Question

The volume of a balloon containing an ideal gas is 3.78 l at 1.05 atm pressure. what would the volume be at 2.75 atm with constant temperature and molar amount

Answer 1

1.44 L

Step-by-step explanation:

According with Ideal gas law, it can be represented the dependencies between temperature, pressure, volume and molar amount of a ideal gas as follows:

`PV=nRT`

Where:

`P=Pressure\\ V=Volume\\n=Molar~amount\\T=Temperature\\R=Ideal~gas~Constant`

In this case the gas is contained in a balloon with a initial pressure, volume, molar amount and temperature (
`P_(1), V_(1), n_(1),T_(1)`) and changes to a second state with a final pressure, volume, molar amount and temperature (
`P_(2), V_(2), n_(2),T_(2)`). As we know, R is the Ideal Gas Constant and do not change with the state changes, then it is possible to obtain the equation:

`(P_(1)V_(1))/(n_(1)T_(1))=R=(P_(2)V_(2))/(n_(2)T_(2))`

But the state change proceed at constant temperature and molar amount, then
`T_(1)=T_(2)` and
`n_(1)=n_(2)` and replacing in the previews equation we obtain:

`P_(1)V_(1)=P_(2)V_(2)`

So we can obtain the final Volume as follows:

`(P_(1)V_(1))/(P_(2))=V_(2)= (1.05(atm)*3.78(L))/(2.75(atm))}=1.44(L)`

Answer 2

The volume that will be at 2.75 atm with constant temperature and molar amount is calculated using bolyes law formula that is P1V1=P2V2
P1=1.05 atm
V1=3.78 L
P2=2.75 atm
v2=?

by making V2 the subject of the formula V2= P1V1/P2

= 3.78 L x 1.05 atm/ 2.75 atm =1.44 L