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Chapter 5: Problem 101

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 this problem, when the gas expands from one flask to the other, no work is performed during the expansion as there is no external force acting on the gas due to the vacuum in the evacuated flask. As the flasks are perfectly insulated, there is no exchange of heat between the system and its surroundings. Consequently, according to the first law of thermodynamics, the change in internal energy, \(\Delta E\), is 0 for this process.

Step by step solution

01

(a) Determine if work is performed during the expansion

We need to determine if work is performed during the expansion of the gas from one flask to the other. When the valve is opened, the gas expands into the evacuated flask. During this process, no external force is applied on the gas, since the expansion occurs into the vacuum. Consequently, no work is performed on the surroundings by the gas. Therefore, the work performed during the expansion is zero.
02

(b) Explain the reason for the conclusion in part (a)

The reason behind the conclusion that no work is performed during the expansion is that there is no external force acting upon the gas during the process. In thermodynamics, work is defined as force applied over a certain distance, which in this case is absent due to the perfect vacuum in the evacuated flask. Therefore, no force is exerted on the surroundings while the gas expands, and no work is performed as a result.
03

(c) Calculate the value of \(\Delta E\) for the process

Now, we need to determine the value of \(\Delta E\) for the process. In the given problem, the flasks are assumed to be perfectly insulated. This means that there is no exchange of heat between the system and its surroundings. Since there is no exchange of heat and no work done on the surroundings, the first law of thermodynamics applied to this process states that for an internally reversible process, \(\Delta E = q - W\). Since \(q = 0\) (no exchange of heat) and \(W = 0\) (work performed during expansion is zero), we can write the equation as follows: \(\Delta E = 0 - 0\) \(\Delta E = 0\) Thus, the change in internal energy of the process, \(\Delta E\), is 0.

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Most popular questions from this chapter

A 1.50 -g sample of quinone $\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)$ is burned in a bomb calorimeter whose total heat capacity is \(8.500 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter increases from 25.00 to \(29.49^{\circ} \mathrm{C}\). (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of quinone and per mole of quinone?

Identify the force present and explain whether work is being (a) You lift a book off the top performed in the following cases: of a desk. (b) Air is compressed in a bicycle pump.

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When an 18.6-g sample of solid potassium hydroxide dissolves in $200.0 \mathrm{~g}$ of water in a coffee-cup calorimeter (Figure 5.18), the temperature rises from 23.7 to \(44.5^{\circ} \mathrm{C}\). (a) Calculate the quantity of heat (in kJ) released in the reaction. (b) Using your result from part (a), calculate \(\Delta H\) (in kJ/mol KOH) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

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