Answer:
The pressure is 37.56 atm.
Temperature at the end of adiabatic compression is 825.73 K.
Explanation:
Given that
[tex]T_1=20C[/tex]
[tex]P_1=1\ atm[/tex]
[tex]V_1=800\ cm^3[/tex]
[tex]V_2=60\ cm^3[/tex]
So r=800/60=13.33
γ=1.4
For process 1-2
[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]
[tex]\dfrac{T_2}{273+20}=13.33^{1.4-1}[/tex]
[tex]T_2=825.73\ K[/tex]
[tex]\dfrac{P_2}{P_1}=r^\gamma[/tex]
[tex]\dfrac{P_2}{1}=13.33^{1.4}[/tex]
[tex]P_2=37.56\ atm[/tex]
So
The pressure is 37.56 atm.
Temperature at the end of adiabatic compression is 825.73 K.
