Question

Dry ice is solid carbon dioxide. A \(0.050-\mathrm{g}\) sample of dry ice is placed in an evacuated 4.6-L vessel at \(30^{\circ} \mathrm{C}\). Calculate the pressure inside the vessel after all the dry ice has been converted to \(\mathrm{CO}_{2}\) gas.

Short answer

The pressure inside the vessel is approximately 0.00606 atm after all the dry ice is converted to CO_2 gas.

Step-by-step solution

Convert grams to moles

First, we need to convert the mass of dry ice (solid CO_2) into moles. The molar mass of CO_2 is approximately 44.01 g/mol. We can use the formula: \ \[\text{moles of } \mathrm{CO}_2 = \frac{\text{mass of } \mathrm{CO}_2}{\text{molar mass of } \mathrm{CO}_2}\] \ Therefore, \[\text{moles of } \mathrm{CO}_2 = \frac{0.050 \text{ g}}{44.01 \text{ g/mol}} \approx 0.001136 \text{ moles}\].

Convert temperature to Kelvin

The ideal gas law requires temperature in Kelvin. To convert Celsius to Kelvin, use the formula: \ \[ T(K) = T(\degree C) + 273.15 \] \ Given, \(T = 30 \degree C\), so: \( T = 30 + 273.15 = 303.15 \text{ K}\).

Use the ideal gas law equation

Next, apply the ideal gas law \(PV = nRT\) to find the pressure \(P\). We know: - Volume \(V = 4.6\text{ L}\) (convert to cubic meters for SI units: \(4.6 \times 10^{-3}\text{ m}^3\)),- Moles \(n \approx 0.001136\) (from Step 1), - Temperature \(T = 303.15\) K (from Step 2),- \(R = 0.0821\) L·atm/mol·K. \ Rearrange the equation for \(P\): \ \[ P = \frac{nRT}{V} \] \ \( P = \frac{0.001136 \times 0.0821 \times 303.15}{4.6} \).

Calculate the pressure

Plug the values into the equation: \ \[ P = \frac{0.001136 \times 0.0821 \times 303.15}{4.6} \approx 0.00606 \text{ atm} \]. \ This is the pressure inside the vessel after all the dry ice has converted to gas.

Key concepts

Understanding Molar Mass

Molar mass is a fundamental concept used to convert between the mass of a substance and the number of moles. It is defined as the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For carbon dioxide (CO\(_2\)), its molar mass can be calculated using the atomic masses of its constituent elements: carbon (C) and oxygen (O).

Carbon has an atomic mass of approximately 12.01 g/mol, and oxygen has an atomic mass of about 16.00 g/mol. CO\(_2\) consists of one carbon atom and two oxygen atoms, so:

• Molar mass of CO\(_2\) = 12.01 + 2(16.00) = 44.01 g/mol.

Knowing the molar mass allows us to convert a given mass of CO\(_2\) to moles using the formula:
\[\text{{moles of CO}}_2 = \frac{{\text{{mass of CO}}_2}}{{\text{{molar mass of CO}}_2}}\]
By performing this conversion, we bridge the gap between qualitative observations and quantitative measurements.

Converting Temperatures to Kelvin

Temperature conversion is critical for calculations involving the ideal gas law, as the formula requires temperature to be in Kelvin. The Kelvin scale is an absolute temperature scale starting from absolute zero, the theoretically lowest temperature possible.

Converting Celsius to Kelvin is straightforward. Simply add 273.15 to the Celsius temperature:

• \[ T(K) = T(\degree C) + 273.15 \]

This conversion ensures that any temperature used in thermodynamic calculations is positive, which aligns with the requirement for absolute temperature scales.

For instance, if you have a temperature of \(30 \degree C\):
\[T(K) = 30 + 273.15 = 303.15 \text{ K}\].
This step ensures accurate and reliable results when using formulas like the ideal gas law.

Calculating Pressure Using Ideal Gas Law

Pressure calculation using the ideal gas law involves understanding how gases behave under different conditions. The ideal gas law is expressed as \[ PV = nRT \], where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant, \(0.0821 \text{ L·atm/mol·K}\).
• \(T\) is the temperature in Kelvin.

To find the pressure \(P\), the equation can be rearranged to:
\[ P = \frac{nRT}{V} \]
By substituting the known values into this equation, you can determine the pressure exerted by the gas in a container. For example, using \(n = 0.001136\), \(R = 0.0821\), \(T = 303.15\, K\), and \(V = 4.6\, L\), the pressure can be calculated as:
\[ P = \frac{0.001136 \times 0.0821 \times 303.15}{4.6} \approx 0.00606 \text{ atm} \].
This method provides a fundamental insight into how changing volume, temperature, or moles of gas influences pressure.