Chapter 12, Problem 59AE

(a)

Interpretation Introduction

Interpretation:

Pressure of gas if volume is doubled at constant pressure and moles has to be determined.

Concept Introduction:

Boyle’s law represents inverse relation of volume of fixed mass of ideal gas with its pressure at fixed temperature.

Mathematically relation for Boyle’s law is as follows:

  $V\propto \frac{1}{P}$

Or,

  $P_1{V}_1=P_2{V}_2$

Here,

$P_1$ is initial pressure.

$P_2$ is final pressure.

$V_1$ is initial volume.

$V_2$ is final volume.

Answer to Problem 59AE

Final pressure of gas is halved $208\text{ L}$.

Explanation of Solution

Expression to calculate final pressure of gas is as follows:

  $P_2=\frac{P_1{V}_1}{V_2}$        (1)

Substitute $\text{10 L}$ for $V_1$, $945\text{ torr}$ for $P_1$, and $20\text{ L}$ for $V_2$ in equation (1).

  $\begin{array}{c}{P}_2=\frac{\left(945\text{ torr}\right)\left(\text{10 L}\right)}{\text{20 L}}\\ =470\text{ L}\end{array}$

Hence final pressure of gas is halved $208\text{ L}$.

(b)

Interpretation Introduction

Interpretation:

Pressure of gas if volume is doubled at constant pressure and moles has to be determined.

Concept Introduction:

Gay-lussac’s law represents direct relation of pressure of fixed mass of ideal gas with its temperature at fixed volume.

Mathematically relation for Gay-lussac’s law is as follows:

  $P\propto T$

Or,

  $P_1{T}_2=P_2{T}_1$

Here,

$P_1$ is initial pressure.

$P_2$ is final pressure.

$T_1$ is initial temperature.

$T_2$ is final temperature.

Answer to Problem 59AE

Final pressure of gas is doubled $208\text{ L}$.

Explanation of Solution

Expression to calculate final pressure of gas is as follows:

  $P_2=\frac{P_1{T}_2}{T_1}$        (2)

Conversion factor to convert $\text{30 }°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =\text{30 }°\text{C}+273\\ =303\text{K}\end{array}$

Substitute $303\text{K}$ for $T_1$, $945\text{ torr}$ for $P_1$, and $606\text{K}$ for $T_2$ in equation (2).

  $\begin{array}{c}{P}_2=\frac{\left(945\text{ torr}\right)\left(606\text{K}\right)}{303\text{K}}\\ =1890\text{ torr}\end{array}$

Hence final pressure of gas is doubled $1890\text{ torr}$.

(c)

Interpretation Introduction

Interpretation:

Pressure of gas if moles of gas is cut in half at constant volume and temperature has to be determined.

Concept Introduction:

Gay-lussac’s law represents direct relation of pressure of fixed mass of ideal gas with its temperature at fixed volume.

Mathematically relation for Gay-lussac’s law is as follows:

  $P\propto T$

Or,

  $P_1{T}_2=P_2{T}_1$

Here,

$P_1$ is initial pressure.

$P_2$ is final pressure.

$T_1$ is initial temperature.

$T_2$ is final temperature.

Answer to Problem 59AE

Final pressure of gas is doubled $208\text{ L}$.

Explanation of Solution

Expression to calculate final pressure of gas at constant volume and temperature is as follows:

  $P_2=\frac{P_1{n}_2}{n_1}$        (3)

Substitute $0.5\text{ mol}$ for $n_1$, $945\text{ torr}$ for $P_1$, and $0.25\text{ mol}$ for $n_2$ in equation (3).

  $\begin{array}{c}{P}_2=\frac{\left(945\text{ torr}\right)\left(0.25\text{ mol}\right)}{0.5\text{ mol}}\\ =470\text{ torr}\end{array}$

Hence final pressure of gas is halved $470\text{ torr}$.

(d)

Interpretation Introduction

Interpretation:

Pressure of container if 1 moles of ${\text{N}}_2$ gas is added to container at same volume and temperature has to be determined.

Concept Introduction:

Ideal gas law represents an equation to relate its volume and pressure with its temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$        (4)

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 59AE

Final pressure of container is $\text{2835 torr}$.

Explanation of Solution

Rearrange expression (1) to calculate pressure as follows:

  $P=\frac{nRT}{V}$        (5)

Substitute $10\text{ L}$ for $V$, $1\text{ mol}$ for $n$, $62.364\text{ L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $303\text{K}$ for $T$ in equation (5).

  $\begin{array}{c}P=\left(\frac{\left(1\text{ mol}\right)\left(62.364\text{ L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(303\text{K}\right)}{10\text{ L}}\right)\\ =1890\text{ torr}\end{array}$

Total pressure can be calculated as follows:

  $\begin{array}{c}{P}_{\text{T}}=\left(945+1890\right)\text{ torr}\\ =\text{2835 torr}\end{array}$

Hence final pressure of container is $\text{2835 torr}$.