Question 76: 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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