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An arctic weather balloon is filled with 20.9L of helium gas inside a prep shed. The temperature inside the shed is 13 degree C. The balloon is then taken outside, where the temperature is -9 degree C. Calculate the new volume of the balloon. You may assume the pressure on the balloon stays constant at exactly 1 atm. Round your answer to significant digits.

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Willchemistry

hey there!

The combined gas equation is  :

P1V1/T1 = P2V2/T2  

Since Pressure is constant for this question it will cancel out leaving  

V1/T1 = V2/T2  

where  :

V1 = initial volume = 20.9 L  

T2 = initial temp (must be in Kelvin) = 286 K  

V2 = final volume = ? L  

T2 = final temp = 264 K  

Solve for V2  

V2 = V1T2 / T1  

= 20.9 L * 264 K / 286 K  

= 19.29 L  

= 19.3 L (3 sig digits)

Hope this helps!

Eduard22sly

The new volume of the balloon when taken outside where the temperature is –9 °C is 19.3 L

From the question given above, the following data were obtained:

Initial volume (V₁) = 20.9 L

Initial temperature (T₁) = 13 °C = 13 + 273 = 286 K

Final temperature (T₂) = –9 °C = –9 + 273 = 264 K

Pressure = constant

Final volume (V₂) =?

The final volume (i.e new volume) of the balloon can be obtained by using the charles' law equation as illustrated below:

[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} \\\\\frac{20.9}{286} = \frac{V_{2}}{264} \\\\[/tex]

Cross multiply

V₂ × 286 = 20.9 × 264

V₂ × 286 = 5517.6

Divide both side by 286

[tex]V_{2} = \frac{5517.6 }{286} \\\\[/tex]

V₂ = 19.3 L

Therefore, the new volume of the balloon is 19.3 L

Learn more: https://brainly.com/question/16927784

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