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erik [133]
2 years ago
11

A hot air ballon contains 270L of helium at 24C and 845 mmHg. WHat will be the volume of the ballon at -50c and pressure of 0.73

5atm?
Chemistry
1 answer:
Julli [10]2 years ago
7 0

Using the ideal gas law,

<em>PV</em> = <em>nRT</em>

where <em>R</em> = 0.08206 L•atm/(mol•°K), solving for <em>n</em> gives

<em>n</em> = <em>PV</em>/(<em>RT</em>)

<em>n</em> = (845 mmHg) (270 L) / ((0.08206 L•atm/(mol•°K)) (24 °C))

Convert the given temperature to °K and the given pressure to atm:

24 °C = (273.15 + 24) °K ≈ 297.2 °K

(845 mmHg) × (1/760 atm/mmHg) ≈ 1.11 atm

Then the balloon contains

<em>n</em> = (1.11 atm) (270 L) / ((0.08206 L•atm/(mol•°K)) (297.2 °K))

<em>n</em> ≈ 12.3 mol

of He.

Solve the same equation for <em>V</em> :

<em>V</em> = <em>nRT</em>/<em>P</em>

<em />

Convert the target temperature to °K:

-50 °C = (273.15 - 50) °K = 223.15 °K

Then the volume under the new set of conditions is

<em>V</em> = (12.3 mol) (0.08206 L•atm/(mol•°K)) (223.15 °K) / (0.735 atm)

<em>V</em> ≈ 306 L

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