Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Chow asked in 科學及數學化學 · 1 decade ago

gas law_F6 CHEM 問題

2 Answers

Rating
  • 1 decade ago
    Favorite Answer

    3.

    (b)

    Pressure on sea level

    = 1 atm

    Pressure at the seabed

    = 1 + 1 x (20/10)

    = 3 atm

    (c)(i)

    Consider the gas in the cylinder on the sea level:

    Volume, V1 = 0.02 m3

    Pressure, P1 = 18 atm

    Temperature, T1 = 273 + 27 = 300 K

    Consider the gas in the cylinder at the seabed just before the diver inflates the balloon:

    Volume, V2 = 0.02 m3 (unchanged)

    Pressure, P2 = ?

    Temperature, T2 = 273 + 21 = 294 K

    PV = nRT

    P/T = nR/V

    Since n, R and V are constant, thus P/T = constant

    P1/T1 = P2/T2

    18/300 = P2/294

    Gas pressure in the cylinder just before the diver inflates the balloon, P2

    = 294 x (18/300)

    = 17.64 atm

    (c)(ii)

    Consider the gas originally in the cylinder used to inflate the balloon:

    Volume, V3 = 0.02 m3

    Partial pressure, P3 = ? atm

    Temperature, T = 294 K

    Consider the gas in the inflated balloon:

    Volume, V4 = 0.01 m3

    Pressure, P4 = 3 atm

    Temperature, T = 294 K (unchanged)

    PV = nRT

    Since n, R and T are constant, thus PV = constant

    P3V3 = P4V4

    P3 x 0.02 = 3 x 0.01

    P3 = 1.5 atm

    Gas pressure in the cylinder just before the diver inflates the balloon

    = 17.64 - 1.5

    = 16.14 atm

    c(iii)

    According to Dalton's law of partial pressure,

    fraction of the gas originally in the cylinder used to inflate the balloon

    = 1.5/17.64

    = 0.085

  • 1 decade ago

    我還是小學4年級生,我吳柏儒5識!!!

Still have questions? Get your answers by asking now.