Q. 8.44 - An Introduction to General, Organic, and Biological Chemistry

Question

A sample of argon gas has a volume of $735 mL$ at a pressure of $1.20 atm$ and a temperature of $112 °C$. What is the final volume of the gas, in milliliters, when the pressure and temperature of the gas sample are changed to the following if the amount of the gas does not change?

a. $658 mmHg and 281 K$

b. $0.55 atm and 75°C$

c. $15.4 atm and -15°C$

Step-by-Step Solution

Answer

a. The argon gas's final volume is $744.21 mL$.

b. The argon gas's final volume is  $1449.52 mL$

c. The argon gas's final volume is  $38.38 mL$.

Part (a) step 1: Given Information

We need to find the final volume of the argon gas. 

Part (a) step 2: Simplify

Consider:

initial pressure $P_{1 }=1.2 atm$

initial volume $P_2=658 mmHg=658÷760 atm=0.865 atm$

final pressure $V_1=735 mL$

initial temperature $T_1=112°C=(112+273)K=385 K$

final temperature $T_2=281 K$

Now, finding the final volume  $V_2$:

combined gas law:

$\frac{P_1\times V_1}{T_1}=\frac{P_2\times V_2}{T_2}$

$$\begin{gathered} V_2=\left(\frac{P_1\times V_1}{T_1}\right)\times \left(\frac{T_2}{P_2}\right) \\ {V}_2=\left(\frac{\left(1.2 atm\right)\times 735 mL}{385 K}\right)\times \left(\frac{281 K}{0.865 atm}\right) \\ {V}_2=744.21 mL \end{gathered}$$

Part (b) step 1: Given Information

We need to find the final volume of the argon gas. 

Part (b) step 2: Simplify

Consider:

initial pressure $P_{1 }=1.2 atm$

final pressure $P_2=0.55 atm$

initial volume $V_1=735 mL$

initial temperature $T_1=112°C=(112+273)K=385 K$

final temperature $T_2=348 K$

Now, finding the final volume $V_2$ :

combined gas law:

$\frac{P_1\times V_1}{T_1}=\frac{P_2\times V_2}{T_2}$

$$\begin{gathered} V_2=\left(\frac{P_1\times V_1}{T_1}\right)\times \left(\frac{T_2}{P_2}\right) \\ {V}_2=\left(\frac{\left(1.2 atm\right)\times 735 mL}{385 K}\right)\times \left(\frac{348 K}{0.55 atm}\right) \\ {V}_2=1449.52 mL \end{gathered}$$

Part (c) step 1: Given Information

We need to find the final volume of the argon gas.

Part (c) step 2: Simplify

Consider:

initial pressure $P_{1 }=1.2 atm$

final pressure $P_2=15.4 atm$

initial volume $V_1=735 mL$

initial temperature $T_1=112°C=(112+273)K=385 K$

final temperature $T_2=-15°C=(-15+273)K=258K$

Now, finding the final volume  $V_2$:

combined gas law:
$\frac{P_1\times V_1}{T_1}=\frac{P_2\times V_2}{T_2}$

$$\begin{gathered} V_2=\left(\frac{P_1\times V_1}{T_1}\right)\times \left(\frac{T_2}{P_2}\right) \\ {V}_2=\left(\frac{\left(1.2 atm\right)\times 735 mL}{385 K}\right)\times \left(\frac{258K}{15.4 atm}\right) \\ {V}_2=38.38 mL \end{gathered}$$