Q. 8.33 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the final pressure, in millimeters of mercury, for each of the following, if $V$ and  $n$ do not change:
a. A gas with an initial pressure of $1200$ Torr at $155°C$ is cooled to $0°C$
b. A gas in an aerosol can at an initial pressure of $1.40 atm$ at $12°C$ is heated to $35°C$.

Step-by-Step Solution

Answer

Part(a) Final pressure is $765.4 mmHg$

Part(b) Final pressure is $1149.9 mmHg$

Part(a) Step 1 : Given information

We need to find final pressure, if volume and number of moles are constant.

Part(a) Step 2 : Simplify

As we know, $$\begin{gathered} 1Torr = 1 mmHg \\ 0°C=273K \end{gathered}$$

Using Gay-Lussac's law : $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$$\begin{gathered} P_1=1200 Torr =1200 mmHg \\ {T}_1=155°C=428K \\ {T}_2=0°C=273K \end{gathered}$$
therefore,

$$\begin{gathered} \frac{1200 \mathrm{mmHg}}{428 \mathrm{K}}=\frac{{\mathrm{P}}_2}{273 \mathrm{K}} \\ {\mathrm{P}}_2=765.4 \mathrm{mmHg} \end{gathered}$$

Substituting the values, We get final pressure is $765.4 mmHg$

Part(b) Step 1 : Given information

We need to find final pressure, if volume and number of moles are constant.

Part(b) Step 2 : Simplify

As we know, $$\begin{gathered} 1atm = 760 mmHg \\ 0°C=273K \end{gathered}$$

Using Gay-Lussac's law : $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$$\begin{gathered} P_1=1.4 atm =1064 mmHg \\ {T}_1=12°C=285K \\ {T}_2=35°C=308K \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{1064 \mathrm{mmHg}}{285 \mathrm{K}}=\frac{{\mathrm{P}}_2}{308 \mathrm{K}} \\ {\mathrm{P}}_2=1149.9 \mathrm{mmHg} \end{gathered}$$

Substituting the values, We get final pressure is $1149.9 mmHg$