Question

Consider a gas at 1.00 atm in a 5.00-L container at 20.0 °C. What pressure does the gas exert when transferred to a volume of 2.00 L at 43 °C?

Answer 1

The pressure is 2.696 atm

Step-by-step explanation:

Boyle's law relates pressure and volume inversely proportionally and is expressed mathematically as:

P * V = k

Charles's law consists of the relationship between the volume and temperature of a certain amount of ideal gas, which is maintained at a constant pressure, by means of a proportionality constant that is applied directly and is expressed mathematically as follows:

`(V)/(T) =k`

Gay-Lussac's law indicates that the gas pressure is directly proportional to its temperature and is expressed mathematically as:

`(P)/(T) =k`

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

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

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

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

In this case, you know:

• P1=1 atm
• V1= 5 L
• T1= 20 °C=293° K (Being 0°C=273°K)
• P2=?
• V2= 2 L
• T2= 43 °C= 316°K

Replacing:

`(1 atm*5 L)/(293K) =(P2*2L)/(316K)`

Solving:

`P2=(1 atm*5 L)/(293K) *(316 K)/(2L)`

P2= 2.696 atm

The pressure is 2.696 atm