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A balloon is inflated to a volume of 8.0 L on a day when the atmospheric pressure is 1.013 bar . The next day, a storm front arrives, and the atmospheric pressure drops to 0.968 bar . Assuming the temperature remains constant, what is the new volume of the balloon, in liters

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sebassandin

Answer:

[tex]V_2=8.4L[/tex]

Explanation:

Hello there!

In this case, according to the definition of the Boyle's law, which describes de pressure-volume behavior as an inversely proportional relationship, it is possible for us to write:

[tex]P_1V_1=P_2V_2[/tex]

Thus, since we are given the initial pressure and temperature, and the final pressure, we are able to calculate the final volume as shown below:

[tex]baV_2=\frac{P_1V_1}{P_2}\\\\V_2=\frac{8.0L*1.013bar}{ 0.968bar}\\\\V_2=8.4L[/tex]

Regards!

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