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Chapter 18: Problem 2

A gas enclosed in a cylinder by means of a piston that can move without friction is warmed, and 1000 J of heat enters the gas. Assuming that the volume of the gas is constant, the change in the internal energy of the gas is a) 0 . b) 1000 J. c) -1000 J. d) none of the above.

Short Answer

Expert verified
Question: A gas enclosed in a cylinder is heated, and 1000 J of heat is added to the system. The volume of the gas remains constant. Find the change in internal energy of the gas using the First Law of Thermodynamics. Choose the correct answer among the following options: a) -1000 J b) 1000 J c) 500 J d) 0 J Answer: b) 1000 J

Step by step solution

01

First Law of Thermodynamics

The First Law of Thermodynamics, which is the conservation of energy, states that the change in internal energy of the system ΔU equals the heat added to the system Q minus the work done by the system W. Mathematically, it is represented as: ΔU = Q - W
02

Identifying the given values and conditions

We are given that 1000 J of heat enters the gas, so Q = 1000 J. Moreover, since the volume of the gas remains constant, this means there is no work done (W = 0).
03

Applying the First Law of Thermodynamics

Now, let's plug in the given values into the First Law of Thermodynamics equation: ΔU = Q - W ΔU = (1000 J) - (0 J)
04

Calculating the change in internal energy

By subtracting the work done from the heat entering the gas, we can determine the change in internal energy: ΔU = 1000 J - 0 J ΔU = 1000 J
05

Identifying the correct answer

Comparing our calculated change in internal energy with the given options, the correct answer is: b) 1000 J

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Most popular questions from this chapter

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - \(\mathrm{L}\) plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of \(6.00 \mathrm{~cm}\), a height of \(12.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm}\). Use \(205 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of aluminum. Assume that the 2.00 - \(\mathrm{L}\) bottle of soda has a diameter of \(10.0 \mathrm{~cm}\), a height of \(25.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm} .\) Use \(0.100 \mathrm{~W} /(\mathrm{mK})\) as the thermal conductivity of plastic.

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