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Chapter 11: Problem 91

Liquid butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is stored in cylinders to be used as a fuel. The normal boiling point of butane is listed as \(-0.5^{\circ} \mathrm{C}\). (a) Suppose the tank is standing in the sun and reaches a temperature of \(35^{\circ} \mathrm{C}\). Would you expect the pressure in the tank to be greater or less than atmospheric pressure? How does the pressure within the tank depend on how much liquid butane is in it? (b) Suppose the valve to the tank is opened and a few liters of butane are allowed to escape rapidly. What do you expect would happen to the temperature of the remaining liquid butane in the tank? Explain. (c) How much heat must be added to vaporize \(250 \mathrm{~g}\) of butane if its heat of vaporization is \(21.3 \mathrm{~kJ} / \mathrm{mol} ?\) What volume does this much butane occupy at 755 torr and \(35^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) The pressure in the tank would be greater than atmospheric pressure due to the higher temperature causing butane to be in vapor form and exerting more pressure. The pressure depends on the amount of liquid butane present in the tank. (b) The temperature of the remaining liquid butane would decrease due to adiabatic cooling when some butane escapes. (c) The heat required to vaporize \(250 \mathrm{~g}\) of butane is \(91.59 \mathrm{~kJ}\), and the volume it occupies at 755 torr and \(35^{\circ} \mathrm{C}\) is approximately \(108.92 \mathrm{L}\).

Step by step solution

01

Identify the condition of butane in the tank

Since the temperature of the tank is \(35^{\circ} \mathrm{C}\), which is higher than the normal boiling point (-0.5°C) of butane, the substance is in the vapor phase.
02

Determine pressure in the tank compared to atmospheric pressure

When butane is in vapor form, it exerts pressure on the walls of the container by the gas molecules colliding with the walls. Since the temperature is higher, the butane molecules have more kinetic energy and move faster, resulting in an increased pressure in the tank compared to atmospheric pressure.
03

Dependence of pressure on the amount of liquid butane

As the amount of liquid butane decreases, the amount of vapor in the tank also decreases. This results in a decrease of pressure on the walls of the tank. In general, the pressure within the tank is dependent on the amount of liquid butane present. ##Part (b)##
04

Predict the result on temperature when some butane escapes

When some butane is allowed to escape, the pressure inside the tank is reduced. As a result, the remaining liquid butane undergoes adiabatic cooling, and the temperature of the remaining liquid butane decreases. ##Part (c)##
05

Calculate heat required to vaporize 250 g of butane

Using the heat of vaporization provided (\(21.3 \mathrm{~kJ} / \mathrm{mol}\)), we can calculate the heat required: First, convert 250 g of butane to moles by dividing by the molar mass of butane: \((\dfrac{250 \mathrm{~g}}{58.12~ \mathrm{g/mol}})= 4.3 \mathrm{~mol}\). Now, multiply the moles by heat of vaporization to find the heat required : \((4.3 \mathrm{~mol} \times 21.3 \mathrm{~kJ/mol}) = 91.59 \mathrm{~kJ}\).
06

Calculate the volume of vaporized butane

To calculate the volume of vaporized butane at 755 torr and \(35^{\circ} \mathrm{C}\), use the Ideal Gas Law equation: \(PV = nRT\). First, convert the temperature to Kelvin: \(35^{\circ} \mathrm{C} + 273.15 = 308.15 \mathrm K\). Convert the pressure to atm: \(\dfrac{755 \mathrm{~torr}}{760 \mathrm{~torr/atm}} = 0.993 \mathrm{~atm}\). Now, rearrange the Ideal Gas Law equation and substitute the values: \(V = \dfrac{nRT}{P} = \frac{4.3 \mathrm{~mol} \times 0.0821 \dfrac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}\times 308.15 \mathrm K}{0.993 \mathrm{~atm}} = 108.92 \mathrm{L}\). So, the volume of the vaporized butane is approximately 108.92 L.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boiling Point of Butane
Understanding the boiling point of butane is crucial when considering its applications and storage. Butane, with the chemical formula \( \mathrm{C}_{4} \mathrm{H}_{10} \), is a hydrocarbon that changes from liquid to gas at its boiling point of \( -0.5^\circ \mathrm{C} \). At temperatures above this point, butane exists as a vapor, and when contained, this vapor exerts pressure on its container. This concept is especially relevant in situations where butane tanks are subjected to environmental temperature changes, such as being left in the sun.

When a butane tank’s temperature increases to \(35^\circ \mathrm{C} \), significantly above its boiling point, the butane within transitions to a gaseous state. The pressure inside the container will be greater than atmospheric pressure due to the increased kinetic energy and frequency of gas molecule collisions against the tank walls. However, as the amount of liquid butane in the tank decreases, the vapor space increases, leading to a drop in tank pressure. This relationship between vapor pressure and temperature is a fundamental aspect of the physical behavior of butane and other similar substances.
Vapor Pressure
Vapor pressure is an essential concept when dealing with volatile substances like butane. It refers to the pressure that is exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. Butane's vapor pressure increases with temperature because the molecules gain more kinetic energy, which enables more of them to escape from the liquid phase into the gaseous phase. As a result, the frequency and intensity of molecular collisions against the container's walls increase, raising the vapor pressure.

