a. 700 mmHg
Given:
[tex]P_{1} = \text{1.5 atm} × \frac{\text{760 mmHg}}{\text{1 atm}} = \text{1140 mmHg}[/tex]
[tex]V_{1} = \text{20 L}[/tex]
[tex]P_{2} = \text{700 mmHg}[/tex]
Required:
[tex]V_{2}[/tex]
Equation:
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
Solution:
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
[tex]V_{2} = V_{1} × \frac{P_{1}}{P_{2}}[/tex]
[tex]V_{2} = \text{20 L} × \frac{\text{1140 mmHg}}{\text{700 mmHg}}[/tex]
[tex]\boxed{V_{2} = \text{33 L}}[/tex]
Answer:
[tex]V_{2} = \text{33 L}[/tex]
b. 2 atm
Given:
[tex]P_{1} = \text{1.5 atm}[/tex]
[tex]V_{1} = \text{20 L}[/tex]
[tex]P_{2} = \text{2 atm}[/tex]
Required:
[tex]V_{2}[/tex]
Equation:
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
Solution:
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
[tex]V_{2} = V_{1} × \frac{P_{1}}{P_{2}}[/tex]
[tex]V_{2} = \text{20 L} × \frac{\text{1.5 atm}}{\text{2 atm}}[/tex]
[tex]\boxed{V_{2} = \text{15 L}}[/tex]
Answer:
[tex]V_{2} = \text{15 L}[/tex]
c. 105 kPa
Given:
[tex]P_{1} = \text{1.5 atm} × \frac{\text{101.325 kPa}}{\text{1 atm}} = \text{151.9875 kPa}[/tex]
[tex]V_{1} = \text{20 L}[/tex]
[tex]P_{2} = \text{105 kPa}[/tex]
Required:
[tex]V_{2}[/tex]
Equation:
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
Solution:
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
[tex]V_{2} = V_{1} × \frac{P_{1}}{P_{2}}[/tex]
[tex]V_{2} = \text{20 L} × \frac{\text{151.9875 kPa}}{\text{105 kPa}}[/tex]
[tex]\boxed{V_{2} = \text{29 L}}[/tex]
Answer:
[tex]V_{2} = \text{29 L}[/tex]
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