Chapter 11: Problem 46
Due to the environmental concern of fluorocarbons as refrigerants, a refrigerant based on a mixture of hydrocarbons was used as a replacement. It is a patented blend of ethane, propane, butane, and isobutane. Isobutane has a normal boiling point of \(-12^{\circ} \mathrm{C}\). The molar specific heat of liquid phase and gas phase isobutane are \(129.7 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) and \(95.2 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) respectively. The heat of vaporization for this compound is \(21.3 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(25.0 \mathrm{~g}\) of isobutane from a liquid at \(-50^{\circ} \mathrm{C}\) to a gas at \(40^{\circ} \mathrm{C}\).
Short Answer
Step by step solution
Calculate Molar Mass of Isobutane
Convert Mass to Moles
Calculate Heat for Temperature Increase in Liquid Phase
Calculate Heat for Vaporization
Calculate Heat for Temperature Increase in Gas Phase
Calculate Total Heat Required
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Specific Heat Capacity
For isobutane, the problem specifies the specific heat capacities for both its liquid and gaseous states:
- Liquid phase: 129.7 J/mol-K
- Gas phase: 95.2 J/mol-K
- For the liquid phase: \( q = m \cdot C_{\text{liquid}} \cdot \Delta T \)
- For the gas phase: \( q = m \cdot C_{\text{gas}} \cdot \Delta T \)
Phase Change
For isobutane in the original problem, the critical phase change is vaporization—the transition from liquid to gas. This occurs at its boiling point, -12°C, and requires a specific amount of heat known as the heat of vaporization.
The heat of vaporization for isobutane is specified as 21.3 kJ/mol. This means that each mole of isobutane requires 21.3 kJ of energy to change from liquid to gas at the boiling point. The formula used for the heat required for vaporization is:
- \( q = n \cdot \Delta H_{\text{vap}} \)
Molar Mass
For isobutane, the chemical formula is \( C_4H_{10} \). To find its molar mass, sum the atomic masses of all the atoms contained in the molecule:
- Carbon (C): 12.01 g/mol, with four atoms contributing to a total of \( 4 \times 12.01 \) g/mol.
- Hydrogen (H): 1.008 g/mol, with ten atoms contributing to a total of \( 10 \times 1.008 \) g/mol.
- \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \)
Boiling Point
For isobutane, the boiling point is given as -12°C. This is the temperature at which isobutane will begin to vaporize. Understanding the boiling point is crucial for thermodynamic calculations, as it indicates when a phase change from liquid to gas will occur, and lets us apply the heat of vaporization at the correct moment in these calculations.
When a substance reaches its boiling point, any heat input will be used solely to transform the liquid into a gas, rather than increasing the temperature. This temperature plateau during a phase change is a significant characteristic of thermodynamics. The boiling point helps define the conditions under which a substance will change its phase, which is vital for any process involving heating or cooling of materials.