Chemistry
lilbellaa
2017-08-19 21:42:15
What temperature does a 0.25 L cylinder containing 0.10 mol of helium gas need to be cooled in order for the pressure to be 253.25 kPa? (Given: R = 8.314 L∙kPa/K∙mol
ANSWERS
hr28
2017-08-20 01:54:20

On this item we will use the formula of the ideal gas law which is: PV = nRT  P = 253.25 kPa  V = 0.25 L  n = 0.10 mol  R = 8.314 kPaLmol^-1K^-1  T = temp in Kelvin = ? K  Work out for T  T = PV / nR  = 253.25 kPa x 0.25 L / 0.10 mol x 8.314kPaLmol^-1K^-1  Temperature = 76.15 K 

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