Question

question 1
0.75 pts
if the pressure of 1.50 l of a gas at 100 °c decreases from 0.500 atm to 0.115 atm, what is the final volume? assume temperature and amount remains constant.
0.345 l
3.45 l
0.652 l
1.50 l
6.52 l
question 2
0.25 pts
in the previous question, what happened to the pressure and the volume?
the volume increased and the pressure decreased.
the volume decreased and the pressure decreased.
the volume decreased and the pressure increased.
the volume increased and the pressure increased.

Explanation:

Question 1

Step1: Identify the gas law

Since temperature and amount of gas are constant, we use Boyle's Law, which states that \( P_1V_1 = P_2V_2 \), where \( P_1 \) is the initial pressure, \( V_1 \) is the initial volume, \( P_2 \) is the final pressure, and \( V_2 \) is the final volume.

Step2: List the given values

\( P_1 = 0.500 \, \text{atm} \), \( V_1 = 1.50 \, \text{L} \), \( P_2 = 0.115 \, \text{atm} \), and we need to find \( V_2 \).

Step3: Rearrange Boyle's Law to solve for \( V_2 \)

From \( P_1V_1 = P_2V_2 \), we get \( V_2=\frac{P_1V_1}{P_2} \).

Step4: Substitute the values into the formula

\( V_2=\frac{0.500 \, \text{atm} \times 1.50 \, \text{L}}{0.115 \, \text{atm}} \)
First, calculate the numerator: \( 0.500\times1.50 = 0.75 \)
Then, divide by the denominator: \( V_2=\frac{0.75}{0.115}\approx6.52 \, \text{L} \)

In Question 1, the initial pressure was \( 0.500 \, \text{atm} \) and the final pressure was \( 0.115 \, \text{atm} \), so pressure decreased. The initial volume was \( 1.50 \, \text{L} \) and the final volume was approximately \( 6.52 \, \text{L} \), so volume increased. This is consistent with Boyle's Law (inverse relationship between pressure and volume at constant temperature and amount of gas).

Answer:

6.52 L (corresponding to the option "6.52 L")

Question 2