A block of dry ice (solid CO₂, density = 1.56 g/mL) of dimensions 25.0 cm × 25.0 cm × 25.0 cm is left to sublime (i.e., to pass from the solid phase to the gas phase) in a closed chamber of dimensions 4.00 m × 5.00 m × 3.00 m. Once the sublimation is complete, what is the partial pressure of carbon dioxide in the chamber at 25°C? (R = 0.08206 L • atm/K • mol, 1 atm = 760 mmHg) A) 172 mmHg B) 107 mmHg C) 0.225 mmHg D) 0.172 mmHg E) 14.4 mmHg

Question

Quizplus · Accepted Answer

To find the partial pressure of CO2 in the chamber after sublimation, we first calculate the moles of CO2 produced from the sublimation of dry ice. The volume of the dry ice block is 25.0 cm × 25.0 cm × 25.0 cm = 15625 cm³ or 15.625 L. Given the density of dry ice is 1.56 g/mL, the mass of the dry ice block is 15625 cm³ × 1.56 g/cm³ = 24375 g. The molar mass of CO2 is approximately 44.01 g/mol, so the moles of CO2 are 24375 g / 44.01 g/mol ≈ 554 moles.Using the ideal gas law PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. The volume of the chamber is 4.00 m × 5.00 m × 3.00 m = 60 m³ or 60000 L. The temperature is given as 25°C, which is 298 K.Rearranging the ideal gas law to solve for P gives P = nRT/V. Substituting the known values gives P = (554 moles × 0.08206 L•atm/K•mol × 298 K) / 60000 L ≈ 0.225 atm. Converting this pressure to mmHg (since 1 atm = 760 mmHg) gives 0.225 atm × 760 mmHg/atm ≈ 172 mmHg.

A Block of Dry Ice (Solid CO2, Density = 1

A Block of Dry Ice (Solid CO₂, Density = 1