A gas collected when the temperature is 11 degrees and the pressure is 710 mm HG measures 14.8 mL. Calculate the volume of the gas at 20 degrees and 740 mm Hg

Answer 1

First, we need to calculate the moles of gas at 11 degrees and 710 mm Hg:

P1V1 = P2V2

(710 mm Hg)(14.8 mL) = (x)(22.4 L) [since 1 mole of gas at STP occupies 22.4 L]

x = (710 mm Hg * 14.8 mL) / 22.4 L
x ≈ 468.33 mmHg mL / L

Next, we will use the combined gas law to find the new volume at the new temperature and pressure:

P1V1 / T1 = P2V2 / T2

(710 mm Hg)(14.8 mL) / 284 K = (740 mm Hg)(x) / 293 K

468.33 mmHg mL / L / 284 K = 740 mmHg * x / 293 K

(468.33 / 284) = (740 * x) / 293

1.6487126 = 2.525171 x

x ≈ 1.64 mL

Therefore, the volume of the gas at 20 degrees and 740 mm Hg is approximately 1.64 mL