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Chapter 20: Problem 26

It is given that \(1.000 \mathrm{~g}\) of water becomes \(1676 \mathrm{~cm}^{3}\) of steam at \(100.0{ }^{\circ} \mathrm{C}\) and atmospheric pressure. How much work is done by the vapor when \(1.000\) g of water is converted to steam at atmospheric pressure?

Short Answer

Expert verified
The work done is approximately 169.5 joules.

Step by step solution

01

Identify the Known Variables

First, identify the known variables from the problem:1. Mass of water, \( m = 1.000 \text{ g} \).2. Volume of steam, \( V = 1676 \text{ cm}^3 \).3. Atmospheric pressure, \( P = 1.013 \times 10^5 \text{ Pa} \).To solve the problem, we need to calculate the work done by the vapor during the vaporization process.
02

Convert Units Where Necessary

The volume is given in cubic centimeters (\( \text{cm}^3 \)), and we need it in cubic meters (\( \text{m}^3 \)) since pressure is in Pascals (\( \text{Pa} \)).Use the conversion: \( 1 \text{ cm}^3 = 10^{-6} \text{ m}^3 \)So, \( V = 1676 \text{ cm}^3 = 1676 \times 10^{-6} \text{ m}^3 = 1.676 \times 10^{-3} \text{ m}^3 \).
03

Use the Work Equation for a Gas

The work done by a gas under constant pressure is given by the formula:\[ W = P \Delta V \]Where:- \( W \) is the work done,- \( P \) is the pressure (\( 1.013 \times 10^5 \text{ Pa} \)),- \( \Delta V \) is the change in volume, which equals the volume of steam since we assume initial volume of water is negligible compared to steam.Thus, \( \Delta V = 1.676 \times 10^{-3} \text{ m}^3 \).
04

Calculate the Work Done

Substitute the known values into the work equation:\[ W = 1.013 \times 10^5 \text{ Pa} \times 1.676 \times 10^{-3} \text{ m}^3 \]Calculate the multiplication:\[ W = 169.5 \text{ J} \]Thus, the work done by the vapor is approximately 169.5 joules.

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Key Concepts

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

Work Done by a Gas
When a substance like water turns into steam, it expands significantly in volume. This expansion means work is done by the gas (in this case, steam) as it pushes out against its surroundings. The concept of work done by a gas is essential in understanding many thermodynamic processes.

The work done is calculated using the equation:
  • \( W = P \Delta V \)
where:
  • \( W \) is the work done,
  • \( P \) is the pressure of the system,
  • \( \Delta V \) is the change in volume.

In this specific exercise, since the gas is expanding against atmospheric pressure, it does work on the environment. This is why knowing both pressure and volume change is critical to solving such problems. The amount of work done offers a glimpse into the energy transferred during the vaporization process.
Vaporization Process
The vaporization process is a phase transition where a liquid, like water, transforms into a gas or vapor. It's a key concept in thermodynamics. At the molecular level, this transition involves the breaking of intermolecular forces, allowing the molecules to enter a gaseous state.

During this process:
  • Heat energy is absorbed by the liquid.
  • The energy breaks the bonds holding the liquid molecules together.
  • The molecules then disperse and spread out, forming a vapor.

In this exercise, vaporization occurs at a constant atmospheric pressure of \(1.013 \times 10^5 \text{ Pa}\). The energy absorbed during vaporization becomes the work done by the gas as it changes the water into a much larger volume of steam without changing its temperature. Understanding this process gives insight into how energy is transferred and transformed during phase changes.
Unit Conversion
In thermodynamics, quantities like volume and pressure need to be consistently measured in the correct units to apply the laws and equations accurately. Unit conversion ensures consistency and correctness in calculations.

For instance:
  • Volume in the exercise is given in cubic centimeters (\( ext{cm}^3 \)). However, our work equation requires volume in cubic meters (\( ext{m}^3 \)).
  • To convert, it's essential to note that \( 1 ext{ cm}^3 = 10^{-6} ext{ m}^3 \).

By converting \(1676 ext{ cm}^3\) to cubic meters by multiplying by \(10^{-6}\), we get \(1.676 \times 10^{-3} ext{ m}^3\). This conversion is crucial for applying pressure in Pascals (\( ext{Pa} \)) correctly. Understanding unit conversion helps prevent errors in calculations and ensures accurate results.
Pressure and Volume Relationship
The relationship between pressure and volume is a cornerstone of thermodynamics, especially when looking at the behavior of gases. Under constant temperature, the relationship follows Boyle's Law which states that the pressure of a gas is inversely proportional to its volume.

