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.
11 Gases
The Ideal Gas Law Derivations
11 Gases
The Ideal Gas Law Derivations: Videos & Practice Problems
Topic summary
The Ideal Gas Law Derivations come from rearranging the ideal gas law, \(PV=nRT\) , when a problem involves two different values for the same variable. This usually means there are two pressures, two volumes, two mole amounts, or two temperatures, so a new relationship must be derived from the original equation rather than using it in its basic form.
The process is to identify which variables change, cross out the quantities that stay constant, and algebraically rearrange the equation so only the changing variables remain. This leads to common forms such as \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\) , \(P_1V_1=P_2V_2\) , and \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\) . A key rule is that temperature must be converted to Kelvin for calculations.
The Ideal Gas Law Derivations are a convenient way to solve gas calculations involving 2 sets of the same variables.
Ideal Gas Law Derivations
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Concept
The Ideal Gas Law Derivations
Video duration:
47sThe Ideal Gas Law Derivations Video Summary
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Example
The Ideal Gas Law Derivations Example
Video duration:
4mThe Ideal Gas Law Derivations Example Video Summary
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.
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Problem
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
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Problem
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
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Problem
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
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When a problem involves two different sets of variables such as pressures, volumes, moles, or temperatures, we derive new equations from the Ideal Gas Law by first identifying which variables change and which remain constant. The original law is . We then algebraically rearrange it to isolate only the changing variables. For example, if the amount of gas and the gas constant remain constant, and we have two states, the equation can be written as . This method allows us to solve problems involving changes in conditions without using the full ideal gas law each time.
In Ideal Gas Law derivations, temperature must always be expressed in Kelvin to ensure accuracy. When dealing with two different states, temperatures and must be converted from Celsius or Fahrenheit to Kelvin by adding 273.15. This is crucial because the gas law is based on absolute temperature. For example, if you are using the relationship , both temperatures must be in Kelvin to correctly relate the volumes and temperatures of the two states.
Common forms of the Ideal Gas Law derivations arise when certain variables remain constant, allowing simplification. For example, if the amount of gas and temperature are constant, Boyle's Law applies: . If pressure and amount are constant, Charles's Law applies: . If volume and amount are constant, Gay-Lussac's Law applies: . These forms help solve problems involving two states without using the full ideal gas law.
Crossing out constant variables when deriving equations from the Ideal Gas Law simplifies the problem by focusing only on the variables that change between two states. Since the ideal gas law is , if certain variables like the amount of gas or the gas constant remain unchanged, they can be canceled out on both sides of the equation. This leads to simpler relationships such as when temperature and moles are constant. This approach makes calculations more straightforward and reduces the chance of errors.
When solving problems involving two different pressures and temperatures, you use the derived form of the Ideal Gas Law that relates these variables while keeping other variables constant. For example, if volume and moles are constant, the relationship is . To solve, convert temperatures to Kelvin, plug in known values for , , and , then solve for the unknown pressure . This method efficiently handles changes in pressure and temperature without needing the full ideal gas law.
