During an adiabatic process on a gas having γ = 1.4 , its pressure...

For an adiabatic process, the relation between pressure and volume is given by:

$P V^{\gamma} = \text{constant}$

Taking the natural logarithm on both sides:

$\ln P + \gamma \ln V = \text{constant}$

Differentiating both sides for small changes, we get:

$\dfrac{\Delta P}{P} + \gamma \dfrac{\Delta V}{V} = 0$

$\Rightarrow -\dfrac{\Delta V}{V} = \dfrac{1}{\gamma} \dfrac{\Delta P}{P}$

Multiplying by $100$ to express in terms of percentage:

-\dfrac{\Delta V}{V} \times 100 = \dfrac{1}{\gamma} \left(\dfrac{\Delta P}{P} \times100 \right)

Given that the pressure increases by $0.5\%$ (\dfrac{\Delta P}{P} \times$100 = 0.5$)$and \gamma = 1.4$, the percentage decrease in volume is:

-\dfrac{\Delta V}{V} \times$100 =$\dfrac{0.5}{1.4} = \dfrac{5}{14} \approx$0.357\%$

Rounding off to two decimal places, the volume decreases by approximately $0.36\%.$