Answer the following problem through the use of Boyle's Law. Each problem is equivalent to 5 points. 1. A balloon contains 9.20 L of He. The pressure is reduced to 2.00 atm and the balloon expands to occupy a volume of 20.2 L. What was the initial pressure exerted on the balloon? 2. A tank of nitrogen has a volume of 10.0 L and a pressure of 760.0 mmHg. Find the volume of the nitrogen when its pressure is changed to 220.0 mmHg while the temperature is held constant. 3. Oxygen gas inside a 5 L gas tank has a pressure of 200 atm. Provided that the temperature remains constant, how much pressure is needed to reduce by 1/2? Answer the following problem through the use of Charles' Law. Each problem is equivalent to 5 points. 4. A tank contains 3.5 L of helium gas at 40 °C. What will be the volume of the tank after heating it and its content of 50 °C temperature at constant pressure? 5. A 200 ml. glass filled with air is placed into water up-side-down while at 9.0°C. The water is heated to 80°C. How much air bubbles out from under the glass?
Answer (1)
[tex]\begin{gathered}\begin{gathered}{\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} \\\end{gathered}\end{gathered}[/tex]Let's work through the problems: Problem 1: Boyle's Law: [tex]P_1V_1 = P_2V_2[/tex] Solution: [tex]P_1 \cdot 9.20 = 2.00 \cdot 20.2[/tex] [tex]P_1 \cdot 9.20 = 40.4[/tex] [tex]P_1 = \frac{40.4}{9.20} \approx 4.39 \text{ atm}[/tex] Problem 2: Boyle's Law: [tex]P_1V_1 = P_2V_2[/tex] Solution: [tex]760.0 \cdot 10.0 = 220.0 \cdot V_2[/tex] [tex]7600 = 220.0 \cdot V_2[/tex] [tex]V_2 = \frac{7600}{220.0} \approx 34.55 \text{ L}[/tex] Problem 3: Boyle's Law: [tex]P_1V_1 = P_2V_2[/tex] Solution: [tex]P_2 = \frac{1}{2} \cdot 200 = 100 \text{ atm}[/tex] Problem 4: Charles' Law: [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex] Solution: [tex]T_1 = 40 + 273.15 = 313.15 \text{ K}[/tex] [tex]T_2 = 50 + 273.15 = 323.15 \text{ K}[/tex] [tex]\frac{3.5}{313.15} = \frac{V_2}{323.15}[/tex] [tex]V_2 = 3.5 \cdot \frac{323.15}{313.15} \approx 3.61 \text{ L}[/tex] Problem 5: Charles' Law: [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex] Solution: [tex]T_1 = 9 + 273.15 = 282.15 \text{ K}[/tex] [tex]T_2 = 80 + 273.15 = 353.15 \text{ K}[/tex] [tex]\frac{200}{282.15} = \frac{V_2}{353.15}[/tex] [tex]V_2 = 200 \cdot \frac{353.15}{282.15} \approx 250 \text{ mL}[/tex][tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\begin{gathered} \boxed{\begin{array}{cc} \sf \footnotesize \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ···\: ʚ\: \: \: \: \: \: Hope\:it\:helps \: \: \: \: \: ɞ \:··· \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\footnotesize\:\# CarryOnLearning \\ \sf\footnotesize ૮₍´˶ \: • \: . \: • \: ⑅ ₎ა \: \leadsto \footnotesize\sf\color{purple} \underline{Study\:Well!}\end{array}}\end{gathered}[/tex][tex]\sf\color{green}{⊱⋅ ────────────────────── ⋅⊰}[/tex][tex]\large\qquad\qquad\qquad\tt MARCH/17/2025 [/tex]
