Question

A sample of argon gas has a volume of $735mL$at a pressure of $1.20atm$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. $658mmHgand281K$

b. $0.55atmand75°C$

c.$15.4atmand-15°C$

Short answer

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

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

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

Step-by-step solution

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$P_1=1.2atm$

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

final pressure$V_1=735mL$

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

final temperature$T_2=281K$

Now, finding the final volume $V_2$:

combined gas law:

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

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

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$P_1=1.2atm$

final pressure$P_2=0.55atm$

initial volume$V_1=735mL$

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

final temperaturelocalid="1652723876868" $T_2=348K$

Now, finding the final volume $V_2$:

combined gas law:

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

localid="1652723921426" $$\begin{gathered} V_2=\left(\frac{P_1\times V_1}{T_1}\right)\times \left(\frac{T_2}{P_2}\right) \\ {V}_2=\left(\frac{\left(1.2atm\right)\times 735mL}{385K}\right)\times \left(\frac{348K}{0.55atm}\right) \\ {V}_2=1449.52mL \end{gathered}$$

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 $P_1=1.2atm$

final pressure$P_2=15.4atm$

initial volume$V_1=735mL$

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

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

Now, finding the final volume $V_2$:

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

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