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Chapter 6: Problem 40

A piston performs work of \(210 . \mathrm{L} \cdot\) atm on the surroundings, while the cylinder in which it is placed expands from \(10 . \mathrm{L}\) to 25 \(\mathrm{L}\) . At the same time, 45 \(\mathrm{J}\) of heat is transferred from the surroundings to the system. Against what pressure was the piston working?

Short Answer

Expert verified
The piston was working against a pressure of approximately 14.00 atm.

Step by step solution

01

Identify the given parameters from the exercise

Work = 210 L路atm Initial volume (V鈧�) = 10 L Final volume (V鈧�) = 25 L
02

Find the volume change during the process

We need to find the change in volume during the expansion. 螖V = V鈧� - V鈧� 螖V = 25 L - 10 L 螖V = 15 L
03

Convert the work from L路atm to J

To find the pressure working against the piston, we need the work to be in Joules. We can use the conversion factor: 1 L路atm = 101.33 J. Work = 210 L路atm 脳 101.33 J/L路atm Work 鈮� 21279.3 J
04

Use the work expression to find the pressure against the piston

Now, we can use the work expression to find the pressure against the piston: Work = Pressure 脳 螖V Pressure = Work / 螖V Pressure 鈮� 21279.3 J / 15 L Pressure 鈮� 1418.62 J/L However, the pressure is usually given in atm. We can use the conversion factor: 1 J/L = 0.00986923 atm. Pressure 鈮� 1418.62 J/L 脳 0.00986923 atm/J路L Pressure 鈮� 14.00 atm Against what pressure was the piston working? The piston was working against a pressure of approximately 14.00 atm.

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Key Concepts

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

Work in Thermodynamics
In thermodynamics, one of the essential tasks is understanding how work is involved when a system undergoes change. When we refer to work in this context, we're talking about the energy transfer involved due to a force acting over a distance.
In our piston exercise, the work is stated as 210 L路atm, indicating that a certain amount of energy has been transferred as the piston moved, compressing or expanding the gas inside the cylinder.
  • This energy transfer is crucial to find out how much work is done by or on the system.
  • In our case, the piston performs work on the surroundings, meaning the system expends energy to make this movement possible.
  • The value of work can be converted into different units to make it compatible with calculations involving other physical quantities, as seen when converting to joules in the solution: 1 L路atm is equivalent to 101.33 J.
It is important to grasp that work done on or by a gas often results in pressure acting against volume changes, leading us to explore how pressure and volume play together.
Pressure and Volume Relationship
Pressure and volume are intricately linked in thermodynamic systems. When considering the behavior of gases within a cylinder, understanding this relationship is pivotal.
In our example, the piston works against an external pressure as it expands, shifting from an initial 10 L to a final 25 L.
  • In such processes, the concept of \[ \Delta V = V_2 - V_1 \]comes into play, representing the change in volume (15 L in this case).
  • When the volume of a gas changes, the pressure it exerts on its surroundings can change, too, if the temperature and the amount of gas remain constant. This concept is modeled by the ideal gas law, but in this exercise, we directly deal with work equations.
  • Expressing the work done by the piston involves the equation \[\text{Work} = \text{Pressure} \times \Delta V \]
This equation helps derive the pressure against which the work is done once the work value is known. It showcases how closely pressure and volume tie into the mechanical processes occurring in thermodynamics.
Heat Transfer in Gas Systems
Heat transfer is a vital concept in thermodynamics, entailing the energy exchanged between a system and its surroundings due to temperature differences.
In the exercise, 45 J of heat enters the system, emphasizing the interaction between heat and work in determining a system's state.
  • Heat transfer can affect the internal energy of a system without necessarily doing work. The First Law of Thermodynamics is often articulated as \[ \Delta U = Q - W \],where:
    • \( \Delta U \) is the change in internal energy,
    • \( Q \) represents the heat added to the system, and
    • \( W \) is the work done by the system.
  • Understanding these interactions helps determine the system's net energy change.
  • In our scenario, although only the work and heat transfer are stated initially, these two factors are key to understanding how the energy dynamics within the gas system play out.
Recognizing the interconnected roles of work, pressure-volume changes, and heat transfer is essential in fully grasping the processes happening in thermodynamic systems.

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Most popular questions from this chapter

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) are gaseous fuels with enthalpies of combustion of \(-49.9 \mathrm{kJ} / \mathrm{g}\) and \(-49.5 \mathrm{kJ} / \mathrm{g}\) , respectively. Compare the energy available from the combustion of a given volume of acetylene to the combustion energy from the same volume of butane at the same temperature and pressure.

The energy content of food is typically determined using a bomb calorimeter. Consider the combustion of a \(0.30-\mathrm{g}\) sample of butter in a bomb calorimeter having a heat capacity of 2.67 \(\mathrm{kJ}^{\prime} \mathrm{C}\) . If the temperature of the calorimeter increases from \(23.5^{\circ} \mathrm{C}\) to \(27.3^{\circ} \mathrm{C}\) , calculate the energy of combustion per gram of butter.

Write reactions for which the enthalpy change will be a. \(\Delta H_{\mathrm{f}}^{\circ}\) for solid aluminum oxide. b. the standard enthalpy of combustion of liquid ethanol, $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) . $$ c. the standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid. d. \(\Delta H_{\mathrm{f}}^{\circ}\) for gaseous vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(g)\) e. the enthalpy of combustion of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) f. the enthalpy of solution of solid ammonium bromide.

In a coffee-cup calorimeter, 150.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) HCl is added to 50.0 \(\mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{NaOH}\) to make 200.0 \(\mathrm{g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\) . If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) , calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is 4.184 \(\mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) and assume no heat loss to the surroundings.

The enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed is \(-803 \mathrm{kJ} / \mathrm{mol}\) . Use these data and Hess's law to determine the enthalpy of vaporization for water.
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