How Fast Does Volume Increase When Air Expands Adiabatically?

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SUMMARY

The discussion centers on the adiabatic expansion of air, described by the equation PV1.14 = C, where C is a constant. The user attempts to calculate the rate of volume increase (dV/dt) given a volume of 600 cm3 and a pressure of 80 kPa, decreasing at a rate of 10 kPa/min. The initial calculation yielded an incorrect result of 192.5 cm3/min due to a misinterpretation of the exponent in the equation. The correct exponent is 1.14, not 1.4, which is crucial for accurate calculations.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with calculus, specifically differentiation
  • Knowledge of the ideal gas law and its applications
  • Proficiency in manipulating equations involving constants and variables
NEXT STEPS
  • Review the derivation of the adiabatic process equations
  • Learn how to apply the chain rule in calculus for related rates
  • Study the implications of the ideal gas law on pressure and volume changes
  • Practice solving problems involving differentiation of implicit functions
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Students studying thermodynamics, physics enthusiasts, and anyone involved in engineering or applied mathematics who seeks to understand the dynamics of gas expansion under adiabatic conditions.

shaunanana
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Homework Statement



When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^1.14=C, where C is a constant. Suppose that at a certain instant the volume is 600 cm^3 and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume increasing at this instant?


Homework Equations


dP/dt=-10
we want dV/dt when V=600 and P=80

The Attempt at a Solution


V=(1.4 root)(C/P)
80(600)^1.4=C
C=620157
dV/dt=1/1.4(C/P)^-0.4((cp'-pc')/p^2)dP/dt
=1/1.4(C/P)^-0.4((c10-0)/p^2)(-10)
then i plugged in P=80 and C=620157 to get an answer of 192.5 cm^3/min which was wrong.

Can anyone show we where I went wrong and how to get the proper solution?
=
 
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shaunanana said:

Homework Statement



When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^1.14=C, where C is a constant. Suppose that at a certain instant the volume is 600 cm^3 and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume increasing at this instant?


Homework Equations


dP/dt=-10
we want dV/dt when V=600 and P=80

The Attempt at a Solution

Error in next line. The constant is 1.14, not 1.4. Also, there are square roots, cube roots, fourth roots, and so on, but not 1.4 or 1.14 roots.
shaunanana said:
V=(1.4 root)(C/P)
80(600)^1.4=C
C=620157
dV/dt=1/1.4(C/P)^-0.4((cp'-pc')/p^2)dP/dt
=1/1.4(C/P)^-0.4((c10-0)/p^2)(-10)
then i plugged in P=80 and C=620157 to get an answer of 192.5 cm^3/min which was wrong.

Can anyone show we where I went wrong and how to get the proper solution?
=

PV1.14 = C, so V1.14 = C/P, so V = (C/P)1/1.14
 

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