Question

a sample of krypton gas at 3.50 atm is heated from 20.0 °c to 150.0 °c. if the volume and the amount remains constant, what is the final pressure?
26.3 atm
1.79 atm
2.42 atm
5.05 atm
0.467 atm
question 2
0.25 pts
in the previous question, what happened to the temperature and the pressure?
the temperature decreased and the pressure decreased.
the temperature decreased and the pressure increased.
the temperature increased and the pressure decreased.
the temperature increased and the pressure increased.

Explanation:

Question 1 (Final Pressure Calculation)

Step1: Identify Gas Law

Since volume (\(V\)) and amount (\(n\)) are constant, use Gay-Lussac's Law: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\), where \(T\) is in Kelvin.

Step2: Convert Temperatures to Kelvin

\(T_1 = 20.0^\circ\text{C} + 273.15 = 293.15\,\text{K}\)
\(T_2 = 150.0^\circ\text{C} + 273.15 = 423.15\,\text{K}\)
\(P_1 = 3.50\,\text{atm}\)

Step3: Solve for \(P_2\)

Rearrange the law: \(P_2 = P_1 \cdot \frac{T_2}{T_1}\)
Substitute values: \(P_2 = 3.50\,\text{atm} \cdot \frac{423.15\,\text{K}}{293.15\,\text{K}} \approx 5.05\,\text{atm}\)

In Question 1, temperature increased from \(20.0^\circ\text{C}\) to \(150.0^\circ\text{C}\), and pressure increased from \(3.50\,\text{atm}\) to \(\approx 5.05\,\text{atm}\) (per Gay-Lussac's Law, pressure is proportional to temperature at constant \(V\) and \(n\)).

Answer:

5.05 atm

Question 2 (Temperature-Pressure Relationship)