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Draw a cylinder with a movable piston containing six molecules of a liquid. A pressure of 1 atm is exerted on the piston. Next draw the same cylinder after the liquid has been vaporized. A pressure of one atmosphere is still exerted on the piston. Is work done on the system or by the system?

Short Answer

Expert verified
Answer: During the vaporization process under a constant pressure of 1 atm, work is being done by the system as the molecules change from the liquid state to the vapor state and push the piston up.

Step by step solution

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01

Draw the initial state with liquid molecules

Draw a cylinder with a movable piston at the top. Inside the cylinder, sketch six molecules of liquid. Label this drawing as "Initial State." It's important to have a clear visualization of this scenario where a pressure of 1 atm is exerted on the piston, and the molecules are in a liquid state.
02

Draw the final state with vaporized molecules

Draw another cylinder, similar to the one in Step 1, with the same movable piston at the top. Inside the cylinder, sketch six molecules, but this time, represent them as vaporized (gaseous) molecules. Label this drawing as "Final State." Notice that the piston is pushed up in comparison to the initial state as the vaporized molecules require more space than liquid molecules, exerting a pressure of 1 atm on the piston.
03

Observe and compare piston position

Compare the initial and final state drawings and observe the position of the pistons. In the final state, as the pressure inside the cylinder is still 1 atm, the piston would have been pushed up, giving more space for the vaporized molecules.
04

Determine if work is done on the system, or by the system

Review the change in the piston position as observed in the drawings. Since the piston has moved up against the constant atmospheric pressure, work is done by the system (the liquid molecules vaporizing). The system expands in the presence of constant external pressure, and the piston moves up, which means the system is doing work on pushing the piston up. In conclusion, work is being done by the system during the vaporization process, where the molecules change from the liquid state to the vapor state under a constant pressure of 1 atm.

Key Concepts

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

Work Done by the System
In thermodynamics, work done by the system refers to the energy transfer when a system expands against an external pressure. In our example, we consider a cylinder with a movable piston exerting a pressure of 1 atmosphere (atm). When the liquid molecules inside the cylinder vaporize, they take up more space, causing the piston to move upwards.

Work done by the system can be determined through the formula:
  • Work done, \( W = P \Delta V \)
Here, \( P \) is the constant pressure (in this case, 1 atm), and \( \Delta V \) is the change in volume as the liquid turns to gas.

As the piston rises during the vaporization process, energy is transferred from the system (the substance inside the cylinder) to its surroundings. This work done by the expanding gas pushes the piston upwards, illustrating how a system performs work.
Vaporization Process
The vaporization process is a phase transition where liquid molecules change into vapor or gas. In our example, this process occurs inside a closed cylinder with a piston. Initially, we have liquid molecules under a pressure of 1 atm.

During vaporization:
  • The molecules absorb enough energy to overcome intermolecular forces.
  • Their kinetic energy increases, making them move rapidly and occupy more space.
  • This expansion requires space, hence pushing the piston upwards.
In essence, vaporization involves the input of heat energy, allowing molecules to transition from a close-packed liquid state to a more dispersed gaseous state. During this transformation, the liquid becomes a gas, expanding and doing work on the surroundings by moving the piston in the cylinder.
Constant Pressure
Constant pressure is a condition where the pressure exerted on a system does not change despite volume changes. In our cylinder example, the external pressure of the piston remains constant at 1 atm throughout the process.

When pressure is constant:
  • Work done by or on the system can be easily calculated since pressure doesn't vary.
  • It simplifies the analysis of thermodynamic processes such as vaporization.
Maintaining constant pressure is quite common in phase change processes. As the liquid becomes vapor, the constant pressure is due to the weight of the piston plus atmospheric pressure. Regardless of the volume increase, the pressure over the system remains steady, making calculations straightforward and illustrating a key concept in thermodynamics.

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Most popular questions from this chapter

Determine whether the statements given below are true or false. Consider an endothermic process taking place in a beaker at room temperature. (a) Heat flows from the surroundings to the system. (b) The beaker is cold to the touch. (c) The pressure of the system decreases. (d) The value of \(q\) for the system is positive.

In the late eighteenth century Priestley prepared ammonia by reacting \(\mathrm{HNO}_{3}(g)\) with hydrogen gas. The thermodynamic equation for the reaction is $$ \mathrm{HNO}_{3}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-637 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) when one mole of hydrogen gas reacts. (b) What is \(\Delta H\) when \(10.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}(g)\) is made to react with an excess of steam to form \(\mathrm{HNO}_{3}\) and \(\mathrm{H}_{2}\) gases?

Find (a) \(\Delta E\) when a gas absorbs \(18 \mathrm{~J}\) of heat and has \(13 \mathrm{~J}\) of work done on it. (b) \(q\) when 72 J of work is done on a system and its energy is increased by \(61 \mathrm{~J}\).

Copper is used in building the integrated circuits, chips, and printed circuit boards for computers. When \(228 \mathrm{~J}\) of heat are absorbed by \(125 \mathrm{~g}\) of copper at \(22.38^{\circ} \mathrm{C}\), the temperature increases to \(27.12^{\circ} \mathrm{C}\). What is the specific heat of copper?

Which statement(s) is/are true about bond enthalpy? (a) Energy is required to break a bond. (b) \(\Delta H\) for the formation of a bond is always a negative number. (c) Bond enthalpy is defined only for bonds broken or formed in the gaseous state. (d) Because the presence of \(\pi\) bonds does not influence the geometry of a molecule, the presence of \(\pi\) bonds does not affect the value of the bond enthalpy between two atoms either. (e) The bond enthalpy for a double bond between atoms \(A\) and \(B\) is twice that for a single bond between atoms \(\mathrm{A}\) and \(\mathrm{B}\).

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