Final Temperature and Pressure Increase in Adiabatic Air Compression?

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In an adiabatic compression of air initially at 20°C, the final temperature is calculated to be approximately 866.017 K after a compression factor of 15, using the relationship T2 = T1 * (V1/V2)^(γ-1). The pressure increases by a factor of about 44.31, derived from the equation P1 * V1^γ = P2 * V2^γ. The calculations confirm that both the final temperature and pressure increase are consistent with the principles of adiabatic processes. The use of the ideal gas law further validates the pressure increase factor. Overall, the results appear accurate and well-supported by the relevant equations.
rg2004
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I'd just like to check my answer, it seems too big

Homework Statement


Air initially at 20◦C is compressed by a factor of 15.

What is the final temperature assuming that the compression is adiabatic and \gamma ≈ 1.4 the value of \gamma for air in the relevant range of temperatures? By what factor does the pressure increase?

Homework Equations



PV\gamma=constant
TV\gamma-1=constant

The Attempt at a Solution



V1/15=V2
T1V1\gamma-1=T2V2\gamma-1
293.15*V1\gamma-1=T2*V1/15
293.15*15\gamma-1=T2
293.15*15.4=T2
T2=866.017 kelvinP1V1\gamma=P2V2\gamma
P1V1\gamma=P2V1/15\gamma
P1*15\gamma=P2
thus P2 increased by a factor of 44.31does it look right?
 
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Looks right. You can also use the ideal gas law to find final pressure:

nR = P_1V_1/T_1 = P_2V_2/T_2

P_2/P_1 = T_2V_1/T_1V_2 = \frac{866*15}{293} = 44.3

AM
 
Thanks!
 

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