Question

2.A helium-filled balloon has a

volume of 48.0 L at 25°C and
1.05 atm. What volume will it
have at 0.860 atm and 15°C?

Answer 1

The volume at 0.860 atm and 15°C will be 56.638 L

Step-by-step explanation:

Gay Lussac's law explains the variation in the pressure of a gas by modifying its temperature, keeping the volume constant: it determines that pressure and temperature are directly proportional quantities. In other words, Gay-Lussac's law states that when a gas undergoes a constant volume transformation, the ratio of the pressure exerted by the temperature of the gas remains constant:

`(P)/(T)=k`

On the other hand, Boyle's law determines that the pressure exerted by a gas is inversely proportional to its volume at constant temperature. This is expressed mathematically as:

P*V=k

Finally, Charles's law says that at constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature and is expressed mathematically as:

`(V)/(T)=k`

Combined law equation is the combination of three gas laws called Boyle's, Charles's and Gay-Lussac's law:

`(P*V)/(T)=k`

When you want to study two different states, an initial one and a final one of a gas, you use:

`(P1*V1)/(T1)=(P2*V2)/(T2)`

In this case:

• P1= 1.05 atm
• V1= 48 L
• T1= 25 C= 298 K (being 0 C= 273 K)
• P2= 0.86 atm
• V2= ?
• T2= 15 C= 288 K

Replacing:

`(1.05 atm*48 L)/(298 K)=(0.86atm*V2)/(288 K)`

and solving:

`V2=(288 K)/(0.86 atm) *(1.05 atm*48 L)/(298 K)`

you get:

V2= 56.638 L

The volume at 0.860 atm and 15°C will be 56.638 L