Q. 8.43 - An Introduction to General, Organic, and Biological Chemistry | StudyQuestionHub

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Q. 8.43

Question

A sample of helium gas has a volume of $6.50 L$ at a pressure of $845 m m H g$ and a temperature of $25 ° C$ . What is the final pressure of the gas, in atmospheres, when the volume and temperature of the gas sample are changed to the following if the amount of gas does not change?

a. $1850 m L a n d 325 K$

b. $2.25 L a n d 12 ° C$

c. $12.8 L a n d 47 ° C$

Step-by-Step Solution

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Answer

a. The helium gas's final pressure is $4.18 a t m$ .

b. The helium gas's final pressure is $3.02 a t m$ .

c. The helium gas's final pressure is $0.6 a t m$ .

1Part (a) step 1: Given Information

We need to find the final pressure of the helium gas.

2Part (a) step 2: Simplify

Consider:

initial pressure $P_{1} = \frac{845 m m H g}{760} = 1.11 a t m$

initial volume $3.02 a t m$

initial temperature $0.6 a t m$

final temperature $P_{1} = \frac{845 m m H g}{760} = 1.11 a t m$

final volume $V_{2} = 1850 m L$

Now, finding the final pressure $V_{2}$ :

$\frac{P_{1} \times V_{1}}{T_{1}} = \frac{P_{2} \times V_{2}}{T_{2}}$

$P_{2} = \frac{P_{1} \times V_{1} \times T_{2}}{T_{1} \times V_{2}} = \frac{1.11 a t m \times 6.50 L \times 325 K}{303 K \times 1.85 L} = 4.18 a t m$

3Part (b) step 1: Given Information

We need to find the final pressure of the helium gas.

4Part (b) step 2: Simplify

Consider:

initial pressure $P_{1} = 1.11 a t m$

initial volume $V_{1} = 6.50 L$

initial temperature $T_{1} = 303 K$

final temperature $T_{2} = 12 ° C = (12 + 273) K = 285 K$

final volume $P_{1} = 1.11 a t m$

Now, finding the final pressure $P_{2}$ :

$\frac{P_{1} \times V_{1}}{T_{1}} = \frac{P_{2} \times V_{2}}{T_{2}}$

$P_{2} = \frac{P_{1} \times V_{1} \times T_{2}}{T_{1} \times V_{2}} = \frac{1.11 a t m \times 6.50 L \times 285 K}{303 K \times 2.25 L} = 3.02 a t m$

5Part (c) step 1: Given Information

We need to find the final pressure of the helium gas.

6Part (c) step 2: Simplify

Consider:

initial pressure $P_{1} = 1.11 a t m$

initial volume $V_{1} = 6.50 L$

initial temperature $T_{1} = 303 K$

final temperature $T_{2} = 320 K$

final volume $V_{2} = 12.8 L$

Now, finding the final pressure $P_{2}$ :

$\frac{P_{1} \times V_{1}}{T_{1}} = \frac{P_{2} \times V_{2}}{T_{2}}$

$P_{2} = \frac{P_{1} \times V_{1} \times T_{2}}{T_{1} \times V_{2}} = \frac{1.11 a t m \times 6.50 L \times 320 K}{303 K \times 12.8 L} = 0.6 a t m$

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