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A mountain climber inhales air that has a temperature of −8 °C. If the final volume of air in the lungs is 569 m L at a body temperature of 37 °C, what was the initial volume of air, in milliliters, inhaled by the climber, if the pressure and amount of gas do not change?

Answer :

Answer:

[tex]486.40\text{ mL}[/tex]

Explanation:

Here, we want to get the initial volume of air

From Charles' law, we know thattemperature and presure are directly proportional at consnstant pressure

mathematically:

[tex]\frac{V_1}{T_1}\text{ = }\frac{V_2}{T_2}[/tex]

Where:

V1 is the original volume which is ?

V2 is the final vlume which is 569 mL

T1 is the initial temperature which we would convert to absolute scale by adding 273 K : -8 + 273 = 265 K

T2 is 37 deg Celsius which is 273 + 37 = 310 K

Substituting the values:

[tex]\begin{gathered} \frac{V_1}{265}\text{ = }\frac{569}{310\text{ }} \\ \\ V_1\text{ = }\frac{569\times265}{310} \\ \\ V_1\text{ = 486.40 mL} \end{gathered}[/tex]

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