Question

Air at 20 Celsius and 483 kPa (70 psig) gauge expands to atmospheric pressure (79 kPa). If the process is isentropic, what is the final temperature in Celsius? Present answer using three significant digits.

Answer 1

Answer:

Explanation:

Process is isentropic that means the process is adiabatic .

$PV^\gamma = constant \\P^{(1 -\gamma)}\times T^\gamma = constant$

$\frac{T_1}{T_2} = \frac{ P_1 ^ {(\gamma-1) }}{P_2^{(\gamma-1 )}}$

For air

$\gamma = 1.4$

$\frac{293}{T_2} = \frac{ 483^ {(1.4-1) }}{79^{(1.4-1 )}}$

$\frac{293}{T_2} = 2.06$

$T_2 = 142 K$

= - 131 ⁰C .