In isobaric processes, such as this exercise, which occur at constant pressure:
  • The volume change is directly used to calculate the work done.
  • As volume increases, work is done by the gas because it expands against constant atmospheric pressure.

This exercise highlights the importance of recognizing the conditions under which a gas expands and understanding how work is calculated when volume changes under constant pressure. By grasping this relationship, we can gain deeper insights into energy conversion during physical processes.

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Most popular questions from this chapter

What is the net work output per cycle for the thermodynamic cycle in Fig. \(20-4\) ? We know that the net work output per cycle is the area enclosed by the \(P-V\) diagram. We estimate that in area \(A B C A\) there are 22 squares, each of area $$\left(0.5 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.1 \mathrm{~m}^{3}\right)=5 \mathrm{~kJ}$$ Therefore, Area enclosed by cycle \(\approx(22)(5 \mathrm{~kJ})=1 \times 10^{2} \mathrm{~kJ}\) The net work output per cycle is \(1 \times 10^{2} \mathrm{~kJ}\).

Three kilomoles (6.00 kg) of hydrogen gas at S.T.P. expands isobarically to precisely twice its volume. (a) What is the final temperature of the gas? (b) What is the expansion work done by the gas? ( \(c\) ) By how much does the internal energy of the gas change? ( \(d\) ) How much heat enters the gas during the expansion? For \(\mathrm{H}_{2}, c_{v}=\) \(10.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Assume the hydrogen will behave as an ideal gas. (a) From \(P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}\) with \(P_{1}=P_{2}\), $$T_{2}=T_{1}\left(\frac{V_{2}}{V_{1}}\right)=(273 \mathrm{~K})(2.00)=546 \mathrm{~K}$$ (b) Because 1 kmol at S.T.P. occupies \(22.4 \mathrm{~m}^{3}\), we have \(V_{1}=\) \(67.2 \mathrm{~m}^{3}\). Then $$\Delta W=P \Delta V=P\left(V_{2}-V_{1}\right)=\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(67.2 \mathrm{~m}^{3}\right)=6.8 \mathrm{MJ}$$ (c) To raise the temperature of this ideal gas by \(273 \mathrm{~K}\) at constant volume requires $$\Delta Q=c_{v} m \Delta T=(10.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(6.00 \mathrm{~kg})(273 \mathrm{~K})=16.4 \mathrm{MJ}$$ Because the volume is constant here, no work is done and \(\Delta Q\) equals the internal energy that must be added to the \(6.00 \mathrm{~kg}\) of \(\mathrm{H}_{2}\) to change its temperature from \(273 \mathrm{~K}\) to \(546 \mathrm{~K}\). Therefore, \(\Delta U=16.4 \mathrm{MJ} .\) ( \(d\) ) The system obeys the First Law during the process and so $$ \Delta Q=\Delta U+\Delta W=16.4 \mathrm{MJ}+6.8 \mathrm{MJ}=23.2 \mathrm{MJ} $$

A gas at a pressure of \(2.10 \times 10^{5}\) Pa occupies \(4.98 \times 10^{-3} \mathrm{~m}^{3}\) in a chamber that can change its volume. The gas is at an initial temperature of \(290 \mathrm{~K}\) when it is heated, so that it expands isobarically, thereupon doing \(200 \mathrm{~J}\) of work. Determine the new volume and the final temperature of the gas. [Hint: Use Eq. (20.1) and the Ideal Gas Law applied before and after the expansion.]

Water is boiled at \(100^{\circ} \mathrm{C}\) and \(1.0\) atm. Under these conditions, 1.0 g of water occupies \(1.0 \mathrm{~cm}^{3}, 1.0\) g of steam occupies 1670 \(\mathrm{cm}^{3}\), and \(L_{v}=540 \mathrm{cal} / \mathrm{g}\). Find \((a)\) the external work done when \(1.0\) g of steam is formed at \(100^{\circ} \mathrm{C}\) and \((b)\) the increase in internal energy.

Molecular oxygen having a mass of \(10.0\) g is in a cylinder sealed with a movable piston. The gas is heated from \(0.00^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\) at a constant pressure and expands. Given that \(c_{p}\) for \(\mathrm{O}_{2}\) is \(0.919\) \(\mathrm{kJ} / \mathrm{kg}\), how much heat was received by the gas?
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