Question 716: Thermodynamics & KTG

Question

Three samples $X, Y$, and $Z$ of same gas have equal volumes and temperatures. The volume of each sample is doubled, the process being isothermal for X , adiabatic for Y and isobaric for Z . If the final pressures are equal for the three samples, the ratio of the initial pressures is ( $\gamma=3$ / 2)

Accepted Answer

Given:
Three samples of a gas ($X, Y, Z$) have equal initial volumes ($V$) and temperatures.
Volume is doubled for each sample ($V \rightarrow 2V$) by different processes:
X: Isothermal
Y: Adiabatic ($\gamma = \frac{3}{2}$)
Z: Isobaric
Let the initial pressures be $P_X, P_Y, P_Z$.
Let the final pressures (after the process) for all be $P$ (equal for all).
We have to find $P_X : P_Y : P_Z$.
Step 1: Isothermal Process (Sample X)
For isothermal process:
$$P_X V = P \cdot 2V$$
Solve for $P_X$:
$ P_X = 2P $
Step 2: Adiabatic Process (Sample Y)
For adiabatic:
$$P_Y V^\gamma = P (2V)^\gamma$$
Plug in $\gamma = \frac{3}{2}$:
$ P_Y V^{3/2} = P \cdot (2V)^{3/2} $
$ P_Y = P \cdot \frac{(2)^{3/2}}{1} $
$ P_Y = P \cdot 2^{3/2} $
$ 2^{3/2} = (\sqrt{2})^3 = 2 \sqrt{2} $
$ P_Y = 2\sqrt{2} ~P $
Step 3: Isobaric Process (Sample Z)
For isobaric, pressure is constant during the process, so:
$ P_Z = P $
Step 4: Ratio
$ P_X : P_Y : P_Z = 2P : 2\sqrt{2}~P : P $
$ = 2 : 2\sqrt{2} : 1 $
Answer:
Option B
$\boxed{2: 2\sqrt{2}: 1}$

Answer

$1: \sqrt{2}: 2 \sqrt{3}$

Answer

$2: 2 \sqrt{2}: 1$

Answer

$3: 3 \sqrt{3}: 1$

Answer

$5: 5 \sqrt{5}: 1$

Thermodynamics & KTG

Three samples X, Y, and Z of same gas have equal volumes and temperatures. The volume of each sample is doubled, the process being isothermal for X , adiabatic for Y and isobaric for Z . If the final pressures are equal for the three samples, the ratio of the initial pressures is ( \gamma=3 / 2)

Choose the correct answer: