistry?

The ideal gas law is significant in chemistry because it provides a simple yet powerful equation to predict the behavior of gases under various conditions. It combines several gas laws (Boyle's, Charles's, and Avogadro's laws) into one comprehensive formula:

$PV=nRT$

This law helps chemists understand and calculate how gases will respond to changes in pressure, volume, temperature, and the number of moles. It is crucial for various applications, including chemical reactions, industrial processes, and environmental science.