Chapter 12, Problem 88AE

(a)

Interpretation Introduction

Interpretation:

Volume of $\text{0}\text{.510 mol}$ of gas at $\text{47 }°\text{C}$ and $\text{1}\text{.6 atm}$ has to be determined.

Concept Introduction:

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

  $PV=nRT$        (1)

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 88AE

Volume of $\text{0}\text{.510 mol}$ of gas at $320\text{K}$ and $\text{1}\text{.6 atm}$ is $8.37\text{ L}$.

Explanation of Solution

Rearrange expression (1) to calculate volume of gas as follows:

  $V=\frac{nRT}{P}$        (2)

Conversion of $\text{47 }°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =\text{47 }°\text{C}+273\\ =320\text{K}\end{array}$

Substitute $\text{0}\text{.510 mol}$ for $n$, $\text{1}\text{.6 atm}$ for $P$, $0.0821\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $320\text{K}$ for $T$ in equation (2).

  $\begin{array}{c}V=\frac{\left(\text{0}\text{.510 mol}\right)\left(0.0821\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(320\text{K}\right)}{\text{1}\text{.6 atm}}\\ =8.37\text{ L}\end{array}$

Hence volume of $\text{0}\text{.510 mol}$ of gas at $320\text{K}$ and $\text{1}\text{.6 atm}$ is $8.37\text{ L}$.

(b)

Interpretation Introduction

Interpretation:

Grams of ${\text{CH}}_4$ in $16.0\text{ L}$ gas at $\text{27 }°\text{C}$ and $\text{1}\text{.6 atm}$ have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 88AE

Grams of ${\text{CH}}_4$ in $16.0\text{ L}$ gas at $\text{27 }°\text{C}$ and $\text{1}\text{.6 atm}$ are $8.2\text{ g}$.

Explanation of Solution

Rearrange expression (1) to calculate value of $n$ as follows:

  $n=\frac{PV}{RT}$        (3)

Conversion factor to convert $27°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =27°\text{C}+273\\ =300\text{K}\end{array}$

Substitute $16.0\text{ L}$ for $V$, $600\text{ torr}$ for $P$, $62.364\text{ L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $300\text{K}$ for $T$ in equation (3).

  $\begin{array}{c}n=\frac{\left(600\text{ torr}\right)\left(16.0\text{ L}\right)}{\left(62.364\text{ L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\\ =0.51\text{ mol}\end{array}$

Mass of gas can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(\text{Moles}\right)\left(\text{Molar mass}\right)\\ =\left(\text{0}\text{.51 mol}\right)\left(16.04\text{ g/mol}\right)\\ =8.2\text{ g}\end{array}$

Hence grams of ${\text{CH}}_4$ in $16.0\text{ L}$ gas at $\text{27 }°\text{C}$ and $\text{1}\text{.6 atm}$ are $8.2\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

Density of ${\text{CO}}_2$ gas at $4.00\text{ atm}$ and $20.0°\text{C}$ has to be determined.

Concept Introduction:

Density of gas represents mass of gas present in unit volume. Expression of density is as follows:

  $d\left(\text{g/L}\right)=\frac{\text{mass}\left(\text{g}\right)}{\text{volume}\left(\text{L}\right)}$

Here, $d$ is density of gas.

Answer to Problem 88AE

Density of ${\text{CO}}_2$ gas at $4.00\text{ atm}$ and $20.0°\text{C}$ is $7.31\text{ g/L}$.

Explanation of Solution

Rearrange expression (1) to calculate $n\text{/}V$ value as follows:

  $\frac{n}{V}=\frac{P}{RT}$        (4)

Conversion of $20°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =20°\text{C}+273\\ =293\text{K}\end{array}$

Substitute $4.00\text{ atm}$ for $P$, $0.0821\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $293\text{K}$ for $T$ in equation (4) to calculate $n\text{/}V$ value for $\text{He}$ gas.

  $\begin{array}{c}\frac{n}{V}\left(\text{mol/L}\right)=\frac{\left(4.00\text{ atm}\right)}{\left(0.0821\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(293\text{K}\right)}\\ =0.166\text{ mol/L}\end{array}$

Value $0.166\text{ mol/L}$ can be converted into to $\text{g/L}$ as follows:

  $\begin{array}{c}\text{Density}\left(\text{g/L}\right)=\left(0.166\text{ mol/L}\right)\left({\text{Molar mass of CO}}_2\right)\\ =\left(0.166\text{ mol/L}\right)\left(44.01\text{ g/mol}\right)\\ =7.31\text{ g/L}\end{array}$

Hence density of ${\text{CO}}_2$ gas at $4.00\text{ atm}$ and $20.0°\text{C}$ is $7.31\text{ g/L}$.

(d)

Interpretation Introduction

Interpretation:

Molar mass of gas that has $2.58\text{ g/L}$ density at $27°\text{C}$ and $1.00\text{ atm}$ has to be determined.

Concept Introduction:

Combination of three laws Charles' law, Boyle's law, and Gay-Lussac's law form combine law. This law used to determine change in pressure, volume and temperature if two of them got varied.

Mathematically relation for combined law is as follows:

  $\frac{PV}{T}=k$

Or,

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Here,

$P_1$ is initial pressure.

$P_2$ is final pressure.

$T_1$ is initial temperature.

$T_2$ is final temperature.

$V_1$ is initial volume.

$V_2$ is final volume.

Answer to Problem 88AE

Molar mass of gas is $70.4\text{ g/mol}$.

Explanation of Solution

Expression to calculate final volume of gas at STP is as follows:

  $V_2=\frac{V_1{T}_2}{T_1}$        (5)

Conversion factor to convert $27°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =27°\text{C}+273\\ =300\text{K}\end{array}$

Substitute $1.00\text{ L}$ for $V_1$, $300\text{K}$ for $T_1$, $273\text{K}$ for $T_2$ in equation (5).

  $\begin{array}{c}{V}_2=\frac{\left(1.00\text{ L}\right)\left(273\text{K}\right)}{\left(300\text{K}\right)}\\ =0.91\text{ L}\end{array}$

Therefore, volume of gas at STP is $0.91\text{ L}$.

Conversion factor for conversion of $\text{g/L}$ to $\text{g/mol}$ at STP is as follows:

  $\left(\frac{22.4\text{ L}}{1\text{ mol}}\right)$

Therefore, molar mass of $2.58\text{ g}$ gas with $0.91\text{ L}$ volume at STP can be calculated as follows:

  $\begin{array}{c}\text{Molar mass}\left(\text{g/mol}\right)=\left(\frac{2.58\text{ g}}{0.91\text{ L}}\right)\left(\frac{22.4\text{ L}}{1\text{ mol}}\right)\\ =70.4\text{ g/mol}\end{array}$

Hence molar mass of gas is $70.4\text{ g/mol}$.