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 can it be rearranged for different variables?

The ideal gas law is a fundamental equation in chemistry that relates pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas. It is expressed as $PV=nRT$, where R is the gas constant. This equation can be rearranged to solve for any one variable if the others are known. For example, to find volume, rearrange to $V=\frac{\mathrm{nRT}}{P}$. Rearranging is especially useful when comparing two states of a gas where some variables change. By setting the two states equal, you can derive relationships like $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ when pressure and moles are constant. This flexibility makes the ideal gas law a powerful tool in solving gas-related problems.

How do you use the ideal gas law derivations when two variables change in a problem?

When two variables change in a gas problem, the ideal gas law derivations help simplify calculations by comparing initial and final states. Start with the ideal gas law $PV=nRT$. If some variables remain constant, they can be canceled out. For example, if the amount of gas (n) and gas constant (R) stay the same, you can write $P_1{V}_1/T_1=P_2{V}_2/T_2$. This equation relates the initial and final pressures, volumes, and temperatures. By identifying which variables change and which remain constant, you can rearrange the equation to solve for the unknown. Remember to convert temperatures to Kelvin for accuracy.

What are the common derived equations from the ideal gas law for comparing two states?

Common derived equations from the ideal gas law are used to compare two states of a gas when certain variables change. These include:

• $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ (Charles's Law) when pressure and moles are constant.
• $P_1{V}_1=P_2{V}_2$ (Boyle's Law) when temperature and moles are constant.
• $\frac{V_1}{n_1}=\frac{V_2}{n_2}$ when pressure and temperature are constant.
• $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ when volume and moles are constant.

These relationships allow you to solve problems involving changes in gas conditions by focusing only on the variables that change.

Why must temperature be converted to Kelvin in ideal gas law calculations?

Temperature must be converted to Kelvin in ideal gas law calculations because the Kelvin scale is an absolute temperature scale starting at absolute zero, where molecular motion theoretically stops. The ideal gas law $PV=nRT$ requires temperature in Kelvin to maintain proportionality and accuracy. Using Celsius or Fahrenheit would lead to incorrect results because these scales do not start at absolute zero and can have negative values, which are physically meaningless in gas law calculations. Kelvin ensures that temperature values are always positive and directly proportional to the average kinetic energy of gas particles, making the calculations consistent and reliable.

How do you solve a problem involving two different pressures and temperatures using the ideal gas law derivations?

To solve a problem with two different pressures and temperatures, start with the ideal gas law $PV=nRT$. If volume and moles remain constant, you can use the derived equation $\frac{P_1}{T_1}=\frac{P_2}{T_2}$. Rearrange to solve for the unknown variable, for example, $P_2=P_1\cdot \frac{T_2}{T_1}$. Remember to convert temperatures to Kelvin before substituting values. This approach simplifies the problem by focusing on the changing variables and using the proportional relationship between pressure and temperature.