Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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·atm Initial volume (V₁) = 10 L Final volume (V₂) = 25 L
02

Find the volume change during the process

We need to find the change in volume during the expansion. ΔV = V₂ - V₁ ΔV = 25 L - 10 L ΔV = 15 L
03

Convert the work from L·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·atm = 101.33 J. Work = 210 L·atm × 101.33 J/L·atm Work ≈ 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 × ΔV Pressure = Work / ΔV Pressure ≈ 21279.3 J / 15 L Pressure ≈ 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 ≈ 1418.62 J/L × 0.00986923 atm/J·L Pressure ≈ 14.00 atm Against what pressure was the piston working? The piston was working against a pressure of approximately 14.00 atm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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·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·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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the dissolution of \(\mathrm{CaCl}_{2} :\) $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ} $$ An 11.0 -g sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 \(\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g} .\)

Some automobiles and buses have been equipped to burn propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .\) Compare the amounts of energy that can be obtained per gram of \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\) and per gram of gasoline, assuming that gasoline is pure octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l) .\) See Example \(6.11 .\) ) Look up the boiling point of propane. What disadvantages are there to using propane instead of gasoline as a fuel?

Which of the following substances have an enthalpy of formation equal to zero? a. \(C l_{2}(g)\) b. \(\mathrm{H}_{2}(g)\) c. \(\mathrm{N}_{2}(l)\) d. \(\mathrm{Cl}(g)\)

Give the definition of the standard enthalpy of formation for a substance. Write separate reactions for the formation of \(\mathrm{NaCl}\) , \(\mathrm{H}_{2} \mathrm{O}, \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) and \(\mathrm{PbSO}_{4}\) that have \(\Delta H^{\circ}\) values equal to \(\Delta H_{\mathrm{f}}^{\circ}\) for each compound.

In a coffee-cup calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) NaOH and 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{HCl}\) are mixed. Both solutions were originally at \(24.6^{\circ} \mathrm{C}\) . After the reaction, the final temperature is \(31.3^{\circ} \mathrm{C}\) . Assuming that all the solutions have a density of 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) and a specific heat capacity of \(4.18 \mathrm{J} / \mathrm{C} \cdot \mathrm{g},\) calculate the enthalpy change for the neutralization of \(\mathrm{HCl}\) by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free