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\(\cdot\) Boiling Water at High Pressure. When water is boiled at a pressure of \(2.00 \mathrm{~atm},\) the heat of vaporization is \(2.20 \times 10^{6} \mathrm{~J} / \mathrm{kg}\) and the boiling point is \(120^{\circ} \mathrm{C}\). At this pressure, \(1.00 \mathrm{~kg}\) of water has a volume of \(1.00 \times 10^{-3} \mathrm{~m}^{3},\) and \(1.00 \mathrm{~kg}\) of steam has a volume of \(0.824 \mathrm{~m}^{3}\). (a) Compute the work done when \(1.00 \mathrm{~kg}\) of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.

Short Answer

Expert verified
The work done when 1.00 kg of steam is formed at this temperature is \(1.67 \times 10^{5} J\). The increase in internal energy of the water is \(2.03 \times 10^{6} J\).

Step by step solution

01

Calculate the Change in Volume

First, we need to calculate the change in volume when water is converted into steam. This can be done by subtracting the initial volume of water (1 kg) from the final volume of steam (1 kg), which gives us \(\Delta V = V_{\text{steam}} - V_{\text{water}} = 0.824 m^3 - (1.00 \times 10^{-3}) m^3 = 0.823 m^3\).
02

Calculate the Work Done

To find the work done, we use the pressure-volume work formula given by \(W = P \Delta V\), where the pressure P is given as 2 atm. However, we need to convert this pressure to Pascals (Pa) as 1 atm equals \(1.013 \times 10^{5}\) Pa. Therefore, the work done is \(W = (2.00 \times 1.013 \times 10^{5} Pa) \times 0.823 m^3 = 1.67 \times 10^{5} J\).
03

Calculate the Heat Added

The heat added to the system equals the heat of vaporization times the mass of water, \( Q = m \Delta H_v = (1.00 kg) \times (2.20 \times 10^{6} J/kg) = 2.20 \times 10^{6} J\).
04

Calculate the Increase in Internal Energy

Lastly, we can find the increase in internal energy (\(\Delta U\)) using the formula \(\Delta U = Q - W\). Substituting our calculated values, we get \(\Delta U = 2.20 \times 10^{6} J - 1.67 \times 10^{5} J =2.03 \times 10^{6} J\).

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Key Concepts

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

Heat of Vaporization
Understanding the heat of vaporization is critical when studying the transformation of water into steam. This term refers to the amount of energy required to convert one kilogram of a substance from a liquid to a gaseous state at its boiling point.
For water boiled at a pressure of 2.00 atm, the heat of vaporization is given as 2.20 x 106 J/kg. To visualize the significance of this value, imagine it as the energy equivalent needed to run a small electric heater for nearly an hour - all being absorbed by a single kilogram of water to vaporize. This substantial energy input doesn't increase the temperature but rather provides the necessary energy for water molecules to overcome their intermolecular forces and transition from a dense liquid to a more dispersed gaseous state.
Work Done by Steam
The phrase 'work done by steam' might conjure images of old locomotives powered by steam engines, yet the principle applies to any steam generating system. Work in a thermodynamic sense is the energy transferred by the system to its surroundings and for steam, this happens when it expands.
In our exercise, the work done when steam is formed at a high temperature is calculated by considering the change in volume it undergoes when it transitions from water to steam. Steam doing work on its environment is an essential principle in engines and turbines, where this expansion is utilized to move pistons or blades, converting thermal energy into mechanical energy.
Internal Energy
Within the fascinating field of thermodynamics, internal energy is the sum of all the energy stored by the molecules of a system. It encompasses kinetic energy due to their movement and potential energy resulting from the intermolecular forces between them.
In our scenario, the increase in internal energy of water when it turns into steam is deduced by considering the work done by the steam and the heat added to it. The key is recognizing that while heat increases the energy of the system, the work done by the system, as it expands, actually reduces its internal energy. Thus, the net change in internal energy is the balance between these two amounts.
Pressure-Volume Work
Dive into the concept of pressure-volume work, which occurs when there is a volume change under constant pressure. Imagine a balloon expanding or contracting; the work done by or on the balloon is an example of pressure-volume work.
In mathematical terms, we represent it with the formula: W = P ΔV. Essentially, to compute the work when steam is formed from water, such as in the exercise, one multiplies the constant pressure the steam is under by the change in its volume as it turns from liquid to gas. Since work is energy transferred, understanding pressure-volume work is fundamental to solving many thermodynamic problems revolving around heat engines and refrigerators where gases expand and compress.

