Question

1. a gas in a sealed container has a starting pressure of 100kpa and a temperature of 30°c. if the pressure increases to 150kpa, what would the new temperature be? (remember that you need to do this in kelvin: k = °c + 273)

answer:________

1. a gas in a sealed container has a starting volume of 4l and a temperature of 100°c. if the volume decreases to 2l, what would the new temperature be? (remember that you need to do this in kelvin: k = °c + 273)

answer:________

1. draw a particle view model that represents two gasses, one with a pressure of 100kpa and one with a pressure of 50kpa. label your models.
2. draw a particle view model that represents two gasses, one with a temperature of 50°c and one with a temperature of 100°c. label your models.

Explanation:

Step1: Convert initial temperature to Kelvin for question 5

For the first - gas law problem (question 5), we use Gay - Lussac's law ($\frac{P_1}{T_1}=\frac{P_2}{T_2}$). First, convert the initial temperature $T_1$ from Celsius to Kelvin. Given $T_1 = 30^{\circ}C$, then $T_1=30 + 273=303K$, $P_1 = 100kPa$, and $P_2 = 150kPa$.

Step2: Solve for $T_2$ in question 5

From $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, we can re - arrange to get $T_2=\frac{P_2T_1}{P_1}$. Substitute the values: $T_2=\frac{150\times303}{100}=454.5K$. Convert back to Celsius: $T_2=454.5 - 273 = 181.5^{\circ}C$.

Step3: Convert initial temperature to Kelvin for question 6

For the second gas - law problem (question 6), we use Charles's law ($\frac{V_1}{T_1}=\frac{V_2}{T_2}$). Convert the initial temperature $T_1$ from Celsius to Kelvin. Given $T_1 = 100^{\circ}C$, then $T_1=100 + 273 = 373K$, $V_1 = 4L$, and $V_2 = 2L$.

Step4: Solve for $T_2$ in question 6

From $\frac{V_1}{T_1}=\frac{V_2}{T_2}$, we re - arrange to get $T_2=\frac{V_2T_1}{V_1}$. Substitute the values: $T_2=\frac{2\times373}{4}=186.5K$. Convert back to Celsius: $T_2=186.5 - 273=-86.5^{\circ}C$.

Answer:

1. $181.5^{\circ}C$
2. $- 86.5^{\circ}C$
3. For drawing a particle - view model of two gases with pressures 100kPa and 50kPa: The gas with 100kPa can be represented with more densely packed particles (more particles per unit volume) compared to the gas with 50kPa, assuming the same volume and temperature for both. Label one model as "Gas with 100kPa" and the other as "Gas with 50kPa".
4. For drawing a particle - view model of two gases with temperatures $50^{\circ}C$ and $100^{\circ}C$: The gas with $100^{\circ}C$ can be represented with particles moving more vigorously (longer arrow lengths to represent higher kinetic energy) compared to the gas with $50^{\circ}C$. Label one model as "Gas at $50^{\circ}C$" and the other as "Gas at $100^{\circ}C$".