Chapter 9, Problem 63QP

Interpretation Introduction

Interpretation:

The missing final values in the table are to be determined.

Concept Introduction:

The combined gas law states that the temperature $\begin{array}{cc}\text{solubility}& \text{Temperature}\\ \text{12}\text{.8}& \text{17}\\ \text{21}\text{.0}& \text{23}\\ \text{25}\text{.0}& \text{31}\\ \text{30}\text{.5}& \text{39}\\ \text{37}\text{.0}& \text{43}\\ \text{39}\text{.0}& \text{51}\end{array}$ is directly proportional to pressure $\left(T\right)$ and volume $\left(P\right)$ of a fixed amount of gas. When the temperature of a gas increases, the pressure and volume of the fixed amount of gas also increase and vice versa.

$\left(V\right)$ …… (1)

Here $\begin{array}{c}T\propto PV\\ \frac{PV}{T}=k\\ \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}\end{array}$ , $\left(T_1\right)$ , and $\left(P_1\right)$ are the initial temperature, pressure, and volume of the gas.

$\left(V_1\right)$ , $\left(V_2\right)$ , and $\left(P_2\right)$ are the final temperature, pressure, and volume of the gas.

Answer to Problem 63QP

Solution:

The final values are presented in the table below.

$\left(T_2\right)$

Explanation of Solution

The calculation for determining the final volume of the gas.

The conversion of temperature from Celsius to kelvin is shown below.

$\begin{array}{|cccc|}\hline V_1& 2.5\text{ L}& 125\text{ L}& 455\text{ mL}\\ P_1& 0.5\text{ atm}& 0.250\text{ atm}& 200\text{ torr}\\ T_1& 20°\text{C}& 25°\text{C}& 300\text{ K}\\ V_2& 1.2\text{ L}& 62\text{ L}& 200\text{ mL}\\ P_2& 760\text{ torr}& 100\text{ torr}& 910.23\text{ torr}\\ T_2& 0°\text{C}& -195.3°\text{C}& 327°\text{C}\\ \hline\end{array}$

For $\text{K}=°\text{C}+273.15$

$20°\text{C}$

For $\begin{array}{c}\text{K}=20°\text{C}+\text{273}\text{.15}\\ =293.15\text{ K}\end{array}$

$0°\text{C}$

The conversion of pressure units from $\begin{array}{c}\text{K}=0°\text{C}+273.15\\ =273.15\text{ K}\end{array}$ to $\text{torr}$ is shown below.

$\text{atm}$

For $\begin{array}{c}1\text{ atm}=760\text{ torr}\\ 1=\frac{1\text{ atm}}{760\text{ torr}}\end{array}$

$760\text{ torr}$

Substitute $\begin{array}{c}\text{760 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times 760\text{ torr}\\ =1\text{ atm}\end{array}$ for $0.50\text{ atm}$ , $P_1$ for $2.50\text{ L}$ , $V_1$ for $293.15\text{ K}$ , $T_1$ for $1\text{ atm}$ and $P_2$ for $273.15\text{ K}$ in equation (1).

$T_2$

The calculation for determining the final temperature of the gas.

The conversion of temperature from Celsius to kelvin is shown below.

$\begin{array}{c}\frac{0.50\text{ atm}\times 2.5\text{ L}}{293.15\text{ K}}=\frac{1\text{ atm}\times V_2}{273.15\text{ K}}\\ V_2=\frac{0.50\text{ atm}\times 2.5\text{ L}}{293.15\text{ K}}\times \frac{273.15\text{ K}}{1\text{ atm}}\\ =1.165\text{ L}\simeq \text{1}\text{.2 L}\end{array}$

For $\text{K}=°\text{C}+273.15$

$25°\text{C}$

The conversion of pressure from $\begin{array}{c}\text{K}=25°\text{C}+\text{273}\text{.15}\\ =298.15\text{ K}\end{array}$ to $\text{torr}$ is shown below.

$\text{atm}$

For $\begin{array}{c}1\text{ atm}=760\text{ torr}\\ 1=\frac{1\text{ atm}}{760\text{ torr}}\end{array}$ ,

$100\text{ torr}$

Substitute $\begin{array}{c}\text{100 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times 100\text{ torr}\\ =0.132\text{ atm}\end{array}$ for $0.250\text{ atm}$ , $P_1$ for $125\text{ L}$ , $V_1$ for $298.15\text{ K}$ , $T_1$ for $100\text{ atm}$ and $P_2$ for $62\text{ L}$ in equation (1).

$V_2$

The conversion of temperature from kelvin to Celsius is shown below.

$\begin{array}{c}\frac{0.250\text{ atm}\times 125\text{ L}}{298.15\text{ K}}=\frac{0.132\text{ atm}\times 62\text{ L}}{T_2}\\ T_2=\frac{0.132\text{ atm}\times 62\text{ L}}{0.250\text{ atm}\times 125\text{ L}}\times 298.15\text{ K}\\ =78.08\text{ K}\end{array}$

For $°\text{C}=\text{K}-273.15$

$78.08\text{ K}$

The calculation for determining the final pressure of the gas.

The conversion of temperature from Celsius to kelvin is shown below.

$\begin{array}{c}°\text{C}=78.08\text{ K}-\text{273}\text{.15}\\ =-195.3°\text{C}\end{array}$

For $\text{K}=°\text{C}+273.15$

$327°\text{C}$

Substitute $\begin{array}{c}\text{K}=327°\text{C}+\text{273}\text{.15}\\ =600.15\text{ K}\end{array}$ for $200\text{ torr}$ , $P_1$ for $455\text{ mL}$ , $V_1$ for $300\text{ K}$ , $T_1$ for $600.15\text{ K}$ and $T_2$ for $200\text{ mL}$ in equation (1).

$V_2$

Conclusion

The final temperature, final volume and final pressure foreach of the cases are calculated above.