When discussing the storage tank in the exercise, this phenomenon explains why the pressure within a butane tank would be higher on a hot day. Such conditions can influence the safety and functionality of butane storage, necessitating pressure-release valves and other safety measures. This concept of vapor pressure is not only vital for understanding how to safely store butane but also plays a pivotal role in numerous industrial processes where control of pressure is synonymous with control of chemical reactions and mechanisms.
Heat of Vaporization
The heat of vaporization is the amount of energy required to convert a substance from liquid to gas without changing its temperature. For butane, this value is \(21.3 \mathrm{~kJ} / \mathrm{mol} \). It quantifies the energy needed to break the intermolecular forces that hold the liquid molecules together.

In the exercise context, calculating the heat to vaporize \(250 \mathrm{~g}\) of butane requires converting the mass to moles and then using the heat of vaporization to determine the total energy needed. The resulting calculation involved simple mole conversion and multiplication to find that \(91.59 \mathrm{~kJ}\) of heat would be necessary. Furthermore, when considering the change of state from liquid to gas from a physical perspective, this heat absorption will cause the butane to occupy a much larger volume as a vapor, as calculated using the Ideal Gas Law in the solution. This concept plays a significant role in designing heating and cooling systems, particularly when dealing with substances that change phases in the operational temperature range.

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Most popular questions from this chapter

Ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) is the major component of antifreeze. It is a slightly viscous liquid, not very volatile at room temperature, with a boiling point of \(198^{\circ} \mathrm{C}\). Pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right),\) which has about the same molecular weight, is a nonviscous liquid that is highly volatile at room temperature and whose boiling point is \(36.1^{\circ} \mathrm{C}\). Explain the differences in the physical properties of the two substances.

The following data present the temperatures at which certain vapor pressures are achieved for dichloromethane \(\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)\) and methyl iodide \(\left(\mathrm{CH}_{3} \mathrm{I}\right)\) : $$ \begin{array}{lllll} \text { Vapor Pressure } & & & & \\ \text { (torr): } & 10.0 & 40.0 & 100.0 & 400.0 \\ \hline T \text { for } \mathrm{CH}_{2} \mathrm{Cl}_{2}\left({ }^{\circ} \mathrm{C}\right): & -43.3 & -22.3 & -6.3 & 24.1 \\ T \text { for } \mathrm{CH}_{3} \mathrm{I}\left({ }^{\circ} \mathrm{C}\right): & -45.8 & -24.2 & -7.0 & 25.3 \end{array} $$ (a) Which of the two substances is expected to have the greater dipole-dipole forces? Which is expected to have the greater dispersion forces? Based on your answers, explain why it is difficult to predict which compound would be more volatile. (b) Which compound would you expect to have the higher boiling point? Check your answer in a reference book such as the CRC Handbook of Chemistry and Physics. (c) The order of volatility of these two substances changes as the temperature is increased. What quantity must be different for the two substances in order for this phenomenon to occur? (d) Substantiate your answer for part (c) by drawing an appropriate graph.

The critical temperatures \((\mathrm{K})\) and pressures \((\mathrm{atm})\) of a series of halogenated methanes are as follows: $$ \begin{array}{lcccc} \text { Compound } & \mathbf{C C l}_{3} \mathbf{F} & \mathbf{C C l}_{2} \mathbf{F}_{2} & \mathbf{C C I F}_{3} & \mathbf{C F}_{4} \\ \hline \text { Critical temperature } & 471 & 385 & 302 & 227 \\ \text { Critical pressure } & 43.5 & 40.6 & 38.2 & 37.0 \end{array} $$ (a) List the intermolecular forces that occur for each compound. (b) Predict the order of increasing intermolecular attraction, from least to most, for this series of compounds. (c) Predict the critical temperature and pressure for \(\mathrm{CCl}_{4}\) based on the trends in this table. Look up the experimentally determined critical temperatures and pressures for \(\mathrm{CCl}_{4}\), using a source such as the CRC Handbook of Chemistry and Physics, and suggest a reason for any discrepancies.

Name the phase transition in each of the following situations and indicate whether it is exothermic or endothermic: (a) When ice is heated, it turns to water. (b) Wet clothes dry on a warm summer day. (c) Frost appears on a window on a cold winter day. (d) Droplets of water appear on a cold glass of beer.

The boiling points, surface tensions, and viscosities of water and several alchohols are as follows: $$ \begin{array}{lrcc} & \begin{array}{l} \text { Boiling } \\ \text { Point }\left({ }^{\circ} \mathbf{C}\right) \end{array} & \begin{array}{l} \text { Surface } \\ \text { Tension }\left(\mathbf{J} / \mathbf{m}^{2}\right) \end{array} & \begin{array}{l} \text { Viscosity } \\ (\mathbf{k g} / \mathbf{m}-\mathbf{s}) \end{array} \\ \hline \text { Water, } \mathrm{H}_{2} \mathrm{O} & 100 & 7.3 \times 10^{-2} & 0.9 \times 10^{-3} \\ \text {Ethanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} & 78 & 2.3 \times 10^{-2} & 1.1 \times 10^{-3} \\ \text {Propanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 97 & 2.4 \times 10^{-2} & 2.2 \times 10^{-3} \\ n \text { -Butanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 117 & 2.6 \times 10^{-2} & 2.6 \times 10^{-3} \\\ \text {Ethylene glycol, } \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 197 & 4.8 \times 10^{-2} & 26 \times 10^{-3} \end{array} $$ (a) For ethanol, propanol, and \(n\) -butanol the boiling points, surface tensions, and viscosities all increase. What is the reason for this increase? (b) How do you explain the fact that propanol and ethylene glycol have similar molecular weights \((60\) versus \(62 \mathrm{amu}),\) yet the viscosity of ethylene glycol is more than 10 times larger than propanol? (c) How do you explain the fact that water has the highest surface tension but the lowest viscosity?
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