Hello!
I have added 15 L of air to a balloon at sea level (1.0 atm). If I take the balloon with me to Denver, where the air pressure is 0.85 atm, what will the new volume of the balloon be?
- We have the following data:
V1 (initial volume) = 15 L
P1 (initial pressure) = 1.0 atm
P2 (final pressure) = 0.85 atm
V2 (final volume) = ? (in ml)
- We have an isothermal transformation, its temperature remains constant, the volume and pressure of the gas are inversely proportional, that is, if the gas pressure in the container increases, so its volume decreases. Applying the data to the formula of Boyle's Law, we have:
[tex]P_1*V_1 = P_2*V_2[/tex]
[tex]1.0*15 = 0.85*V_2[/tex]
[tex]15 = 0.85\:V_2[/tex]
[tex]0.85\:V_2 = 15[/tex]
[tex]V_2 = \dfrac{15}{0.85}[/tex]
[tex]V_2 = 17.647... \to \boxed{\boxed{V_2 \approx 17.65\:L}}\:\:\:\:\:\:\bf\purple{\checkmark}[/tex]
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