Giancoli's Physics: Principles with Applications, 7th Edition

13

Temperature and Kinetic Theory

Change chapter13-1: Atomic Theory

13-2: Temperature and Thermometers

13-4: Thermal Expansion

13-5: Gas Laws; Absolute Temperature

13-6 and 13-7: Ideal Gas Law

13-8: Ideal Gas Law in Terms of Molecules; Avogadro's Number

13-9: Molecular Interpretation of Temperature

13-11: Real Gases; Phase Changes

13-12: Vapor Pressure and Humidity

13-13: Diffusion

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.

Problem 23

Q

A

$1.25 \textrm{ m}^3$

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Initially at standard temperature and pressure you have *p1* times *v1*, pressure 1 times volume 1, equals *n R* times *t1*, this is the number of moles times the ideal gas constant times the initial temperature. And then you have the same ideal gas law applying in the second case after the temperature rises and the pressure rises. And we can solve for *v2* because we know that *p1 v1* over *t1* is *n R*. You can divide both sides of this by *t1*. And you have the exact same quantity of gas in the container after you've squished it. And so *p2 v2* over *t2* is also equal to *n R*. And so that makes these two things equal to each other *p1 v1* over *t1* equals *p2 v2* over *t2*. So, we can multiply both sides by *t2* over *p2*. And then leaving us with *v2* on the right end and then this *t2* over *p2* gets multiplied by the stuff on the left. Then switch the sides around. So, we have *v2* equals *t2* times *p1 v1* over *p2* times *t1*. And the only thing to keep in mind is that temperatures always have to be expressed in kelvin and pressures are absolute pressures. So, the STP conditions standard temperature and pressure, the pressure is not 0, it's 1 atmosphere, that's the absolute pressure. The gauge pressure is 0 but the absolute pressure is 1 atmosphere. And we multiply that by 3.5 times 10 to the, or, sorry, times 10 to the nothing, 3.5 meters cubed for the initial volume. And then the temperature is 38 degrees Celsius plus 273.15 to conversion into kelvin. And divide that by 3.2 atmospheres. And we can leave these pressures in units of atmospheres because the only important thing is that they're the same unit so that they cancel, no need to change them into pascals. And then the temperature at STP conditions is 273 kelvin. And this makes 1.25 cubic meters as the final volume.

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