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Most popular questions from this chapter

An air pump has a cylinder \(0.250 \mathrm{~m}\) long with a movable piston. The pump is used to compress air from the atmosphere (at absolute pressure \(1.01 \times 10^{5} \mathrm{~Pa}\) ) into a very large tank at \(3.80 \times 10^{5} \mathrm{~Pa}\) gauge pressure. (For air, \(\left.C_{V}=20.8 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} .\right)\) (a) The piston begins the compression stroke at the open end of the cylinder. How far down the length of the cylinder has the piston moved when air first begins to flow from the cylinder into the tank? Assume that the compression is adiabatic. (b) If the air is taken into the pump at \(27.0^{\circ} \mathrm{C},\) what is the temperature of the compressed air? (c) How much work does the pump do in putting \(20.0 \mathrm{~mol}\) of air into the tank?

A quantity of air is taken from state \(a\) to state \(b\) along a path that is a straight line in the \(p V\) -diagram (Fig. \(\mathbf{P 1 9 . 3 5}\) ). (a) In this process, does the temperature of the gas increase, decrease, or stay the same? Explain. (b) If \(\quad V_{a}=0.0700 \mathrm{~m}^{3}\), \(V_{b}=0.1100 \mathrm{~m}^{3}, p_{a}=1.00 \times 10^{5} \mathrm{~Pa}\) and \(p_{b}=1.40 \times 10^{5} \mathrm{~Pa},\) what is the work \(W\) done by the gas in this process? Assume that the gas may be treated as ideal.

In a hospital, pure oxygen may be delivered at 50 psi (gauge pressure) and then mixed with \(\mathrm{N}_{2} \mathrm{O}\). What volume of oxygen at \(20^{\circ} \mathrm{C}\) and 50 psi (gauge pressure) should be mixed with \(1.7 \mathrm{~kg}\) of \(\mathrm{N}_{2} \mathrm{O}\) to get a \(50 \% / 50 \%\) mixture by volume at \(20^{\circ} \mathrm{C} ?\) (a) \(0.21 \mathrm{~m}^{3} ;\) (b) \(0.27 \mathrm{~m}^{3}\); (c) \(1.9 \mathrm{~m}^{3} ;\) (d) \(100 \mathrm{~m}^{3}\).

A gas undergoes two processes. In the first, the volume remains constant at \(0.200 \mathrm{~m}^{3}\) and the pressure increases from \(2.00 \times 10^{5} \mathrm{~Pa}\) to \(5.00 \times 10^{5} \mathrm{~Pa}\). The second process is a compression to a volume of \(0.120 \mathrm{~m}^{3}\) at a constant pressure of \(5.00 \times 10^{5} \mathrm{~Pa}\). (a) In a \(p V\) -diagram, show both processes. (b) Find the total work done by the gas during both processes.

A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at \(1.00 \times 10^{5} \mathrm{~Pa}\) and \(300 \mathrm{~K}\) and occupies a volume of \(1.50 \mathrm{~L}\). The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature at \(300 \mathrm{~K}\). This continues until the pressure reaches \(2.50 \times 10^{4} \mathrm{~Pa}\). In the second process, the gas is compressed at constant pressure until it returns to its original volume of \(1.50 \mathrm{~L}\). Assume that the gas may be treated as ideal. (a) In a \(p V\) -diagram, show both processes. (b) Find the volume of the gas at the end of the first process, and the pressure and temperature at the end of the second process. (c) Find the total work done by the gas during both processes. (d) What would you have to do to the gas to return it to its original pressure and temperature?

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