Ch.5 (Gases): Additional Practice
1) Use the ideal gas law to show the mathematical relationship between volume and temperature. Clearly show your work and what stays constant!
PV = nRT V1/T1 = nR/P = k (constant)so: V1/T1 = k
PV = nRT V2/T2 = nR/P = k (constant)so: V2/T2 = k
V1/T1 = k = V2/T2soV1/T1 = V2/T2
2) Use your answer from #1 to find the volume of a gas at 25°C if it was initially at 10.°C and had a volume of 4.38L. Assume constant P and n.
*Put all T in K!
4.38L/283K = V2/298K V2 = 4.61L
3) A gas has large values for a and b in the Van der Waals equation.
a) Does this gas exhibit more ideal or real behavior? Why?
The gas exhibits more real behavior (and less ideal behavior). This is because according to the Van der Waals equation, the larger the correction factors are (a & b), the more the pressure and volume will need to be corrected from the ideal behavior. Specifically, a large a value means there must be more attraction between molecules (previously ignored by the ideal gas law), which results in lower pressure. A large b value means there must be more volume taken up by the gas particles (previously ignored by the ideal gas law), which results in lower volume.
b) What is a possible identity for this gas?
Many identities would work here, as long as the gas particles are larger and are polar (which we know must be true from the large a and be correction factors). Examples of gases that make sense: NH3, SO2, and SF4.
You DO NOT want to put down small, nonpolar gases as these would exhibit more ideal behavior and have smaller correction factors (ex: He, H2, Ne).
c) How could the behavior of the gas be made to exhibit more ideal behavior?
We know that gases tend to approach ideal behavior at lower pressures and higher temperatures. Therefore, gases that behave as real gases can transition to more ideal behavior by lowering the pressure (which can be done by increasing the volume) and increasing the temperature.
4) Consider the following information (both gases are at a pressure of 1.00atm):
Flask A: O2 at 273K
Flask B: H2 at 298K
a) Which gas would have the greatest average KE? Why?
KE = 3/2RT Since KE depends only on T, the flask with the gas at a higher temperature would have the greater average KE. This is H2 in flask B.
b) Which gas would have the greatest average velocity? Why?
urms = √(3RT/mmolar) Since velocity depends on both T and the molar mass, both must be taken into account. Gases that have smaller molar masses and are at higher temperatures will have greater velocities. Since H2 has a smaller molar mass and is at a higher temperature, this is the gas that would have the greater average velocity.
c) Which gas has the greatest density? Why?
D = Pmmolar/RT Since the density depends on P, molar mass, and T, all need to be taken into account. P is constant, so you don’t need to worry about this effecting the densities of the gases. Thus, you need to focus on the portion mmolar/T since these are different for each gas. Since the molar mass is larger for oxygen and the temperature is lower, this would give you a larger density (the quotient of mmolar/T will be larger than for hydrogen).
5) Consider the following reaction: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
What volume of CO2 will be produced if 7.09g CH4 fully reacts with excess oxygen at 32°C and 725 torr?
7.09g CH4/1 x 1molCH4/16.05gCH4 x 1molCO2/1molCH4 = 0.442mol CO2
PV = nRT solve for V: V = nRT/P
V = [(0.442mol)(62.36Ltorr/molK)(305K)]/725torr
V = 11.6L
*Note: Since the conditions ARE NOT at STP you cannot use 22.4L = 1mol here to solve more quickly for volume! You must use the ideal gas law!
6) (#75 from book- pg. 222) Hydrogen azide, HN3, decomposes on heating by the following unbalanced reaction: HN3 (g) N2 (g) + H2 (g)
If 3.0atm of pure NH3 is decomposed initially, what is the final total pressure in the reaction container? What are the partial pressures of nitrogen and hydrogen gas? Assume the volume and temperature of the reaction container are constant.
Balanced Equation: 2HN3 (g) 3N2 (g) + H2 (g)
Since only moles affect P here (V & T are constant), you can make the following stoichiometric relationship: 2mol gas = 3atm.
To find partial pressures, multiply the moles of each gas of the products by the ratio formed above:
3molN2/1 x 3atm/2mol = 4.5atm for N2
1molH2/1 x 3atm/2mol = 1.5atm for H2