/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 101 Consider a system consisting of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a system consisting of the following apparatus, in which gas is confined in one flask and there is a vacuum in the other flask. The flasks are separated by a valve. Assume that the flasks are perfectly insulated and will not allow the flow of heat into or out of the flasks to the surroundings. When the valve is opened, gas flows from the filled flask to the evacuated one. (a) Is work performed during the expansion of the gas? (b) Why or why not? (c) Can you determine the value of \(\Delta E\) for the process?

Short Answer

Expert verified
In summary, for this adiabatic process involving gas expansion into a vacuum: (a) No work is performed during the expansion since the gas doesn't have to push against any external pressure. (b) This explains why no work is performed during the process. (c) Consequently, the change in internal energy \(\Delta E\) for the process is 0.

Step by step solution

01

Understanding the type of process

The initial state has gas confined in one flask and a vacuum in the other, separated by a valve. It is given that the flasks are perfectly insulated, meaning this is an adiabatic process - there is no exchange of heat between the system (gas) and its surroundings.
02

Determining Work Performed

In an adiabatic process, no heat exchange occurs between the system and its surroundings. When the valve is opened, the gas expands from the filled flask to the evacuated one. However, during this expansion process, the gas is not pushing against any external pressure, as it just fills the vacuum in the other flask. Therefore, no work is performed during the expansion of the gas.
03

Explaining Why No Work Is Performed

Work is performed when a force acts over a certain displacement. In our case, when the gas expands, it does not have to overcome any force or push against any external pressure since the other flask is initially a vacuum. As a result, there is no work done during the expansion of the gas.
04

Determining the Change in Internal Energy (ΔE)

For an adiabatic process, the first law of thermodynamics states that the change in internal energy, \(\Delta E\), is equal to the work done on the system (\(W\)). In our case, we concluded in step 3 that no work is performed during the expansion. Therefore, the change in internal energy is: \[ \Delta E = W \] In this situation, we have \(W = 0\), which means the change in internal energy is also zero: \[ \Delta E = 0 \] So, the change in internal energy for this process is zero. To conclude, (a) No work is performed during the expansion of the gas. (b) This is because the gas is expanding into a vacuum and doesn't have to push against any external pressure. (c) The value of \(\Delta E\) for the process is 0.

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 91Ó°ÊÓ!

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

Suppose an Olympic diver who weighs 52.0 kg executes a straight dive from a 10-m platform. At the apex of the dive, the diver is 10.8 m above the surface of the water. (a) What is the potential energy of the diver at the apex of the dive, relative to the surface of the water? (b) Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed, in m/s, will the diver enter the water? (c) Does the diver do work on entering the water? Explain.

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) \(H\) is a state function, but \(q\) is not a state function. Explain. (c) For a given process at constant pressure, \(\Delta H\) is positive. Is the process endothermic or exothermic?

At the end of \(2012,\) global population was about 7.0 billion people. What mass of glucose in kg would be needed to provide 1500 Cal/person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) according to the following thermochemical equation: $$\begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H^{\circ}=&-2803 \mathrm{kJ} \end{aligned}$$

In a thermodynamic study, a scientist focuses on the properties of a solution in an apparatus as illustrated. A solution is continuously flowing into the apparatus at the top and out at the bottom, such that the amount of solution in the apparatus is constant with time. (a) Is the solution in the apparatus a closed system, open system, or isolated system? (b) If the inlet and outlet were closed, what type of system would it be?

(a) What is the electrostatic potential energy (in joules) between two protons that are separated by 62 pm? (b) What is the change in potential energy if the distance separating the two is increased to 1.0 nm? (c) Does the potential energy of the two particles increase or decrease when the distance is increased to 1.0 nm?

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.