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

Question

Jaycee Curtis

Chemistry

11 August, 17:02

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

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Francis Wilkerson · Answer 1

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