Question

A monoatomic gas at pressure ‘P’ having volume ‘V’ expands isothermally to a volume 2 V and then adiabatically to a volume 16 V. The final pressure of the gas is \(\gamma = \frac{5}{3}\)

• A. \(\frac{P}{64}\)
• B. \(\frac{P}{128}\)
• C. \(\frac{P}{8}\)
• D. \(\frac{P}{32}\)

Hint:

In problems with multiple thermodynamic processes, handle each step sequentially. For adiabatic processes, remember \(TV^{\gamma-1} = \text{constant}\) or \(PV^{\gamma} = \text{constant}\). Here using \(PV^{\gamma}\) is direct.

Answer 1

The Correct Option is A

Step 1: Follow the gas.
A monatomic gas ($\gamma = \frac{5}{3}$) starts at pressure $P$ and volume $V$. First it expands isothermally to $2V$, then adiabatically to $16V$. We want the final pressure.

Step 2: Do the isothermal step.
For an isothermal change $PV$ stays constant. So $P \cdot V = P_2 \cdot 2V$, giving $P_2 = \dfrac{P}{2}$ at volume $2V$.

Step 3: Set up the adiabatic step.
For an adiabatic change $PV^{\gamma}$ stays constant: $P_2 V_2^{\gamma} = P_3 V_3^{\gamma}$, with $V_2 = 2V$ and $V_3 = 16V$.

Step 4: Write the pressure ratio.
$P_3 = P_2 \left(\dfrac{V_2}{V_3}\right)^{\gamma} = \dfrac{P}{2}\left(\dfrac{2V}{16V}\right)^{5/3} = \dfrac{P}{2}\left(\dfrac{1}{8}\right)^{5/3}$.

Step 5: Crunch the power.
Since $8 = 2^3$, we have $\left(\frac{1}{8}\right)^{5/3} = 2^{-5} = \dfrac{1}{32}$.

Step 6: Multiply through.
$P_3 = \dfrac{P}{2} \times \dfrac{1}{32} = \dfrac{P}{64}$, which is option (A).
\[ \boxed{P_3 = \frac{P}{64}} \]