Question 751: Thermodynamics & KTG

Question

Two gases A and B having same initial state ( $\mathrm{P}, \mathrm{V}, \mathrm{n}, \mathrm{T}$ ). Now gas A is compressed to $\frac{\mathrm{V}}{8}$ by isothermal process and other gas B is compressed to $\frac{\mathrm{V}}{8}$ by adiabatic process. The ratio of final pressure of gas $A$ and $B$ is (Both gases are monoatomic, $\gamma=5 / 3$)

Accepted Answer

For isothermal process (Gas A):
In an isothermal process, the product of pressure and volume remains constant:
$ P_1 V_1 = P_2 V_2  $
Given:
$ P_0 \cdot (8V_0) = P_2 \cdot V_0  $
Solving this relation gives:
$ P_2 = 8 P_0 $
For adiabatic process (Gas B):
In an adiabatic process, $ PV^\gamma = 	ext{constant} $.
Thus:
$ \frac{P_2}{P_1} = (8)^\gamma $
Therefore:
$ P_2 = (8)^\gamma P_0 $
The ratio of the final pressures of gas B to gas A is:
$ \frac{(P_2)_B}{(P_1)_A} = \frac{(8)^\gamma P_0}{8 P_0} = (8)^{\gamma-1} $
Given $\gamma = \frac{5}{3}$:
$ \frac{(P_2)_B}{(P_1)_A} = 8^{2/3} = \sqrt[3]{64} = 4 $
Thus, the inverse ratio is:
$ \frac{(P_1)_A}{(P_2)_B} = \frac{1}{4} $

Answer

$\frac{1}{8}$

Answer

$\frac{1}{4}$

Answer

$\frac{1}{64}$

Answer

$\frac{1}{12}$

Thermodynamics & KTG

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