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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 19 / Problem 21E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Solved: A cylinder contains 0.0100 mol of helium at T =

Chapter 19, Problem 21E

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QUESTION:

Problem 20E

A cylinder contains 0.0100 mol of helium at T = 27.0o C. (a) How much heat is needed to raise the temperature to 67.0o C while keeping the volume constant? Draw a pV-diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.0o C to 67.0o C? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?

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QUESTION:

Problem 20E

A cylinder contains 0.0100 mol of helium at T = 27.0o C. (a) How much heat is needed to raise the temperature to 67.0o C while keeping the volume constant? Draw a pV-diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.0o C to 67.0o C? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?

ANSWER:

Solution 21E

Introduction

The first process is a constant volume process and the second one is a constant pressure process. So to calculate the energy change in the first process, we will use the specific molar heat at constant volume and in the second process we will use the specific molar heat at constant pressure.

We also have to draw PV diagram for both process.

Step 1

The molar specific heat of nitrogen gas at constant volume, as given in the table 19.1, is 20.76 J/mol.K.

If $\Delta{}T$ is the change in temperature and $\Delta{}Q$ is the supplied heat in a constant volume process, then we can write

Here here $n$ is the mol and ${C}_{V}$ is the specific heat at constant volume. Now, in the given problem we have:

$\Delta{}Q=645$ J

$n=0.185$ mol

Hence the change in energy is

Now the initial temperature of the gas was 780 K, hence the final temperature of the is $780\ K+168\ K=948\ K$

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