Question

A cylinder with a moveable piston contains 0.502 mol of gas and has a volume of 275 mL What will its volume be if an additional 0.198 mol of gas is added to the cylinder

Answer 1

Here we have two case

First case

n=number of moles = 0.502 mol and volume = 275 mL

Now apply the law of ideal gas equation.

$PV=nRT\\ P\times 275mL=0.502\times RT....eqn (1)$

Second Case

n=number of moles = 0.198+0.502 mol = 0.7 mol

and volume = 275 mL

Now apply the law of ideal gas equation.

$PV=nRT\\ P\times V=(0.502+0.198)\times RT\\ P\times V=(0.7)\times RT....eqn (2)\\$

now divide equation 1 by equation 2 , we get

$\\ \frac{P\times 275}{P\times V}=\frac{0.502\times RT}{0.7\times RT}\\ \frac{ 275}{ V}=\frac{0.502}{0.7}\\ \frac{ 275}{ V}=0.71\\ V=\frac{275}{0.71}\\ V=387.3mL$

Hence the volume will be 387.3 mL.