Question

A piece of dry ice \(\left(\mathrm{CO}_{2}(s)\right)\) has a mass of \(22.50 \mathrm{~g} .\) It is dropped into an evacuated 2.50 -L flask. What is the pressure in the flask at \(-4^{\circ} \mathrm{C}\) ?

Short answer

Answer: The pressure in the flask at -4°C is 4.56 atm.

Step-by-step solution

Convert mass of \(\mathrm{CO}_{2}\) to moles

To convert the mass of \(\mathrm{CO}_{2}\) into moles, we need to divide the mass by its molar mass. The molar mass of \(\mathrm{CO}_{2}\) is \(44.01\mathrm{~g/mol}\). Given mass of \(\mathrm{CO}_{2}\) is \(22.50\mathrm{~g}\): Number of moles (n) = \(\frac{Mass}{Molar\:mass}\) = \(\frac{22.50\:g}{44.01\:g/mol}\) = \(0.511\:mol\)

Convert temperature to Kelvin

The ideal gas law uses temperature in Kelvin. The given temperature is \(-4^{\circ} \mathrm{C}\): Temperature (T) = \(-4^{\circ} \mathrm{C} + 273.15 = 269.15\: K\)

Use the ideal gas law to find the pressure

Recall the ideal gas law formula: \(PV = nRT\), 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. We will use the gas constant R = \(0.0821\ \mathrm{L\cdot atm/(mol\cdot K)}\). We already found: - n = \(0.511\:mol\) - T = \(269.15\:K\) - V = \(2.50\:L\) We can now plug in the values and solve for P: \((P)(2.50\mathrm{~L}) = (0.511\mathrm{~mol})(0.0821\ \mathrm{L\cdot atm/(mol\cdot K)})(269.15\mathrm{~K})\) P = \(\frac{(0.511\:mol)(0.0821\:L\:atm/(mol\:K))(269.15\:K)}{2.50\:L}\) = \(4.56\:atm\) The pressure in the flask at \(-4^{\circ} \mathrm{C}\) is \(4.56\:atm\).