mm Hg 5) What pressure (mm Hg) is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder whose volume is 26.0 liters? P₂ = P₁v₁ P₁V₁=lava 7.540+m V₂ Tatm=760mm Hg 5730 mmHg 7.54 at 7.54 atm/760mmitty latm 1.00atm x 196.0L. 26.04 V₁=196.0L P₁ = 19+m V₂=2610L P2=5720 mm Hg
mm Hg 5) What pressure (mm Hg) is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder whose volume is 26.0 liters? P₂ = P₁v₁ P₁V₁=lava 7.540+m V₂ Tatm=760mm Hg 5730 mmHg 7.54 at 7.54 atm/760mmitty latm 1.00atm x 196.0L. 26.04 V₁=196.0L P₁ = 19+m V₂=2610L P2=5720 mm Hg
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can you help me with the number 5 question? Can you explain step by step including the formula (Boyle’s law)? I need to plug in the fraction to give the correct answer.
![**Gas Laws Worksheet**
**Charles’s Law (temperature, volume)**
1) A 550.0 mL sample of nitrogen gas is warmed from 77°C to 86°C. Find its new volume if the pressure remains constant.
- Formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
- \( V_1 = 550.0 \, \text{mL}, \, V_2 = ? \, \text{mL} \)
- \( T_1 = 77^\circ \text{C} = 350 \, \text{K}, \, T_2 = 86^\circ \text{C} = 359 \, \text{K} \)
Calculation:
\[
V_2 = \left(\frac{550.0 \times 359}{350}\right) = 564 \, \text{mL}
\]
2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0°C?
- Formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
- \( V_1 = 1.00 \, \text{L} \)
- \( T_1 = 273 \, \text{K}, \, T_2 = 333 + 273 = 606 \, \text{K} \)
Calculation:
\[
V_2 = \left(\frac{1.00 \times 606}{273}\right) = 2.22 \, \text{L}
\]
**Boyle’s Law (pressure, volume)**
3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm.
- Formula: \( P_1V_1 = P_2V_2 \)
- \( P_1 = 63.0 \, \text{atm}, \, V_1 = 338 \, \text{L} \)
- \( P_2 = 1.00 \, \text{atm}, \, V_2 = ? \)
Calculation:
\[
V_2 = \left](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea82e0d0-a0f2-48c1-893f-44877b2a5c1c%2Ffcedb7af-fb6f-4d90-87a4-5b086c71fc0d%2Fydbaxq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Gas Laws Worksheet**
**Charles’s Law (temperature, volume)**
1) A 550.0 mL sample of nitrogen gas is warmed from 77°C to 86°C. Find its new volume if the pressure remains constant.
- Formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
- \( V_1 = 550.0 \, \text{mL}, \, V_2 = ? \, \text{mL} \)
- \( T_1 = 77^\circ \text{C} = 350 \, \text{K}, \, T_2 = 86^\circ \text{C} = 359 \, \text{K} \)
Calculation:
\[
V_2 = \left(\frac{550.0 \times 359}{350}\right) = 564 \, \text{mL}
\]
2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0°C?
- Formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
- \( V_1 = 1.00 \, \text{L} \)
- \( T_1 = 273 \, \text{K}, \, T_2 = 333 + 273 = 606 \, \text{K} \)
Calculation:
\[
V_2 = \left(\frac{1.00 \times 606}{273}\right) = 2.22 \, \text{L}
\]
**Boyle’s Law (pressure, volume)**
3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm.
- Formula: \( P_1V_1 = P_2V_2 \)
- \( P_1 = 63.0 \, \text{atm}, \, V_1 = 338 \, \text{L} \)
- \( P_2 = 1.00 \, \text{atm}, \, V_2 = ? \)
Calculation:
\[
V_2 = \left
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