Step 1 – Amount of pentane. $n=\dfrac{10.0\text{ g}}{72.15\text{ g/mol}}=0.1386\text{ mol}$.

Step 2 – What happens at 30 °C, below the normal boiling point (36.1 °C)? At this temperature pentane cannot be a gas at high pressure — some of it will condense to liquid whenever the container is small enough. As long as both liquid and vapor are present, the pressure in the container is fixed at the equilibrium vapor pressure of pentane at 30 °C (roughly 600–620 mmHg, close to but a bit below 1 atm since 30 °C is a little below the boiling point). This gives a flat (constant-pressure) region at small volumes, where liquid is still present.

Step 3 – When all the liquid has evaporated. Once the volume is large enough that all 0.1386 mol of pentane exists as vapor, the system behaves as an ideal gas and further increases in $V$ cause $P$ to drop according to $PV=nRT$.

Step 4 – Check the numbers. If all the pentane were vapor:
$$P=\frac{nRT}{V}=\frac{(0.1386)(62.36)(303)}{V}$$
- At $V=8\text{ L}:\ P\approx 328\text{ mmHg}$ — matches the low-pressure tail of the curve.
- At $V=2\text{ L}$, the ideal-gas formula would predict $P\approx1311\text{ mmHg}$, but this exceeds the vapor pressure of pentane at 30 °C, so instead the pentane simply condenses and the pressure stays pinned at the vapor pressure (~600–620 mmHg) rather than rising that high.

So the correct curve must start flat (constant $P$ = vapor pressure) at small $V$, then bend downward and follow the $1/V$ ideal-gas curve once all the liquid has vaporized — this is exactly graph (C).