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9.3: Two State Ideal Gas Problems

  • Page ID
    284093
  • Two State Approach

    If you have a gaseous system and then perturb it by changing one of more variables the other variables adjust to that change. The two state approach is a quick way to solve these problems. The idea is simple, you have a relationship PV=nRT, so at one state, P1V1 =n1RT1 ​​​​and P2V2=n2RT2 ​​​, so we can set up the equivalence:

    \[ \frac{P_{1}V_{1}}{P_{2}V_{2}}=\frac{n_{1}RT_{1}}{n_{2}RT_{2}}=\frac{n_{1}T_{1}}{n_{2}T_{2}}\]

    so

    \[\frac{P_{1}V_{1}}{P_{2}V_{2}}= \frac{n_{1}T_{1}}{n_{2}T_{2}} \]

    (note: the constant (R) cancels out)

    Warning: T must be in Kelvin, as the equation PV=nRT requires T to be in Kelvin, and even though R cancels in the two state approach, T still needs to be in Kelvin.

    When solving these problems it is wise to make a table and write down what you are given. If something is held constant, it cancels out of the equation.

    State 1 State 2
    T1= T2=
    P1= P2=
    V1= V2=
    n1= n2=

    Example \(\PageIndex{3}\)

    What is the volume of a perfectly elastic balloon if it rises to an altitude where the pressure drops to 0.100 atm and the temperature drops to 173 K if it had a volume of 5.00 liters at sea level where the pressure was 1.00 atm and the temperature was 273K?

    Solution

    (a) Set up two state equation and solve for V1 or V2 (which either is easier, as we have not assigned the data to the states

    \[\frac{P_{1}V_{1}}{P_{2}V_{2}}= \frac{n_{1}T_{1}}{n_{2}T_{2}} \\ \; \\ becomes \\ \; \\ V_1=V_2\left ( \frac{n_1}{n_2} \right )\left ( \frac{T_1}{T_2} \right )\left ( \frac{P_2}{P_1} \right )\]

    (b) Fill our the data table so we properly identify the data with the state.

    TIP, Since we want the volume in the upper atmosphere we will call that state 1, but we could have called sea level state one, in which case we would solve for V2.

    State 1 State 2
    T1= 173K T2= 273K
    P1= 0.100 atm P2= 1.00 atm
    V1= ? V2= 5.00 L
    n1= n2=

    NOTE: This is a closed system and no gas entered or escaped from the balloon and so n1=n2 and they cancel.

    \[V_1=5.00L\cancel{ \left ( \frac{n_1}{n_2} \right )} \underbrace{ \left ( \frac{173 K}{273 K} \right )}_{\color{red}{T \; must \; be \\ in \; Kelvin}} \left ( \frac{1 atm}{0.100 atm} \right )\]

    The solution to this problem is demonstrated in video \(\PageIndex{4}\).

    Video \(\PageIndex{4}\): 5'10" YouTube solving this problem (https://youtu.be/BCYEZv9IQFA)

    Note

    Many textbooks cover the "combined gas law":
    \[\frac{P_1V_2}{T_1}=\frac{P_2V_2}{T_2}\]
    On closer inspection the combined gas law is applying the two-state approach to the ideal gas law where n, the number of molecules is constant, and that was the case we just solved (example 10.3.2). The advantage of the two state approach is that it not only allows us to handle changes in the number of particles, but it is a technique that can be applied to many equations, not just the ideal gas equation.

    If you have a relationship y=mx, you can always say that \(\frac{y_1}{y_2}=\frac{x_2}{x_2}\), and during the second semester this technique will be applied to power functions (Y=AXm) and exponential functions Y=Aexm. It is worth mastering the two state approach now.

     

    Test Yourself 

    Homework: Section 9.3

    Graded Assignment: Section 9.3

    Exercise \(\PageIndex{2}\)

    Solve the following problems using the two state approach.

    1. What pressure would 6.1 moles of a gas in a rigid container have at 20.0 oC have if it had a pressure of 0.45 atm at -45.0 oC?
    2. A pressure tank containing chlorine gas at a pressure of 2.00 atm and a temperature of 40oC is set with a pressure relief valve set to open at a pressure of 10.0 atm. At what temperature will the relief valve open?
    3. A gas is in a sealed cylinder at 25oC with a piston which can expand or contract to change the volume. The initial volume and pressure are 75.0 L and 980.0 torr. A force is applied to the piston and the volume adjusts to a final pressure and temperature of 5.00 atm at 188oC. What is the final volume?
    Answer a

    \[P_{1}=P_{2}\frac{T_{1}}{T_{2}}=\left ( 0.45atm \right )\left ( \frac{293.15K}{228.15K} \right )=0.58atm\]

    Answer b

    \[T_{2}=T_{1}\frac{P_{2}}{P_{1}}= \left ( 313.15K \right )\left ( \frac{10.0atm}{2.00atm} \right )=1570K\]

    Answer c

    \[V_{2}=V_{1}\left ( \frac{P_{1}}{P_{2}} \right )\left ( \frac{T_{2}}{T_{1}} \right )=75.0L\left ( \frac{980torr\left ( \frac{1atm}{760torr} \right )}{5.00atm} \right )\left ( \frac{461.15K}{298K} \right )=29.9L\]

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