Explanation:
As it is given that process is entropic which means that it is reversible adiabatic in nature.
Also, the given data is as follows.
[tex]T_{1}[/tex] = [tex]20 ^{o}C[/tex] = (20 + 273) K = 293 K
Gage pressure = 150 kPa, Atmospheric pressure = 101.3 kPa
Hence, pressure ([tex]P_{1}[/tex]) will be (150 kPa + 101.3 kPa) = 251.3 kPa
and, pressure ([tex]P_{2}[/tex]) will be (325 kPa + 101.3 kPa) = 426.3 kPa
Also, specific heat ratio ([tex]\gamma[/tex])is given as 1.40.
Hence, relation between T and P is as follows.
[tex]T \times P \times (\frac{1 - \gamma}{\gamma})[/tex] = constant
[tex]T_{1}P_{1} \times (\frac{1 - \gamma}{\gamma})[/tex] = [tex]T_{2}P_{2} \times (\frac{1 - \gamma}{\gamma})[/tex]
[tex]T_{2}[/tex] = [tex]T_{1} \times \frac{P_{1}}{P_{2}} \times \frac{-0.4}{1.4}[/tex]
= [tex]293 K \times \frac{251.3kPa}{426.3kPa} \times \frac{-0.4}{1.4}[/tex]
= 340.758 K
Hence, convert this temperature into degree celsius as follows.
[tex](340.758 - 273) ^{o}C[/tex]
= [tex]67.758 ^{o}C[/tex]
Thus, we can conclude that the temperature is [tex]67.758 ^{o}C[/tex].
