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

  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

Answer 2

Answer:

1.44 L

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:

$\frac{P_{1}V_{1}}{n_{1}T_{1}}=R=\frac{P_{2}V_{2}}{n_{2}T_{2}}$

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:

$\frac{P_{1}V_{1}}{P_{2}}=V_{2}= \frac{1.05(atm)*3.78(L)}{2.75(atm)}}=1.44(L)$