The dry ice that can purchased at Fry\'s is nothing more than solid carbon dioxi

Question

The dry ice that can purchased at Fry's is nothing more than solid carbon dioxide. 'Course it's very cold! Suppose a cube of dry ice is placed in the compartment below right which in turn is connected to an open-end glass manometer. The total volume of the compartment including the glass tubing connected to it equals 4, 455mL. Liquid mercury is represented by the black solid color. The light grey shading represents dry air inside the compartment and connective tubing. Initially before the dry ice was added, the mercury levels were level. The outside atmospheric pressure remained constant at 0.978atm. The drawing below represents that happened once the solid dry ice sublimed. The observation below was made two days after the solid dry ice cube was added. The room temperature remained unchanged during this time at 20.7 degree C. Based on this Information compute the mass of dry ice (in milligrams) added to the compartment. in order to receive full, credit the correct calculations with correct units for each step must be dearly shown. Otherwise zero credit will be eased.

Explanation / Answer

MW of dry ice = 44.01 g/mol

Initially mercury levels were equal,

Temperature, T = 20.7 C = 293.85 K

Volume of compartment, V = 4455 mL = 4.455 L

R = 8.31 L.kPa/mol-K

Pressure inside compartment = Atmospheric pressure

P = 0.978 atm

= 0.978 * 101325 Pa

= 99095.85 Pa = 99.096 kPa

Initial moles of gas present, n1 = P * V /(R T)

After 2 days,

Temperature of compartment, T = 20.7 C = 293.85 K

h = 654 mm = 0.654 m

g = 9.81 m/s2

Density of mercury, = 13600 kg/m3

P = g h

= 13600 * 9.81 * 0.654

= 87254.06 Pa

Final pressure, Pf = P + P

= 99095.85 + 87254.06 Pa

= 186349.9 Pa = 186.35 kPa

Final moles of gas present, n2 = Pf * V / (R T)

Moles of CO2 vaporized = n2 – n1

= (Pf – P) V / (R T)

= (186.35 – 99.096) * 4.455 / (8.31 * 293.85)

= 0.159 moles

Mass of dry ice present = 0.159 * 44.01 g

= 7.006 g

= 7006 mg

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