Question

in this simulation, you will be exploring the relationship between temperature and pressure while keeping volume constant. click here to begin. select the explore box. answer the questions as you perform the steps. 1. raise and push down the pressure pump handle to add gas molecules to the container. the initial temperature is 300 k (27°c) and the pressure has a range of 5.4 - 6.2 atm. notice the pressure values change as the particles collide with the wall. 2. click on the bucket to raise the temperature of the container to approximately 400 k (127°c). what happened to the pressure? approximately what pressure range did you observe? 3. click on the bucket to reduce the temperature to approximately 200 k (-73°c). what happened to the pressure? approximately what pressure did you observe? 4. what gas law mathematically shows the relationships between temperature and pressure

Explanation:

Step1: Recall Gay - Lussac's Law

Gay - Lussac's law states that for a given mass of gas at constant volume, the pressure $P$ is directly proportional to the temperature $T$ (in Kelvin), i.e., $\frac{P_1}{T_1}=\frac{P_2}{T_2}$.

Step2: Analyze the initial conditions

The initial temperature $T_1 = 300\ K$ and initial pressure $P_1$ is in the range of $5.4 - 6.2\ atm$.

Step3: Consider temperature change to 400 K

Using Gay - Lussac's law $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, when $T_1 = 300\ K$, $T_2=400\ K$ and $P_1$ is in the range $5.4 - 6.2\ atm$. We can find $P_2$ as $P_2=\frac{T_2}{T_1}P_1$. Substituting the values, $P_2=\frac{400}{300}P_1=\frac{4}{3}P_1$. When $P_1 = 5.4\ atm$, $P_2=\frac{4}{3}\times5.4 = 7.2\ atm$; when $P_1 = 6.2\ atm$, $P_2=\frac{4}{3}\times6.2\approx8.3\ atm$. So the pressure range is $7.2 - 8.3\ atm$.

Step4: Consider temperature change to 200 K

Using $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ again, with $T_1 = 300\ K$, $T_2 = 200\ K$ and $P_1$ in the range $5.4 - 6.2\ atm$. Then $P_2=\frac{T_2}{T_1}P_1=\frac{200}{300}P_1=\frac{2}{3}P_1$. When $P_1 = 5.4\ atm$, $P_2=\frac{2}{3}\times5.4 = 3.6\ atm$; when $P_1 = 6.2\ atm$, $P_2=\frac{2}{3}\times6.2\approx4.1\ atm$. So the pressure range is $3.6 - 4.1\ atm$.

Answer:

When the temperature is raised to 400 K, the pressure range is $7.5 - 8.3\ atm$. When the temperature is decreased to 200 K, the pressure range is $3.5 - 4.3\ atm$. For the gas - law relationship, as the temperature of a gas at constant volume increases, the pressure increases proportionally and vice - versa.