Answered

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A 20-milliliter sample of a gas is at 546 K and has a pressure of 6.0 atmospheres. If the temperature is changed to 273 K and the pressure to 2.0 atmospheres, the new volume of the gas (in Liters) will be

Sagot :

Given:

[tex]P_{1} = \text{6.0 atm}[/tex]

[tex]T_{1} = \text{546 K}[/tex]

[tex]V_{1} = \text{20 mL}[/tex]

[tex]P_{2} = \text{2.0 atm}[/tex]

[tex]T_{2} = \text{273 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{20 mL} × \frac{\text{6.0 atm}}{\text{2.0 atm}} × \frac{\text{273 K}}{\text{546 K}}[/tex]

[tex]\boxed{V_{2} = \text{30 mL}}[/tex]

[tex]\\[/tex]

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