The Ideal Gas Law Derivations: Videos & Practice Problems

The Ideal Gas Law Derivations

The ideal gas law, expressed as \( PV = nRT \), serves as a foundational equation in understanding the relationships between pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas. When faced with problems involving two sets of values for these variables, we can rearrange the ideal gas law to derive new equations that facilitate calculations. In scenarios where you have two pressures and two temperatures, or two volumes with two moles, the ideal gas law can be adapted to compare the initial and final states of a gas. This is particularly useful in applications such as gas expansion or compression, where you might need to find an unknown variable while knowing the others. For example, if you have an initial state represented as \( P_1, V_1, n_1, T_1 \) and a final state as \( P_2, V_2, n_2, T_2 \), you can set up the relationship as follows: \[ \frac{P_1 V_1}{n_1 T_1} = \frac{P_2 V_2}{n_2 T_2} \] This equation allows you to solve for any unknown variable, provided you have the necessary information about the other variables. By understanding how to manipulate the ideal gas law in this way, you can effectively tackle a variety of problems involving gases under different conditions.

The Ideal Gas Law Derivations Example

To solve the problem of how temperature affects the volume of a gas, we can utilize the ideal gas law, which is expressed as PV = nRT. In this scenario, we are given two volumes and one temperature, indicating that we need to derive a new formula based on the ideal gas law.

Since the pressure remains constant, we can focus on the relationship between volume and temperature. The relevant variables are the initial volume V_1 and temperature T_1, as well as the final volume V_2 and the unknown final temperature T_2. The derived formula can be expressed as:

\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\)

In this case, we have:

• V_1 = 8.30 \, \(\text{liters}\)
• V_2 = 5.25 \, \(\text{liters}\)
• T_1 = 202 \, \(\text{°C}\) = 202 + 273.15 = 475.15 \, \(\text{K}\)
• T_2 = ?

Next, we substitute the known values into the derived formula:

\(\frac{8.30}{475.15}\) = \(\frac{5.25}{T_2}\)

Cross-multiplying gives us:

8.30 \(\cdot\) T_2 = 475.15 \(\cdot\) 5.25

Solving for T_2 involves dividing both sides by 8.30:

T_2 = \(\frac{475.15 \cdot 5.25}{8.30}\) \(\approx\) 300.55 \, \(\text{K}\)

Finally, to convert T_2 back to degrees Celsius, we subtract 273.15:

T_2 \(\approx\) 300.55 - 273.15 \(\approx\) 27.40 \, \(\text{°C}\)

Thus, the temperature needed to decrease the volume of sulfur hexachloride gas to 5.25 liters is approximately 27.40 degrees Celsius.

A sample of nitrogen dioxide gas at 130 ºC and 315 torr occupies a volume of 500 mL. What will the gas pressure be if the volume is reduced to 320 mL at 130 ºC?

• A. 0.65 atm
• B. 0.83 atm
• C. 0.92 atm
• D. 1.6 atm
• E. 2.7 atm

Answer: (A)

A cylinder with a movable piston contains 0.615 moles of gas and has a volume of 295 mL. What will its volume be if 0.103 moles of gas escaped?

• A. 0.176 L
• B. 0.217 L
• C. 0.246 L
• D. 0.361 L
• E. 1.28 L

Answer: (C)

On most spray cans it is advised to never expose them to fire. A spray can is used until all that remains is the propellant gas, which has a pressure of 1350 torr at 25 ºC. If the can is then thrown into a fire at 455 ºC, what will be the pressure (in torr) in the can?
a) 750 torr
b) 1800 torr
c) 2190 torr
d) 2850 torr
e) 3300 torr

• A. 750 torr
• B. 1800 torr
• C. 2190 torr
• D. 2850 torr
• E. 3300 torr

Answer: (E)

Here's what students ask on this topic:

What is the ideal gas law and how is it used to derive other gas law equations?

The ideal gas law is a fundamental equation in chemistry that relates pressure (P), volume (V), moles of gas (n), and temperature (T) through the equation $PV=nRT$, where R is the ideal gas constant. This law can be rearranged to derive other gas law equations when dealing with two sets of conditions, such as initial and final states of a gas. By holding some variables constant and comparing the changes in others, we can form relationships like $P_1{V}_1=P_2{V}_2$ (Boyle's Law) or $V_1/T_1=V_2/T_2$ (Charles's Law). These derivations help solve problems where pressure, volume, temperature, or moles change between two states.

How do you handle temperature in ideal gas law derivations when it changes between two states?

When temperature changes between two states in ideal gas law derivations, it is crucial to convert all temperature values to the Kelvin scale before performing any calculations. This is because the ideal gas law requires an absolute temperature scale to maintain proportionality. For example, if you have initial and final temperatures in Celsius, convert them using $T=t+273.15$, where $t$ is the temperature in Celsius. After conversion, you can use the derived relationships such as $P_1/T_1=P_2/T_2$ to solve for unknown variables. Always remember that failing to convert to Kelvin will lead to incorrect results.

What are the common forms of ideal gas law derivations used in problems with changing variables?

Common forms of ideal gas law derivations arise when comparing two states of a gas with changing variables. These include: $P_1{V}_1=P_2{V}_2$ (Boyle's Law, constant n and T), $V_1/T_1=V_2/T_2$ (Charles's Law, constant n and P), $P_1/T_1=P_2/T_2$ (Gay-Lussac's Law, constant n and V), and $V_1/n_1=V_2/n_2$ (Avogadro's Law, constant P and T). These forms are derived by canceling constant terms and isolating the variables that change between the two states.

How do you solve an ideal gas law problem involving two different pressures and temperatures?

To solve an ideal gas law problem with two different pressures and temperatures, start by identifying the variables that change and those that remain constant. If volume and moles are constant, use the derived relationship $P_1/T_1=P_2/T_2$. Ensure temperatures are in Kelvin. Rearrange the equation to solve for the unknown variable, for example, $P_2=P_1*\; (T_2/T_1)$. Plug in the known values and calculate the unknown. This approach simplifies the problem by focusing only on the changing variables and using the ideal gas law derivations.

Why is it important to identify constant variables when using ideal gas law derivations?

Identifying constant variables is essential in ideal gas law derivations because it allows you to simplify the equation and focus on the variables that change between two states. The ideal gas law $PV=nRT$ involves four variables, but in many problems, some remain constant. By canceling these constants, you derive simpler relationships like $P_1{V}_1=P_2{V}_2$ or $V_1/T_1=V_2/T_2$. This makes calculations manageable and accurate. Without identifying constants, you might incorrectly apply the law, leading to errors in solving gas problems.