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Chapter 6: Problem 15

At constant temperature, a sample of helium gas expands from \(922 \mathrm{~mL}\) to \(1.14 \mathrm{~L}\) against a pressure of \(2.33 \mathrm{~atm} .\) Find \(w\) (in \(\mathrm{J}\) ) done by the gas \((101.3 \mathrm{~J}=1 \mathrm{~atm} \cdot \mathrm{L})\)

Short Answer

Expert verified
The work done by the gas is \text{-51.51 J}.

Step by step solution

01

- Convert Initial Volume

First, notice that the initial volume is given in \text{mL}. We need to convert this value to \text{L}. \[ 1 \text{~mL} = 0.001 \text{~L} \] So, the initial volume in liters is: \[ 922 \text{~mL} \times 0.001 \text{~L/mL} = 0.922 \text{~L} \]
02

- Calculate the Change in Volume

Now, calculate the change in volume (\text{ΔV}). \[ \text{ΔV} = \text{Final Volume} - \text{Initial Volume} \] Given the final volume is \text{1.14 L}, and the initial volume is \text{0.922 L}: \[ \text{ΔV} = 1.14 \text{~L} - 0.922 \text{~L} = 0.218 \text{~L} \]
03

- Use the Work Formula

To find the work (w) done by the gas, use the following relation: \[ w = -P \times \text{ΔV} \] Here, P = 2.33 \text{~atm} and \text{ΔV} = 0.218 \text{~L} \[ w = -2.33 \text{~atm} \times 0.218 \text{~L} \] \[ w = -0.50874 \text{~atm} \times \text{L} \]
04

- Convert Work to Joules

Finally, convert the work from \text{atm} \times \text{L} to \text{J}. Given the conversion factor: \[ 1 \text{~atm} \times \text{L} = 101.3 \text{~J} \] \[ w = -0.50874 \text{~atm} \times \text{L} \times 101.3 \text{~J/atm} \times \text{L} \] \[ w = -51.51 \text{~J} \]

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that studies how heat, work, and energy interact. In our example with the helium gas, we are dealing with a concept called **work done by gas**. When the gas expands, it does work on its surroundings. Thermodynamics helps us understand these energy changes. By investigating the principles of thermodynamics, we can predict how the gas will behave under different conditions which is crucial for knowing how much work it can do.
Gas Laws
Gas laws describe how gases behave in terms of these parameters: pressure (P), volume (V), and temperature (T). One of the most important gas laws at constant temperature is Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume. This means if the volume increases, the pressure decreases, as long as the temperature stays constant. In the given exercise, the pressure is constant and our main focus is on how the volume of the gas changes and impacts the work done.
Pressure-Volume Work
Pressure-volume work occurs when a gas expands or compresses, changing its volume. It is calculated using the formula: \[ w = -P \times \Delta V \] Here, **P** is the constant external pressure and **ΔV** is the change in volume. The negative sign indicates work done **by** the gas. If the gas expands, ΔV will be positive, and the work will be negative, meaning the gas loses energy. In our exercise, we converted all measurements to the correct units first, calculated the change in volume, applied the work formula, and finally converted the result to joules. By understanding these steps, one can see how energy is transferred when a gas changes its volume.

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Most popular questions from this chapter

Is \(\Delta H\) positive or negative when \(1 \mathrm{~mol}\) of water vapor condenses to liquid water? Why? How does this value compare with \(\Delta H\) for the vaporization of 2 mol of liquid water to water vapor?

One piece of copper jewelry at \(105^{\circ} \mathrm{C}\) has twice the mass of another piece at \(45^{\circ} \mathrm{C}\). Both are placed in a calorimeter of negligible heat capacity. What is the final temperature inside the calorimeter \((c\) of copper \(=0.387 \mathrm{~J} / \mathrm{g} \cdot \mathrm{K}) ?\)

Diamond and graphite are two crystalline forms of carbon. At 1 atm and \(25^{\circ} \mathrm{C},\) diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the list below, determine \(\Delta H\) for C(diamond) \(\longrightarrow\) C (graphite) (1) C(diamond) \(+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-395.4 \mathrm{~kJ}\) (2) \(2 \mathrm{CO}_{2}(g) \longrightarrow 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) \(\Delta H=566.0 \mathrm{~kJ}\) (3) C(graphite) \(+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-393.5 \mathrm{~kJ}\) (4) \(2 \mathrm{CO}(g) \longrightarrow \mathrm{C}\) ( graphite) \(+\mathrm{CO}_{2}(g) \quad \Delta H=-172.5 \mathrm{~kJ}\)

If you feel warm after exercising, have you increased the internal energy of your body? Explain.

Compounds of boron and hydrogen are remarkable for their unusual bonding (described in Section 14.5 ) and for their reactivity. With the more reactive halogens, for example, diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) forms trihalides even at low temperatures: $$\begin{array}{r}\mathrm{B}_{2} \mathrm{H}_{6}(g)+6 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{BCl}_{3}(g)+6 \mathrm{HCl}(g) \\\\\Delta H=-755.4 \mathrm{~kJ}\end{array}$$ What is \(\Delta H\) per kilogram of diborane that reacts?
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