How Does Adiabatic Expansion Affect Ideal Gas Pressure?

AI Thread Summary
In an adiabatic expansion of an ideal gas, the pressure decreases as the volume increases. The relationship between pressure and volume is expressed by the equation P1V1^gamma = P2V2^gamma. For a 50% increase in volume, V2 is 1.5 times V1, leading to a pressure reduction factor of approximately 2, calculated as P2 = P1/(3/2)^gamma. The setup of the calculation uses the ratio of pressures (P1/P2) equal to the ratio of volumes (V2/V1) raised to the power of gamma, which is a standard approach in thermodynamics to relate changes in state variables. Understanding this relationship clarifies why the pressure decreases as the gas expands adiabatically.
trah22
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Homework Statement


an ideal gas is allowed to expand adiabatically until its volume increases by 50%. By what factor is the pressured reduced?(gamma=5/3)


Homework Equations



P1V1^gamma=P2V2^gamma

The Attempt at a Solution



P1V1^gamma=P2V2^gamma

v2=v1+.5V1=3/2V1

(p1/P2)=(v2/v1)^gamma=(3/2v1/v1)^gamma=3/2^gamma

so P2=P1/(3/2)^gamma=0.51P1 sp P2=.5P1

so reduced by approximately factor of 2

This is actually the correct answer, but i don't understand the why the intial setup of the calculation, is P1/P2=(V2/V1)^gamma, the problem is looking for the reduction of the pressure, could someone explain why your suppose to set up calculation this way?
 
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more specifically why do you put p1 over p2 (p1/p2) equal to v2 over v1 (v2/v1) in solving this problem?
 
trah22 said:
more specifically why do you put p1 over p2 (p1/p2) equal to v2 over v1 (v2/v1) in solving this problem?
?? It is just algebra:

P_1V_1^\gamma = K = P_2V_2^\gamma

\frac{P_1V_1^\gamma}{P_2} = V_2^\gamma

\frac{P_1}{P_2} = \frac{V_2^\gamma}{V_1^\gamma}= \left(\frac{V_2}{V_1}\right)^\gamma

AM
 
lol for some reason i thought something else was going on, thanks anyways:biggrin:
 
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