One can show that the volume of a solid when it is isothermally compressed from a pressure Po to a pressure P is given by V = Voe-k(P-P。) where V and Vo are the volumes at pressures P and Po respectively and is the isothermal compressibility which can be considered constant through the compression process. Use the above expression to calculate the work in J done on a solid cylinder containing 3.39 moles of material when the pressure is raised reversibly from 1.000 atm to 9,937.9 atm at a constant temperature of 20 °C. For this material the following information is known: Molar Mass: 43.45 g mol-1 Density at 20 °C and 1.000 atm: 1.2532 g ml-1 K = 40.816 x 10-5 atm-1 Remarks: • To solve the problem you will need to put either P as a function of V or V as a function of P. Both can be done with the equation given above. • If you choose the last route (easier in my opinion), you will need to switch the integration variable from V to P. You can do this by calculation dV/dP, solving for dV and substituting in the integral for work. • Note that the product KP does not have any units which means that you can evaluate it in whatever units you choose as long as the units are the same (i.e. atm-1 for K and atm for P). • Be careful with the signs. Remember that work done on the system is always positive.
One can show that the volume of a solid when it is isothermally compressed from a pressure Po to a pressure P is given by V = Voe-k(P-P。) where V and Vo are the volumes at pressures P and Po respectively and is the isothermal compressibility which can be considered constant through the compression process. Use the above expression to calculate the work in J done on a solid cylinder containing 3.39 moles of material when the pressure is raised reversibly from 1.000 atm to 9,937.9 atm at a constant temperature of 20 °C. For this material the following information is known: Molar Mass: 43.45 g mol-1 Density at 20 °C and 1.000 atm: 1.2532 g ml-1 K = 40.816 x 10-5 atm-1 Remarks: • To solve the problem you will need to put either P as a function of V or V as a function of P. Both can be done with the equation given above. • If you choose the last route (easier in my opinion), you will need to switch the integration variable from V to P. You can do this by calculation dV/dP, solving for dV and substituting in the integral for work. • Note that the product KP does not have any units which means that you can evaluate it in whatever units you choose as long as the units are the same (i.e. atm-1 for K and atm for P). • Be careful with the signs. Remember that work done on the system is always positive.
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter18: Heat Engines, Entropy, And The Second Law Of Thermodynamics
Section: Chapter Questions
Problem 42P
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Transcribed Image Text:One can show that the volume of a solid when it is isothermally compressed from a pressure Po to a pressure
P is given by
V = Voe-k(P-P。)
where V and Vo are the volumes at pressures P and Po respectively and is the isothermal compressibility
which can be considered constant through the compression process. Use the above expression to calculate
the work in J done on a solid cylinder containing 3.39 moles of material when the pressure is raised reversibly
from 1.000 atm to 9,937.9 atm at a constant temperature of 20 °C. For this material the following information
is known:
Molar Mass: 43.45 g mol-1
Density at 20 °C and 1.000 atm: 1.2532 g ml-1
K = 40.816 x 10-5 atm-1
Remarks:
• To solve the problem you will need to put either P as a function of V or V as a function of P. Both can be
done with the equation given above.
• If you choose the last route (easier in my opinion), you will need to switch the integration variable from
V to P. You can do this by calculation dV/dP, solving for dV and substituting in the integral for work.
• Note that the product KP does not have any units which means that you can evaluate it in whatever
units you choose as long as the units are the same (i.e. atm-1 for K and atm for P).
• Be careful with the signs. Remember that work done on the system is always positive.
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