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), moles of gas (n), and temperature (T) through the equation $PV=nRT$, where R is the gas constant. This law can be rearranged algebraically to solve for any one variable if the others are known. For example, to find volume, you can rearrange it as $V=\frac{\mathrm{nRT}}{P}$. Rearranging is especially useful when dealing with problems involving changes in conditions, such as different pressures or temperatures, allowing you to isolate the variables that change while treating others as constant.

How do you apply the ideal gas law derivations when given two sets of different pressures and temperatures?

When a problem provides two different pressures and temperatures, you use the ideal gas law derivations to relate the initial and final states of the gas. Since the amount of gas and volume might be constant or known, you can use the combined gas law form derived from the ideal gas law: $\frac{\mathrm{P\_1}}{\mathrm{T\_1}}=\frac{\mathrm{P\_2}}{\mathrm{T\_2}}$. This equation allows you to solve for an unknown pressure or temperature by substituting the known values. Remember to convert all temperatures to Kelvin before using the equation to ensure accuracy.

What are the common relationships derived from the ideal gas law for two different states of a gas?

Common relationships derived from the ideal gas law for two states of a gas include equations that relate variables when others are constant. These include: $\frac{\mathrm{V\_1}}{\mathrm{T\_1}}=\frac{\mathrm{V\_2}}{\mathrm{T\_2}}$ (volume and temperature), $\mathrm{P\_1}\mathrm{V\_1}=\mathrm{P\_2}\mathrm{V\_2}$ (pressure and volume), $\frac{\mathrm{P\_1}}{\mathrm{T\_1}}=\frac{\mathrm{P\_2}}{\mathrm{T\_2}}$ (pressure and temperature), and $\frac{\mathrm{V\_1}}{\mathrm{n\_1}}=\frac{\mathrm{V\_2}}{\mathrm{n\_2}}$ (volume and moles). These formulas help solve problems where two variables change between states.

Why must temperature be converted to Kelvin when using the ideal gas law and its derivations?

Temperature must be converted to Kelvin when using the ideal gas law because the Kelvin scale is an absolute temperature scale starting at absolute zero, where molecular motion theoretically stops. The ideal gas law depends on absolute temperature to correctly relate pressure, volume, and moles. Using Celsius or Fahrenheit would lead to incorrect results because these scales do not start at zero molecular motion. Therefore, always convert temperatures to Kelvin by adding 273.15 to the Celsius value before performing calculations.

How do you solve a problem involving two volumes and two mole amounts using the ideal gas law derivations?

When a problem involves two volumes and two mole amounts, you can use the relationship derived from the ideal gas law that relates volume and moles at constant pressure and temperature: $\frac{\mathrm{V\_1}}{\mathrm{n\_1}}=\frac{\mathrm{V\_2}}{\mathrm{n\_2}}$. This equation allows you to find an unknown volume or mole amount by substituting the known values. It assumes that pressure and temperature remain constant during the change. This derivation is useful for understanding how volume changes with the amount of gas.