Q. 8.16 - An Introduction to General, Organic, and Biological Chemistry

Question

The air in a 5.00-L tank has a pressure of 1.20 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

Step-by-Step Solution

Answer

a. The final pressure, in atmospheres, when the air is placed in tanks that have the volume $1.00 L$ is $6.0 atm$

b. The final pressure, in atmospheres, when the air is placed in tanks that have the volume $2500.0 ml$ is $2.40 atm$

c. The final pressure, in atmospheres, when the air is placed in tanks that have the volume $750.0 ml$ is $8.0 atm$

d. The final pressure, in atmospheres, when the air is placed in tanks that have the volume $8.0 L$

is $0.75 atm$

Part(a) Step 1: Given information

We need to find the final pressure for the given data.

Part(a) Step 2: Explanation

As the temperature is constant we can apply Boyle's law

$P1V1=P2V2$

Now given data is,
$P1=1.20 atm$

$V1=5.0 L$

$V2=1.0 L$

Substituting the given values we get,

$P2= \frac{1.20 atm \times 5.0 L}{1.0 L}=6.0 atm$

Part(b) Step 1: Given information

We need to find the final pressure for the given data.

Part(b) Step 2: Explanation

Using Boyle's law

$P1V1=P2V2$

Now,

$$\begin{gathered} P1=1.20 atm \\ V1=5.0 L \\ V2=2500.0 ml \end{gathered}$$

Substituting the given values we get,

$P2= \frac{1.20 atm \times 5 L}{2500 mL}2.40 atm$

Part(c) Step 1: Given information

We need to find the final pressure for the given data.

Part(c) Step 2: Explanation

Using Boyle's law

$P1V1=P2V2$

Now,

$$\begin{gathered} P1=1.20 atm \\ V1=5.0 L \\ V2=750 ml \end{gathered}$$

Substituting the given values we get,

$P2=\frac{1.20 atm\times 5 L}{750 ml}=8.0 atm$

Part(d) Step 1: Given information

We need to find the final pressure for the given data.

Part(d) Step 2: Explanation

Using Boyle's law

$P1V2=P1V2$

Now,

$$\begin{gathered} P1=1.20 atm \\ V1=5.0 L \\ V2=8.0 L \end{gathered}$$

Substituting the given values we get,

$P2=\frac{1.20 atm \times 5 L}{8 L}=0.75 atm$