Question

a piece of solid carbon dioxide, with a mass of 20.2 g, is placed in an otherwise empty 4.35 l container at 29°c. (a) what is the pressure in the container after all the carbon dioxide vaporizes? pressure = atm (b) if 20.2 g of solid carbon dioxide was placed in a similar container already containing air at 715 torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide had vaporized? partial pressure of co₂ = atm total pressure = atm

Explanation:

Step1: Calculate the number of moles of \(CO_2\)

The molar mass of \(CO_2\) is \(M = 44.01\space g/mol\). Using the formula \(n=\frac{m}{M}\), where \(m = 20.2\space g\).

$$n=\frac{20.2\space g}{44.01\space g/mol}\approx0.459\space mol$$

Step2: Convert the temperature to Kelvin

Using the formula \(T=(t + 273.15)\space K\), where \(t = 29^{\circ}C\).

$$T=(29+ 273.15)\space K=302.15\space K$$

Step3: Use the ideal gas law \(PV = nRT\) to find the pressure in part (a)

The ideal gas constant \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(V = 4.35\space L\), \(n = 0.459\space mol\), \(T = 302.15\space K\).
From \(P=\frac{nRT}{V}\),

$$P=\frac{0.459\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times302.15\space K}{4.35\space L}$$

$$P=\frac{0.459\times0.0821\times302.15}{4.35}\space atm$$

$$P\approx2.61\space atm$$

Step4: For part (b), find the partial pressure of \(CO_2\)

The partial pressure of \(CO_2\) is the same as in part (a) (because the amount of \(CO_2\), volume and temperature are the same), so \(P_{CO_2}=2.61\space atm\)
Convert the initial pressure of air from torr to atm. Since \(1\space atm = 760\space torr\), if \(P_{air}=715\space torr\), then \(P_{air}=\frac{715}{760}\space atm\approx0.941\space atm\)

Step5: Use Dalton's law of partial pressures \(P_{total}=P_1 + P_2\)

Here \(P_1 = P_{CO_2}\) and \(P_2=P_{air}\)

$$P_{total}=2.61\space atm+ 0.941\space atm$$

$$P_{total}=3.55\space atm$$

Answer:

(a) Pressure = \(2.61\space atm\)
(b) Partial pressure of \(CO_2=2.61\space atm\), Total pressure = \(3.55\space atm\)