Question

What would be the new volume if the pressure on 600 mL is increased from 90
kPa to 150 kPa?

Answer 1

According to the equations of Mendeleev-Clapeyron for initial and final states:

$p(1)•V(1)=\nu•R•T$

$p(2)•V(2)=\nu•R•T$

We can divide the first equation into the second equation and then we will get this:

$p(1)•V(1)=p(2)•V(2)$

So, $V(2)=(p(1)•V(1))/p(2)$ = (90(kPa)•600(ml))/150(kPa) = 360 ml

Answer: V(2) = 360 ml