Question 80: Thermodynamics & KTG

Question

In an adiabatic process with the ratio of two specific heat, $\gamma=\frac{3}{2}$, pressure is increased by $\frac{2}{3}$%, then decrease in the volume will be

Accepted Answer

In an adiabatic process,
$pV^\gamma$ = constant (k)
where $p$ is pressure, V is volume and $\gamma$ is the ratio of the specific heats.
Given, ${{{C_p}} \over {{C_V}}} = \gamma  = {3 \over 2}$
Since, for adiabatic process $p{V^\gamma } = k$
$ \Rightarrow p{V^{3/2}} = k \Rightarrow \log p + {3 \over 2}\log V = \log k$
$	herefore$ ${{\Delta p} \over p} + {3 \over 2}{{\Delta V} \over V} = 0$
$ \Rightarrow {{\Delta V} \over V} 	imes 100 =  - {2 \over 3}\left( {{{\Delta p} \over p} 	imes 100} ight) =  - {2 \over 3} 	imes {2 \over 3} =  - {4 \over 9}\% $
Negative sign implies that volume decreases by ${4 \over 9}\% $.

Answer

$\frac{4}{9}$%

Answer

$\frac{2}{3}$%

Answer

$4$%

Answer

$\frac{9}{4}$%

Thermodynamics & KTG

In an adiabatic process with the ratio of two specific heat, \gamma=\frac{3}{2}, pressure is increased by \frac{2}{3}%, then decrease in the volume will be

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