Given:
[tex]P_{1} = \text{1.20 atm}[/tex]
[tex]T_{1} = \text{10°C + 273 = 283 K}[/tex]
[tex]V_{1} = \text{0.500 mL}[/tex]
[tex]P_{2} = \text{0.95 atm}[/tex]
[tex]T_{2} = \text{14°C + 273 = 287 K}[/tex]
Unknown:
[tex]V_{2}[/tex]
Solution:
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
[tex]V_{2} = V_{1} × \frac{P_{1}}{P_{2}} × \frac{T_{2}}{T_{1}}[/tex]
[tex]V_{2} = \text{0.500 mL} × \frac{\text{1.20 atm}}{\text{0.95 atm}} × \frac{\text{287 K}}{\text{283 K}}[/tex]
[tex]\boxed{V_{2} = \text{0.641 mL}}[/tex]
[tex]\\[/tex]
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