The air in a 4.50 L tank has a pressure of 1.40 atm . What is the final pressure, in atmospheres, when the air is placed in tanks that have the following volumes, if there is no change in temperature and amount of gas?

a. 1.00 L
b. 2500. mL
c. 750. mL
d. 8.00 L

Respuesta :

Answer:

a. 6.30 atm.

b. 2.52 atm.

c. 8.40 atm

d. 0.788 atm

Explanation:

Hello!

In this case, since the Boyle's law was proposed in order to understand how the volume and pressure are related via an inversely proportional relationship:

[tex]P_2V_2=P_1V_1[/tex]

In order to compute the final pressure we use:

[tex]P_2=\frac{P_1V_1}{V_2}[/tex]

Therefore we proceed by using same volume units as follows:

a.

[tex]P_2=\frac{4.50L*1.40atm}{1.00L}=6.3atm[/tex]

b.

[tex]P_2=\frac{4.50L*1.40atm}{2.5L}=2.52atm[/tex]

c.

[tex]P_2=\frac{4.50L*1.40atm}{0.750L}=8.4atm[/tex]

d.

[tex]P_2=\frac{4.50L*1.40atm}{8.00L}=0.788atm[/tex]

Best regards!