Question 220: Thermodynamics & KTG

Question

The volume of an ideal gas $$(\gamma=1.5)$$ is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is :

Accepted Answer

To find the ratio of the initial pressure to the final pressure of an ideal gas undergoing an adiabatic process, we can use the adiabatic equation for an ideal gas, which relates pressure (P) and volume (V) as follows:
$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}$
Here, $P_{1}$ and $V_{1}$ are the initial pressure and volume, respectively, $P_{2}$ and $V_{2}$ are the final pressure and volume, respectively, and $\gamma$ is the heat capacity ratio of the gas.
Given in the question, $\gamma = 1.5$, $V_{1} = 5$ litres and $V_{2} = 4$ litres. We need to find the ratio $\frac{P_{1}}{P_{2}}$.
Rearranging the adiabatic equation for the ratio $\frac{P_{1}}{P_{2}}$, we get:
$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma} \Rightarrow \frac{P_{1}}{P_{2}} = \left(\frac{V_{2}}{V_{1}}\right)^{\gamma}$
Substituting the given values:
$\frac{P_{1}}{P_{2}} = \left(\frac{V_{2}}{V_{1}}\right)^{\gamma} = \left(\frac{4}{5}\right)^{1.5}$
To calculate the value:
$\frac{P_{1}}{P_{2}} = \left(\frac{4}{5}\right)^{1.5} = \left(\frac{4}{5}\right)^{\frac{3}{2}}$
Simplifying further:
$\frac{P_{1}}{P_{2}} = \left(\frac{2^2}{5}\right)^{\frac{3}{2}} = \left(\frac{2^3}{5^{\frac{3}{2}}}\right) = \frac{8}{5\sqrt{5}}$
Therefore, the ratio of the initial pressure to the final pressure is $\frac{8}{5\sqrt{5}}$, which corresponds to
Option B
.

Answer

$$\frac{4}{5}$$

Answer

$$\frac{8}{5 \sqrt{5}}$$

Answer

$$\frac{2}{\sqrt{5}}$$

Answer

$$\frac{16}{25}$$

Thermodynamics & KTG

The volume of an ideal gas $(\gamma=1.5)$ is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is :

Choose the correct answer: