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.
Trending News
TRY this if you can?--no one already gt its correct answer?
a certain gas at 101.325 kpa and 16deg C whoose volume is 2.83m^3 are compressed into a storage vessel of .31 m^3. Before admission, the storage vessel contained the gas at a pressure and temperature of 137.8kpa and 24degrees Celcius. After the admission, the pressure has increased to 1171.8 kpa.What should be the final temperature of the gas in the vessel in kelvin?
2 Answers
- Anonymous1 decade agoFavorite Answer
convert the degree Celsius into Kelvins (273+C) then apply this to the equation and calculate.
- MadhukarLv 71 decade ago
0.0224 m^3 of gas at 101.3 kpa and 273 K = 1 g-mole
=> 2.83 m^3 at 101.325 kpa and 289 K
= (2.83/0.0224) *(273/289) = 119.3 g-mole
Gas inside the storage vessel was
(0.31/0/024) * (137.8/101.3) * ( 273/297) = 17.3 g-mole
Total gas = 119.3 + 17.3 = 136.6 g-mole
If T = final tenmperature in K,
136.6 * (0.0224) * (101.3/1171.8) * ( T/273) = 0.31
=> 9.68 x 10^(-4) T = 0.31
=> T = 320 K.
