Question

unit 1 summative: matter in motion
1.1 i can explain everyday examples that illustrate the four laws of thermodynamics.
1.2 i can describe how gas particles move and interact in terms of pressure, volume, temperature, and number of moles.
1.3 i can calculate the relationship between pressure, volume, temperature, and number of moles using the ideal gas law pv = nrt
1.4 i can define and apply dalton’s law of partial pressure.
when you inflate a basketball with an air pump, you are adding a mixture of gases, including nitrogen, oxygen, and carbon dioxide. your total air pressure is 0.543 atm, with nitrogen having a partial pressure of 0.424 atm and carbon dioxide having a partial pressure of 0.00592 atm. what is the partial pressure of oxygen in the basketball?
using the pressure of oxygen from above, use the ideal gas law to determine the volume of the basketball. the basketball has a temperature of 27°c and 2.6 moles of gas.
you leave your basketball outside on a hot summer day in texas; the temperature rises to 43.3°c. what do you predict will happen to the volume of the basketball if pressure and number of moles stays the same?
as heat was applied to the basketball, changes with volume happened within the basketball to reach equilibrium. this scenario would be an example of which law of thermodynamics? provide one justification.

Explanation:

Sub - Question 1: Partial Pressure of Oxygen

Step 1: Recall Dalton's Law of Partial Pressures

Dalton's Law states that the total pressure ($P_{total}$) of a gas mixture is the sum of the partial pressures of its individual components. So, $P_{total}=P_{N_2}+P_{O_2}+P_{CO_2}$. We need to solve for $P_{O_2}$, so we can rearrange the formula as $P_{O_2}=P_{total}-P_{N_2}-P_{CO_2}$.

Step 2: Substitute the given values

We know that $P_{total} = 0.543\ atm$, $P_{N_2}=0.424\ atm$, and $P_{CO_2}=0.00592\ atm$. Substituting these values into the formula: $P_{O_2}=0.543 - 0.424 - 0.00592$.
First, calculate $0.543-0.424 = 0.119$. Then, calculate $0.119 - 0.00592=0.11308\ atm$.

Step 1: Recall the Ideal Gas Law

The ideal gas law is $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant ($R = 0.0821\ L\cdot atm/(mol\cdot K)$), and $T$ is temperature in Kelvin. We need to solve for $V$, so we rearrange the formula to $V=\frac{nRT}{P}$.

Step 2: Convert temperature to Kelvin

The temperature is given as $27^{\circ}C$. To convert to Kelvin, we use the formula $T(K)=T(^{\circ}C)+273.15$. So, $T = 27 + 273.15=300.15\ K\approx300\ K$.

Step 3: Substitute the values into the formula

We know that $n = 2.6\ mol$, $R = 0.0821\ L\cdot atm/(mol\cdot K)$, $T = 300\ K$, and from the first sub - question, $P = P_{O_2}=0.11308\ atm$.
Substituting these values into $V=\frac{nRT}{P}$, we get $V=\frac{2.6\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times300\ K}{0.11308\ atm}$.
First, calculate the numerator: $2.6\times0.0821\times300 = 2.6\times24.63 = 64.038$.
Then, divide by the denominator: $V=\frac{64.038}{0.11308}\approx566.3\ L$.

According to Charles's Law, for a fixed amount of gas at constant pressure, the volume of a gas is directly proportional to its absolute temperature ($V\propto T$ when $n$ and $P$ are constant). The temperature of the basketball increases from $27^{\circ}C$ (or $300\ K$) to $43.3^{\circ}C$. First, convert $43.3^{\circ}C$ to Kelvin: $T = 43.3+273.15 = 316.45\ K$. Since the temperature (in Kelvin) increases and pressure ($P$) and number of moles ($n$) are constant, according to Charles's Law, the volume of the gas (and thus the basketball, as it is a flexible container) will increase.

Answer:

The partial pressure of oxygen is $0.113\ atm$ (rounded to three decimal places) or $0.11308\ atm$.

Sub - Question 2: Volume of the Basketball using Ideal Gas Law