Chapter 3, Problem 3B.6E

(a)

Interpretation Introduction

Interpretation:

Amount of argon present in $125\text{mL}$ has to be determined.

Concept Introduction:

An ideal gas contains a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is a theoretical concept. Gases that show perfect elastic collision are practically not possible. At higher $T$ and lower $P$ gases behave almost ideally.

Ideal gas law is an equation of state for any hypothetical gas. Expression for ideal gas equation is as follows:

  $PV=nRT$

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 3B.6E

Amount of argon present in $125\text{mL}$ is $7.7529\text{mol}$.

Explanation of Solution

Ideal gas equation is as follows:

  $PV=nRT$        (1)

Rearrange equation (1) to calculate $n$.

  $n=\frac{PV}{RT}$        (2)

The conversion factor to convert $86\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =86\text{°C}+273.15\\ =359.15\text{K}\end{array}$

Substitute $1.85\text{atm}$ for $P$, $8.205\times {10}^{-2}\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $125\text{mL}$ for $V$ and $359.15\text{K}$ for $T$ in equation (2) to calculate $P$.

  $\begin{array}{c}n=\frac{\left(1.85\text{atm}\right)\left(125\text{mL}\right)\left(\frac{\text{0}\text{.001}\text{L}}{1\text{mL}}\right)}{\left(8.205\times {10}^{-2}\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(359.15\text{K}\right)}\\ =0.00784\text{mol}\end{array}$

Amount of argon present in $125\text{mL}$ is $0.00784\text{mol}$.

(b)

Interpretation Introduction

Interpretation:

Pressure of volume $250\text{mL}$ of ${\text{O}}_2$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.6E

Pressure of volume $250\text{mL}$ of ${\text{O}}_2$ is $0.01469\text{Torr}$.

Explanation of Solution

Expression to calculate moles of ${\text{O}}_2$ is as follows:

  ${\text{moles of O}}_2=\left(\frac{{\text{mass of O}}_2}{\text{molecular weight }\text{}{\text{of O}}_2}\right)$        (3)

Substitute $6.5\text{μg}$ for mass of ${\text{O}}_2$ and $32\text{g/mol}$ for molar mass of ${\text{O}}_2$ in equation (3).

  $\begin{array}{c}{\text{moles of O}}_2=\left(\frac{6.5\text{μg}\left(\frac{{\text{10}}^{-6}\text{g}}{\text{1}\text{μg}}\right)}{32 \text{g/mol}}\right)\\ =2.0312\times {10}^{-7}\text{mol}\end{array}$

Rearrange equation (1) to calculate $P$.

  $P=\frac{nRT}{V}$        (4)

The conversion factor to convert $17\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =17\text{°C}+273.15\\ =290.15\text{K}\end{array}$

Substitute $2.0312\times {10}^{-7}\text{mol}$ for $n$, $8.205\times {10}^{-2}\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $250\text{mL}$ for $V$ and $290.15\text{K}$ for $T$ in equation (4) to calculate $P$.

  $\begin{array}{c}P=\frac{\left(2.0312\times {10}^{-7}\text{mol}\right)\left(8.205\times {10}^{-2}\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(290.15\text{K}\right)}{250\text{mL}\left(\frac{0.001\text{L}}{1\text{mL}}\right)}\\ =1.934\times {10}^{-5}\text{atm}\left(\frac{760\text{Torr}}{1\text{atm}}\right)\\ =0.01469\text{Torr}\end{array}$

Pressure of $250\text{mL}$ of ${\text{O}}_2$ is $0.01469\text{Torr}$.

(c)

Interpretation Introduction

Interpretation:

Mass of ${\text{N}}_2$ at $203\text{K}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.6E

Mass of ${\text{N}}_2$ at $203\text{K}$ is $2.21277\text{g}$.

Explanation of Solution

Substitute $50\text{L}$ for $V$, $62.36\text{L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $20\text{Torr}$ for $P$ and $203\text{K}$ for $T$ in equation (2) to calculate $n$.

  $\begin{array}{c}n=\frac{\left(20\text{Torr}\right)\left(50\text{L}\right)}{\left(62.36\text{L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(203\text{K}\right)}\\ =0.07899\text{mol}\end{array}$

Moles of ${\text{N}}_2$ is $0.07899\text{mol}$.

Expression to calculate moles of ${\text{N}}_2$ is as follows:

  ${\text{moles of N}}_2=\left(\frac{{\text{mass of N}}_2}{\text{molecular weight }\text{}{\text{of N}}_2}\right)$        (5)

Rearrange equation (5) to calculate mass of ${\text{N}}_2$.

  ${\text{mass of N}}_2=\left({\text{moles of N}}_2\right)\left(\text{molecular weight }\text{}{\text{of N}}_2\right)$        (6)

Substitute $0.07899\text{mol}$ for moles of ${\text{N}}_2$ and $28.0134 \text{g/mol}$ for molecular weight of ${\text{N}}_2$ in equation (6).

  $\begin{array}{c}{\text{mass of N}}_2=\left(0.07899\text{mol}\right)\left(28.0134 \text{g/mol}\right)\\ =2.21277\text{g}\end{array}$

Therefore, mass of ${\text{N}}_2$ at $203\text{K}$ is $2.21277\text{g}$.

(d)

Interpretation Introduction

Interpretation:

Volume of $\text{Kr}$ at $2.00\times {10}^2\text{mTorr}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.6E

Volume of $\text{Kr}$ at $2.00\times {10}^2\text{mTorr}$ is $48.5001\text{L}$.

Explanation of Solution

The conversion factor to convert $65\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =65\text{°C}+273.15\\ =338.15\text{K}\end{array}$

Expression to calculate moles of $\text{Kr}$ is as follows:

  $\text{moles of Kr}=\left(\frac{\text{mass of Kr}}{\text{molecular weight }\text{}\text{of Kr}}\right)$        (7)

Substitute $0.0387\text{g}$ for mass of $\text{Kr}$ and $83.798 \text{g/mol}$ for molecular weight of $\text{Kr}$ in equation (7).

  $\begin{array}{c}\text{moles of Kr}=\left(\frac{0.0387\text{g}}{83.798 \text{g/mol}}\right)\\ =0.00046\text{mol}\end{array}$

Rearrange equation (1) to calculate $P$.

  $V=\frac{nRT}{P}$        (8)

Substitute $0.00046\text{mol}$ for $n$, $62.36\text{L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $2.00\times {10}^2\text{mTorr}$ for $P$ and $338.15\text{K}$ for $T$ in equation (8) to calculate $V$.

  $\begin{array}{c}V=\frac{\left(0.00046\text{mol}\right)\left(62.36\text{L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(338.15\text{K}\right)}{2.00\times {10}^2\text{mTorr}\left(\frac{{10}^{-3}\text{Torr}}{1\text{mTorr}}\right)}\\ =48.5001\text{L}\end{array}$

Hence, volume of $\text{Kr}$ is $48.5001\text{L}$.