Question

16. Calculate volume ratio vi for an adiabatic change (Pi,V) → (P,-2 × P1%) of one mole of HsO gas. The gas can be treated as an ideal gas.

Answer 1

Given :

At state 1

Pressure = P1

Volume = V1

At state 2

Pressure = P2 = 2P1

Volume = V 2 , H2O is treated as ideal gas

For an adiabatic process

PV^($\dpi{200} \bg_black \gamma$ ) = constant.........(1)

where $\dpi{200} \bg_black \gamma$ = Heat capacity ratio ( Cp/ Cv)

From eqn 1 , we can write as

P1 *V1^($\dpi{150} \bg_black \gamma$) = P2 * V2^( $\dpi{150} \bg_black \gamma$ )

[V1/V2 ]=[ P2/ P1]^(1/$\dpi{150} \bg_black \gamma$)

V1/ V2 = [ 2 ] ^(1/1.33) { given : P2/ P1= 2 }

( Note: For tri- atomic gas , H2O, $\dpi{150} \bg_black \gamma$ = 1.33 )

V1/V2 = 1.683

Hence volume ratio, V1/ V2 = 1.683