Question 65: Thermodynamics & KTG

Question

In an adiabatic expansion of air, the volume is increased by $$6.2 \%$$. The percentage change in pressure is $$(\gamma=1.4)$$

Accepted Answer

To solve this problem, we can use the relation for an adiabatic process, which is given as:
$$ P V^\gamma = 	ext{constant} $$
Where $$P$$ is the pressure, $$V$$ is the volume, and $$\gamma$$ (gamma) is the heat capacity ratio. In an adiabatic process, as the volume increases, the pressure decreases. We are given that the volume increases by $$6.2\%$$, so we can express the final volume $$V_f$$ in terms of the initial volume $$V_i$$ as:
$$ V_f = V_i + 0.062V_i = 1.062V_i $$
Using the relation for an adiabatic process for both initial and final states, we have:
$$ P_i V_i^\gamma = P_f V_f^\gamma $$
Substituting the expression for $$V_f$$, we get:
$$ P_i V_i^\gamma = P_f (1.062V_i)^\gamma $$
$$ \frac{P_f}{P_i} = \left(\frac{V_i}{1.062V_i}ight)^\gamma $$
$$ \frac{P_f}{P_i} = (1.062)^{-\gamma} $$
Given that $$\gamma = 1.4$$, let's substitute that into the equation:
$$ \frac{P_f}{P_i} = (1.062)^{-1.4} $$
We can now calculate the percentage change in pressure. However, remember that the percentage change is given by:
$$ 	ext{Percentage Change} = \left(\frac{P_f - P_i}{P_i}ight) 	imes 100\% $$
But since we are finding $$ \frac{P_f}{P_i} $$ directly, we need to adjust our formula to:
$$ 	ext{Percentage Change} = \left(\frac{P_f}{P_i} -1 ight) 	imes 100\% $$
Substituting our value for $$ \frac{P_f}{P_i} $$:
$$ 	ext{Percentage Change} = \left((1.062)^{-1.4} -1ight) 	imes 100\% $$
Computing the value $$ (1.062)^{-1.4} $$:
$$ = \left((1.062)^{-1.4} -1ight) 	imes 100\% \approx (-0.0868) 	imes 100\% = -8.68\% $$
Since a negative sign indicates a decrease in pressure, the answer implies that the pressure decreases by approximately 8.68%. Thus, the correct answer is Option A: 8.68.

Answer

8.68

Answer

4.84

Answer

6.48

Answer

2.24

Thermodynamics & KTG

In an adiabatic expansion of air, the volume is increased by $6.2 \%. The percentage change in pressure is (\gamma=1.4)

Choose the correct answer: