Question

Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in a reversible manner. Determine the exit pressure of helium.

Answer 1

The value is
`P_2 = 40.54 \ psla`

Step-by-step explanation:

From the question we are told that

The initial pressure is
`P_1 = 14\ psla`

The initial temperature is
`T_1 = 50 \ F = (50 - 32) * [(5)/(9) ] + 273 = 283 \ K`

The final temperature is
`T_2 = 320 \ F = (320 - 32) * [(5)/(9) ] + 273 =433 \ K`

Generally the equation for adiabatic process is mathematically represented as

`PT^{(\gamma)/(1- \gamma) } = Constant`

=>
`P_1T_1^{(\gamma)/(1- \gamma) } = P_2T_2^{(\gamma)/(1- \gamma) }`

Generally for a monoatomic gas
`\gamma = (5)/(3)`

So

`14 * 283^{((5)/(3) )/(1- [(5)/(3) ]) } =P_2 * 433^{((5)/(3) )/(1- [(5)/(3) ]) }`

=>
`14 * 283^(-2.5) =P_2 * 433^(-2.5)`

=>
`P_2 = 40.54 \ psla`