Safari 9:05 ID periodic table . Google Student iI UCR Portal CH6HW 76% @ucr edu
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Question
Safari 9:05 ID periodic table . Google Student iI UCR Portal CH6HW 76% @ucr edu 01/25/18 This is a Numeric Entry question/It is worth 1 point/You have unlimited attempts/There is no attempt penalty 06 Question (1 point) a See page 275 The severity of a tropical storm is related to the depressed atmospheric pressure at its center. Shown below is a photograph of Typhoon Odessa taken from the space shuttle Discovery in August 1985, when the maximum winds of the storm were about 90 mi/hr and the pressure was 40.0 mbar lower at the center than normal atmospheric pressure. In contrast, the central pressure of Hurricane Andrew was 90.0 mbar lower than its surroundings when it hit south Florida with winds as high as 165 mi/hr 06/14 11 OF 14 QUESTIONS COMPLETED VIEW SOLUTION C TRY AGA
Explanation / Answer
Ans. Given, Pressure at the center = 40.0 mbar LOWER than normal atmospheric pressure.
Normal atmospheric pressure = 1.00 atm
40 mbar = 0.04 atm
So,
Pressure at the center = 1.00 atm – 0.04 atm = 0.96 atm
# Initial conditions: volume of balloon = 48.0 L, Pressure = 1.00 atm
Final conditions: volume of balloon = ? Pressure = 0.96 atm
Boyle’s Law: For a gas at constant temperature confined in a closed vessel, the product of pressure and volume is a constant. That is, PV = Constant - Temperature kept constant’
P1V1 = P2V2 - T, n constant
Since balloon is a closed system, the number of moles remains constant. Since the temperature is not mentioned anywhere, it’s assumed that temperature also remains constant.
Now, using Boyle’s law-
P1V1 (initial condition) = P2V2 (final conditions)
Or, 1.00 atm x 48.0 L = 0.96 atm x V2
Or, V2 = (1.00 atm x 48.0 L) / 0.96 atm
Hence, V2 = 50.0 L
Therefore, volume of balloon at the center of Typhoon Odessa = 50.0 L
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