Question

The following graph correctly reflects log P vs log $\left(\frac{1}{V}\right)$ for a finite amount of gas at a constant temperature - True or False?

• A. True
• B. False

Answer 1

The correct option is A

True

According to Boyle's Law,

$v\propto \frac{1}{p}\Rightarrow P=KV$

Take log on both sides.

Log P = log$\frac{1}{V}$ + log K

Now, comparing it to

Y= mn + c [Straight line]

Therefore, log p and log $\frac{1}{V}$ should be a straight line with slope = k