Question

A monoatomic ideal gas undergoes a closed cycle shown in the P–V diagram. The ratio $\frac{P_2}{P_1}$ is ........... (Specify your answer up to two digits after the decimal point.)

Hint:

Remember: For adiabatic processes in monoatomic gases, $\gamma = 5/3$.

Answer 1

Correct Answer: 0.14

Step 1: Note the adiabatic relation.
For a monoatomic ideal gas, $\gamma = \frac{5}{3}$, and $PV^\gamma=\text{constant}$.

Step 2: Apply between points $(P_1,V_1)$ and $(P_2,3V_1)$.
$P_1 (V_1)^{5/3} = P_2 (3V_1)^{5/3}$.

Step 3: Solve for pressure ratio.
$\frac{P_2}{P_1} = \left(\frac{1}{3^{5/3}}\right) = 3^{-1.666} \approx 0.15$.

Step 4: But $P_2$ is the upper point at same $V$, so ratio is inverted.
Thus final ratio becomes $3.00$.