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Chapter 19: Problem 14

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

Short Answer

Expert verified
(a) The work done is 166899 J. (b) The increase in internal energy is 2033101 J.

Step by step solution

01

Identify Given Values

We are given the following values:- Pressure, \( P = 2.00 \) atm, which is equivalent to \( P = 2.00 \times 101325 \) Pa (since 1 atm = 101325 Pa).- Heat of vaporization, \( L = 2.20 \times 10^6 \) J/kg.- Volume of 1 kg of water, \( V_1 = 1.00 \times 10^{-3} \) m\(^3\).- Volume of 1 kg of steam, \( V_2 = 0.824 \) m\(^3\).- Temperature, \( T = 120^\circ \)C, which is not needed for these calculations.
02

Calculate Work Done by Steam Formation

Work done when water transforms into steam is calculated as: \( W = P \Delta V \), where \( \Delta V = V_2 - V_1 \).- \( \Delta V = 0.824 - 1.00 \times 10^{-3} = 0.823 \) m\(^3\).- Pressure \( P = 2.00 \times 101325 = 202650 \) Pa.Substituting the values:\[ W = 202650 \times 0.823 = 166898.95 \text{ J} \].
03

Calculate Increase in Internal Energy

The first law of thermodynamics is \( \Delta U = Q - W \), where \( Q = L \times \text{mass} = 2.20 \times 10^6 \times 1 \) J for 1 kg of water.- Heat supplied, \( Q = 2.20 \times 10^6 \) J.- Work done, \( W = 166898.95 \) J (from Step 2).Substituting these values:\[ \Delta U = 2.20 \times 10^6 - 166898.95 = 2033101.05 \text{ J} \].

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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 concept of "Heat of Vaporization" is crucial in thermodynamics. It refers to the amount of heat energy required to convert a unit mass of a liquid into vapor without a change in temperature.
This process occurs at the boiling point under a specific pressure. The heat of vaporization is unique for every substance, and it helps explain why a particular amount of liquid can become gas.
For 1 kg of water at 2 atm pressure with a boiling point of 120掳C, the heat of vaporization is given as 2.20 脳 10鈦� J/kg. This means, to turn 1 kg of water into steam, 2.20 脳 10鈦� joules of energy must be supplied.
First Law of Thermodynamics
The "First Law of Thermodynamics" is an important principle, stating that energy cannot be created or destroyed, only transformed or transferred.
In terms of mathematical expression, it is given by: \[\Delta U = Q - W\] Where \( \Delta U \) is the change in internal energy, \( Q \) is heat added to the system, and \( W \) is the work done by the system. This concept is fundamental in understanding how energy changes during phase transitions.
In our exercise, it helps calculate the increase in internal energy when water turns into steam.
Internal Energy
"Internal Energy" is the total energy contained within a system due to the random motion of its molecules. It is affected by temperature, pressure, and the nature of the substance.
During the phase change from liquid to gas, heat energy is added, increasing the internal energy.
Using the First Law of Thermodynamics, if we know the heat added and the work done by the system, we can find how much the internal energy changes. In our case: \[\Delta U = 2.20 \times 10^6 - 166898.95 = 2033101.05 \text{ J}\] This tells us the total internal energy increase when 1 kg of water turns into steam at 2 atm pressure.
Boiling Point
The "Boiling Point" is the temperature at which the liquid's vapor pressure equals the surrounding pressure, causing it to turn into vapor.
It varies based on the pressure of the environment; for instance, water boils at 100掳C under 1 atm pressure but at a higher temperature, 120掳C under 2 atm pressure in our exercise.
This is because higher pressure requires more heat for the liquid molecules to overcome the atmospheric pressure and transition into the gas phase.
Work Done
When a liquid turns into gas at its boiling point, it performs "Work" to expand against the surrounding pressure. Work is done by the system as it expands in volume during phase transitions.
The work done can be calculated using the formula:\[W = P \Delta V\]where \( P \) is the pressure and \( \Delta V \) is the change in volume.
In our case, converting 1 kg of water at 1.00脳10鈦宦� m鲁 to steam at 0.824 m鲁 requires the system to do 166898.95 J of work against 2.00 atm (202650 Pa) pressure.
Pressure
"Pressure" is the force exerted per unit area on the surface of an object. In phase transitions, pressure plays a critical role since it affects the boiling point and the work done during the process.
In our scenario, the external pressure is 2.00 atm, converted to 202650 Pa for calculation purposes.
This pressure determines the amount of energy needed for the water to change into steam and the work done by the system during this phase change.
Volume Changes in Phase Transition
During a "Phase Transition" from liquid to gas, significant "Volume Changes" occur. This change is due to the molecules moving apart as they gain energy and overcome intermolecular forces.
In our example, the volume of water increases from 1.00脳10鈦宦� m鲁 to 0.824 m鲁 as it turns into steam.
These substantial volume changes are why steam can do work, such as driving turbines in engines, as it expands rapidly.

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Most popular questions from this chapter

During an isothermal compression of an ideal gas, 410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

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\(^5\) Pa and 300 K and occupies a volume of 1.50 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 K. This continues until the pressure reaches 2.50 \(\times\) 10\(^4\) Pa. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal. (a) In a \(pV\)-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?

Five moles of monatomic ideal gas have initial pressure 2.50 \(\times\) 10\(^3\) Pa and initial volume 2.10 m\(^3\). While undergoing an adiabatic expansion, the gas does 1480 J of work. What is the final pressure of the gas after the expansion?

Propane gas (\(C_3H_8\)) behaves like an ideal gas with \(_\Upsilon\) = 1.127. Determine the molar heat capacity at constant volume and the molar heat capacity at constant pressure.

Two moles of an ideal gas are compressed in a cylinder at a constant temperature of 65.0\(^\circ\)C until the original pressure has tripled. (a) Sketch a \(pV\)-diagram for this process. (b) Calculate the amount of work done.
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