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\(^\circ\)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

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

Step-by-step solution

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.

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:\[ W = 202650 \times 0.823 = 166898.95 \text{ J} \].

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:\[ \Delta U = 2.20 \times 10^6 - 166898.95 = 2033101.05 \text{ J} \].

Key concepts

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°C, the heat of vaporization is given as 2.20 × 10⁶ J/kg. This means, to turn 1 kg of water into steam, 2.20 × 10⁶ 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°C under 1 atm pressure but at a higher temperature, 120°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 1.00×10⁻³ m³ to steam at 0.824 m³ 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 1.00×10⁻³ m³ to 0.824 m³ 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.