When water is boiled at a pressure of 2.002.00 atm, the heat of vaporization is 2.20×1062.20\$$times10^6 J/kg $$and the boiling point is 120120°C. At this pressure, 1.001.00 kg of water has a volume of 1.00×10−31.00\$$times10^{-3} m3$$, and 1.001.00 kg of steam has a volume of 0.8240.824 m3. Compute the work done when 1.001.00 kg of steam is formed at this temperature.

Ch 19: The First Law of Thermodynamics

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 19: The First Law of Thermodynamics

Problem 14a

Chapter 19, Problem 14a

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 $120$ °C. At this pressure, $1.00$ kg of water has a volume of $1.00 \times 10^{- 3}$ m³, and $1.00$ kg of steam has a volume of $0.824$ m³. Compute the work done when $1.00$ kg of steam is formed at this temperature.

Verified step by step guidance

Identify the formula for work done during a phase change at constant pressure: $ W = P \Delta V $, where $ P $ is the pressure and $ \Delta V $ is the change in volume.

Convert the given pressure from atmospheres to pascals for consistency in SI units. Use the conversion: $ 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} $. Therefore, $ 2.00 \text{ atm} = 2.00 \times 1.013 \times 10^5 \text{ Pa} $.

Calculate the change in volume $ \Delta V $ by subtracting the initial volume of water from the final volume of steam: $ \Delta V = V_{\text{steam}} - V_{\text{water}} = 0.824 \text{ m}^3 - 1.00 \times 10^{-3} \text{ m}^3 $.

Substitute the values of pressure $ P $ and change in volume $ \Delta V $ into the work formula: $ W = (2.00 \times 1.013 \times 10^5 \text{ Pa}) \times (0.824 \text{ m}^3 - 1.00 \times 10^{-3} \text{ m}^3) $.

Simplify the expression to find the work done in joules. This will give you the work done when 1.00 kg of steam is formed at the given temperature and pressure.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a unit mass of a liquid into vapor without a temperature change. For water at 2.00 atm, this value is 2.20 * 10^6 J/kg. It is crucial for understanding the energy involved in phase transitions, particularly when calculating the energy needed to boil water at a given pressure.

Work Done by Expanding Gas

The work done by a gas during expansion can be calculated using the formula W = PΔV, where P is the pressure and ΔV is the change in volume. In this scenario, the work done is the product of the pressure (2.00 atm converted to Pascals) and the volume change from liquid to steam, which is essential for determining the energy associated with the phase change.

Boiling Point and Pressure Relationship

The boiling point of a liquid increases with pressure. At 2.00 atm, water boils at 120°C instead of the standard 100°C at 1 atm. This relationship is crucial for understanding how pressure affects phase changes and the conditions under which water transitions from liquid to steam, impacting calculations of energy and work in thermodynamic processes.

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Textbook Question

A gas in a cylinder is held at a constant pressure of $1.80 \times 10^{5}$ Pa and is cooled and compressed from $1.70$ m³ to $1.20$ m³. The internal energy of the gas decreases by $1.40 \times 10^{5}$ J. Does it matter whether the gas is ideal? Why or why not?

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Textbook Question

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?

Textbook Question

The $p V$ -diagram in Fig. E $19.13$ shows a process $a b c$ involving $0.450$ mol of an ideal gas. How much heat had to be added during the process to increase the internal energy of the gas by $15 comma 000$ J?

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Textbook Question

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 $120$ °C. At this pressure, $1.00$ kg of water has a volume of $1.00$\times$10^{-3}$ m³, and $1.00$ kg of steam has a volume of $0.824$ m³. Compute the increase in internal energy of the water.

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Textbook Question

An ideal gas is taken from $a$ to $b$ on the $p V$ -diagram shown in Fig. E $19.15$ . During this process, $700$ J of heat is added and the pressure doubles. How does the internal energy of the gas at $a$ compare to the internal energy at $b$ ? Be specific and explain.

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Textbook Question

The process $a b c$ shown in the $p V$ -diagram in Fig. E $19.11$ involves $0.0175$ mol of an ideal gas. What was the lowest temperature the gas reached in this process? Where did it occur?

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