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Sublimation of \(1.0 \mathrm{g}\) of dry ice, \(\mathrm{CO}_{2}(\mathrm{s}),\) forms \(0.36 \mathrm{L}\) of \(\mathrm{CO}_{2}(\mathrm{g})\) (at \(-78^{\circ} \mathrm{C}\) and 1 atm pressure). The expanding gas can do work on the surroundings. Calculate the amount of work done on the surroundings using the equation \(w=-P \times \Delta V\) (Note: \(L \times\) atm is a unit of energy; 1 L atm \(=101.3 \mathrm{J}\).)

Short Answer

Expert verified
The work done is approximately \(-36.47 \text{ J}\).

Step by step solution

01

Identify Given Values

We have the following given values:- Volume of gas, \( V = 0.36 \text{ L} \)- Pressure, \( P = 1 \text{ atm} \)
02

Calculate Change in Volume

Assume the initial volume of the solid is negligible. Therefore, \( \Delta V = V_{final} - V_{initial} = 0.36 \text{ L} - 0 \text{ L} = 0.36 \text{ L} \).
03

Apply Work Formula

Use the formula for work: \[ w = -P \times \Delta V \]Substitute the values:\[ w = -(1 \text{ atm}) \times (0.36 \text{ L}) = -0.36 \text{ L atm} \]
04

Convert Work to Joules

Convert the work from L atm to Joules using the conversion factor:\[ 1 \text{ L atm} = 101.3 \text{ J} \]\[ w = -0.36 \text{ L atm} \times 101.3 \text{ J/L atm} = -36.468 \text{ J} \]
05

State Final Answer

The amount of work done on the surroundings is approximately \(-36.47 \text{ J}\).

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Key Concepts

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

Sublimation
Sublimation is a fascinating process where a substance transitions directly from a solid to a gas without passing through the liquid state. This is exactly what happens with dry ice, or carbon dioxide (\(\text{CO}_{2}\)) in solid form. When you see foggy fumes emerging from dry ice, it's the result of sublimation. This process occurs because the molecules in dry ice absorb energy, typically in the form of heat from the surroundings, which allows them to break free from their fixed solid arrangement and become gaseous.Sublimation has practical applications in fields such as chemistry and food preservation. For example:
  • Freeze-drying foods preserves them by removing water content through sublimation.
  • Manufacturers use sublimation printing for high-quality images on fabrics and materials.
Understanding sublimation helps explain interesting phenomenons and showcases the versatility of physical matter changes.
Work Done on Surroundings
In thermodynamics, the concept of work is essential to understanding energy transfer processes. When gas expands, it can do work on its surroundings by pushing against an external pressure. For example, in our dry ice scenario, as the carbon dioxide gas forms and expands, it does work on its surroundings.We use the equation \(w = -P \times \Delta V\) to calculate this work:
  • \(w\) represents the work done (in energy units)
  • \(P\) is the external pressure acting on the system
  • \(\Delta V\) denotes the change in volume
It's important to note that the work done is negative, which signifies energy is leaving the system (as the gas expands). Whenever a system does work on its surroundings, it's losing energy, which in turn reflects as a negative sign in calculations.
Conversion of Units
Converting units is a necessary and vital skill in any scientific calculations. Units must be consistent to ensure accuracy and meaningful results. In our dry ice example, after calculating the work done in \(\text{L atm}\), it's crucial to convert these units to joules.Here’s the conversion process:
  • Recognize that \(1 \text{ L atm}\) is equivalent to \(101.3 \text{ J}\)
  • Multiply the work value in liters atm by this conversion factor
This gives us the work value in joules. This step is indispensable when aligning with standard scientific interpretations and calculations, allowing for universal understanding and application of results.
Ideal Gas Law
The ideal gas law is a fundamental principle that provides the relationship between pressure, volume, temperature, and the number of moles of a gas. Described by the equation \(PV = nRT\), where:
  • \(P\) stands for pressure
  • \(V\) denotes volume
  • \(n\) represents moles of gas
  • \(R\) is the ideal gas constant
  • \(T\) signifies temperature in Kelvin
In studying dry ice sublimation, understanding this law gives insight into how changes in temperature and pressure can affect gas behavior. Although our specific problem primarily dealt with work calculation, it’s essential to comprehend the larger gas behavior context that the ideal gas law provides.Keep in mind, the ideal gas law is an approximation and works best under conditions of low pressure and high temperature, where gases behave more ideally. This framework simplifies many complex calculations and offers invaluable insights into the dynamics of gases.

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Most popular questions from this chapter

A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The bomb had a heat capacity of \(550 \mathrm{J} / \mathrm{K},\) and the calorimeter contained \(650 \mathrm{g}\) of water. Burning \(4.20 \mathrm{g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)\) resulted in a rise in temperature from \(18.5^{\circ} \mathrm{C}\) to \(22.3^{\circ} \mathrm{C} .\) Calculate the enthalpy of combustion of ethanol, in \(\mathrm{kJ} / \mathrm{mol}\).

You want to heat the air in your house with natural gas \(\left(\mathrm{CH}_{4}\right) .\) Assume your house has \(275 \mathrm{m}^{2}\) (about \(2800 \mathrm{ft}^{2}\) ) of floor area and that the ceilings are 2.50 m from the floors. The air in the house has a molar heat capacity of \(29.1 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K} .\) (The number of moles of air in the house can be found by assuming that the average molar mass of air is \(28.9 \mathrm{g} / \mathrm{mol}\) and that the density of air at these temperatures is \(1.22 \mathrm{g} / \mathrm{L} .\). What mass of methane do you have to burn to heat the air from \(15.0^{\circ} \mathrm{C}\) to \(22.0^{\circ} \mathrm{C} ?\)

Identify whether the following processes are exothermic or endothermic. (a) the reaction of \(\mathrm{Na}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\) (b) cooling and condensing gaseous \(N_{2}\) to form liquid \(\mathrm{N}_{2}\) (c) cooling a soft drink from \(25^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\) (d) heating \(\mathrm{HgO}(\mathrm{s})\) to form \(\mathrm{Hg}(\ell)\) and \(\mathrm{O}_{2}(\mathrm{g})\)

What does the term standard state mean? What are the standard states of the following substances at \(298 \mathrm{K}\) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{NaCl}, \mathrm{Hg}, \mathrm{CH}_{4} ?\)

Chloromethane, \(\mathrm{CH}_{3} \mathrm{Cl}\), arises from microbial fermentation and is found throughout the environment. It is also produced industrially, is used in the manufacture of various chemicals, and has been used as a topical anesthetic. How much energy is required to convert \(92.5 \mathrm{g}\) of liquid to a vapor at its boiling point, \(-24.09^{\circ} \mathrm{C} ?\) (The heat of vaporization of \(\mathrm{CH}_{3} \mathrm{Cl}\) is \(21.40 \mathrm{kJ} / \mathrm{mol}\).)

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