Answer to Question #324602 in General Chemistry for denden

Question

A sample of a gas taken at STP was analyzed and found to have a volume of 6.3L. The gas was then placed to a chamber where the pressure is increased by 50% while decreasing the temperature by 25% with respect to its original condition. Calculate the volume increase/decrease of the gas.

Answer 1

pV = nRT

At STP T = 273 K, p = 1 atm.

R = 0.082057 L atm mol^-1K^-1

1 x 6.3 = n x 0.082 x 273

n = (0.082 x 273) / (1 x 6.3) = 3.55 mol

In the new conditions, the equation will be:

1.5 x V = 3.55 x 0.082 x (273 x 0.75)

V = (3.55 x 0.082 x (273 x 0.75)) / 1.5 = 39.74 L

The new volume of the gas will be 39.74 L.