Question

A submarine suddenly breaks up at a depth of 266 m below sea level. The air in the submarine will then be suddenly (adiabatically) compressed. The initial pressure of the air is 1.0 atm, and the initial temperature is 16.0°C.

What is the final pressure?

What is the final temperature?

Answer 1

let

Pi = 1 atm = 1.013*10^5 pa
Ti = 16 C = 16 + 273 = 289 K

final pressure, Pf = Pi + rho*g*h

= 1.013*10^5 + 1000*9.8*266

= 2.71*10^6 Pa (or) 26.7 atm   <<<<<<<<<<---------------Answer

let Tf is the final temperature

for air, gamma = 1.4

In adiabatic process, P^(1 - gamma)*T^gamma = constant

Pf^(1-gamma)*Tf^gamma = Pi^(1-gamma)*Ti^gamma

Tf^gamma = (Pi/Pf)^(1-gamma)*Ti^gamma

Tf = (Pi/Pf)^(1/gamma - 1)*Ti

= (1/26.7)^(1/1.4 - 1)*289

= 739 K

= 739 - 273

= 466 degrees celsius <<<<<<<<<<---------------Answer