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what is the final volume of a 400.0mL gas sample that is subjected to a temperature change from 22.0 C to 30.0 C and a pressure change from standard pressure to 360.0 mm Hg


Sagot :

Given:

[tex]P_{1} = \text{760.0 mmHg}[/tex]

[tex]T_{1} = \text{22.0°C + 273.15 = 295.15 K}[/tex]

[tex]V_{1} = \text{400.0 mL}[/tex]

[tex]P_{2} = \text{360.0 mmHg}[/tex]

[tex]T_{2} = \text{30.0°C + 273.15 = 303.15 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{400.0 mL} × \frac{\text{760.0 mmHg}}{\text{360.0 mmHg}} × \frac{\text{303.15 K}}{\text{295.15 K}}[/tex]

[tex]\boxed{V_{2} = \text{867.3 mL}}[/tex]

[tex]\\[/tex]

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