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, an

Question

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

in progress 0
Cherry 4 years 2021-08-13T07:14:05+00:00 1 Answers 17 views 0

Answers ( )

  1. taiduc
    0
    2021-08-13T07:15:36+00:00
    Reply

    Answer:

    V_2=8.4L

    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:

    P_1V_1=P_2V_2

    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:

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

    Regards!

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )