Chapter 9, Problem 64QP

Interpretation Introduction

Interpretation:

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

Concept Introduction:

The combined gas law states that the temperature $\left(T\right)$ is directly proportional to pressure $\left(P\right)$ and volume $\left(V\right)$ of a fixed amount of gas. When the temperature of the gas increases, the pressure and volume of the fixed amount of gas also increase and vice versa.

$\begin{array}{c}T\propto PV\\ \frac{PV}{T}=k\\ \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}\end{array}$ …… (1)

Here $\left(T_1\right)$ , $\left(P_1\right)$ , and $\left(V_1\right)$ are the initial temperature, pressure, and volume of the gas. $\left(T_2\right)$ , $\left(P_2\right)$ , and $\left(V_2\right)$ are the final temperature, pressure, and volume of the gas.

Answer to Problem 64QP

Solution:

The final values are shown in the table below.

$\begin{array}{|cccc|}\hline V_1& \text{601 mL}& 237\text{ mL}& 1.12\text{ L}\\ P_1& 900\text{ torr}& 75\text{ atm}& 760\text{ torr}\\ T_1& -10°\text{C}& 147\text{ K}& 0°\text{C}\\ V_2& 1200\text{ mL}& \text{474 mL}& 1.33\text{ L}\\ P_2& 468\text{ torr}& 150\text{ atm}& 700\text{ torr}\\ T_2& 0°\text{C}& \text{588 K}& 25°\text{C}\\ \hline\end{array}$

Explanation of Solution

:

The calculation for determining the final pressure of the gas.

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

$\text{K}=°\text{C}+273.15$

For $-10°\text{C}$

$\begin{array}{c}\text{K}=-10°\text{C}+\text{273}\text{.15}\\ =263.15\text{ K}\end{array}$

For $0°\text{C}$

$\begin{array}{c}\text{K}=0°\text{C}+273.15\\ =273.15\text{ K}\end{array}$

Substitute $900\text{ torr}$ for $P_1$ , $601\text{ mL}$ for $V_1$ , $263.15\text{ K}$ for $T_1$ , $273.15\text{ K}$ for $T_2$ and $1200\text{ mL}$ for $V_2$ in equation (1).

$\begin{array}{c}\frac{900\text{ torr}\times 601\text{ mL}}{263.15\text{ K}}=\frac{P_2\times 1200\text{ mL}}{273.15\text{ K}}\\ P_2=\frac{900\text{ torr}\times 601\text{ mL}}{263.15\text{ K}}\times \frac{273.15\text{ K}}{1200\text{ mL}}\\ =\text{468 torr}\end{array}$

The calculation for determining the final temperature of the gas.

Substitute $75.0\text{ atm}$ for $P_1$ , $237\text{ mL}$ for $V_1$ , $147\text{ K}$ for $T_1$ , $150\text{ atm}$ for $P_2$ and $474\text{ mL}$ for $V_2$ in equation (1).

$\begin{array}{c}\frac{\text{75 atm}\times 237\text{ mL}}{147\text{ K}}=\frac{150\text{ atm}\times 474\text{ mL}}{T_2}\\ T_2=\frac{150\text{ atm}\times 474\text{ mL}}{\text{75 atm}\times 237\text{ mL}}\times 147\text{ K}\\ \text{=588 K}\end{array}$

The calculation for determining the final volume of the gas.

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

$\text{K}=°\text{C}+273.15$

For $25°\text{C}$

$\begin{array}{c}\text{K}=25°\text{C}+\text{273}\text{.15}\\ =298.15\text{ K}\end{array}$

For $0°\text{C}$

$\begin{array}{c}\text{K}=0°\text{C}+273.15\\ =273.15\text{ K}\end{array}$

Substitute $760\text{ torr}$ for $P_1$ , $1.12\text{ L}$ for $V_1$ , $273.15\text{ K}$ for $T_1$ , $700\text{ torr}$ for $P_2$ and $298.15\text{ K}$ for $T_2$ in equation (1).

$\begin{array}{c}\frac{760\text{ torr}\times 1.12\text{ L}}{273.15\text{ K}}=\frac{700\text{ torr}\times V_2}{298.15\text{ K}}\\ V_2=\frac{760\text{ torr}\times 1.12\text{ L}}{273.15\text{ K}}\times \frac{298.15\text{ K}}{700\text{ torr}}\\ =1.33\text{ L}\end{array}$

Conclusion

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