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: Do the isothermal step.
For constant temperature, $PV$ stays fixed. Going from $V$ to $2V$: $P\cdot V = P_2\cdot 2V$, so $P_2 = \tfrac{P}{2}$.

Step 2: Set up the adiabatic step.
For an adiabatic change $PV^{\gamma}$ is fixed, with $\gamma = \tfrac53$. Going from $2V$ to $16V$: \[ P_3 = P_2\left(\frac{2V}{16V}\right)^{\gamma} = \frac{P}{2}\left(\frac{1}{8}\right)^{5/3}. \]

Step 3: Evaluate the power.
$\left(\tfrac18\right)^{5/3} = (2^3)^{-5/3} = 2^{-5} = \tfrac{1}{32}$.

Step 4: Multiply.
$P_3 = \tfrac{P}{2}\times\tfrac{1}{32} = \tfrac{P}{64}$. \[ \boxed{\tfrac{P}{64}} \]