Q. 8.49 A sample containing1.50聽moles聽... [FREE SOLUTION]

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Chapter 8: Q. 8.49 (page 276)

A sample containing $1.50 m o l e s o f N e$ gas has an initial volume of $8.00 L$ . What is the final volume, in liters, when each of the following occurs and pressure and temperature do not change?
a. A leak allows one-half of $N e$ atoms to escape.
b. A sample of $3.50 m o l e s o f N e$ is added to the $1.50 m o l e s o f N e$ gas in the container.
c. A sample of $25.0 g$ of $N e$ is added to the $1.50 m o l e s o f N e$ gas in the container.

Short Answer

Expert verified

a. The final volume is $4 L$ .

b. The final volume is role="math" localid="1652778604661" $26.7 L$ .

c. The final volume is $14.4 L$ .

Step by step solution

Part (a) step 1: Given Information

We need to find the final volume.

Part (a) step 2: Simplify

Consider:

$n_{1} = 1.50 m o l e V_{1} = 8.00 L$

Now, calculating the final volume $V_{2}$ :

conversion factor is:

$\frac{6 脳 10^{23} a t o m o f N e}{1 m o l o f N e}$

Contained in $1.50 m o l o f N e$

$1.5 m o l e o f N e 脳 \frac{6 脳 10^{23} a t o m o f N e}{1 m o l o f N e} = 9 脳 10^{23} a t o m s o f N e$

the final volume $V_{2}$ is calculated by Avogadro's law equation:

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Putting the value:

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}} V_{2} = \frac{V_{1} 脳 n_{2}}{n_{1}} V_{2} = \frac{8.00 L 脳 0.75 m o l}{1.50 m o l} V_{2} = \frac{6 L}{1.50 L} V_{2} = 4 L$

Part (b) step 1: Given Information

We need to find the final volume.

Part (b) step 2: Simplify

Consider:

$n_{1} = 1.50 m o l e V_{1} = 8.00 L$

Now, calculating the final volume $V_{2} :$

$n_{2} = 3.50 m o l + 1.50 m o l = 5.00 m o l o f N e$

the final volume $V_{2}$ is calculated by Avogadro's law equation:

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Putting the value:

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}} V_{2} = \frac{V_{1} 脳 n_{2}}{n_{1}} V_{2} = \frac{8.00 L 脳 5.00 m o l}{1.50 m o l} V_{2} = \frac{40 L}{1.50 L} V_{2} = 26.7 L$

Part (c) step 1: Given Information

We need to find the final volume.

Part (c) step 2: Simplify

Consider:

$n_{1} = 1.50 m o l e V_{1} = 8.00 L$

Now, calculating the final volume $V_{2}$ :

conversion factor is:

$\frac{1 m o l o f N e}{20.2 g o f N e}$

$25.0 g o f N e 脳 \frac{1 m o l o f N e}{20.2 g o f N e} = 1.20 m o l o f N e$

$n_{2} = 1.20 m o l o f N e + 1.50 m o l o f N e = 2.70 m o l o f N e$

the final volume $V_{2}$ is calculated by Avogadro's law equation:

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Putting the value:

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}} V_{2} = \frac{V_{1} 脳 n_{2}}{n_{1}} V_{2} = \frac{8.00 L 脳 2.70 m o l}{1.50 m o l} V_{2} = \frac{21.6 L}{1.50 L} V_{2} = 14.4 L$

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Short Answer

Step by step solution

Part (a) step 1: Given Information

Part (a) step 2: Simplify

Part (b) step 1: Given Information

Part (b) step 2: Simplify

Part (c) step 1: Given Information

Part (c) step 2: Simplify

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