Chapter 3, Problem 3B.17E

(a)

Interpretation Introduction

Interpretation:

Volume of sample at $1\text{atm}$ and $298\text{K}$ 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.17E

Volume of sample at $1\text{atm}$ and $298\text{K}$ is $11.82399\text{mL}$.

Explanation of Solution

Ideal gas equation is also expressed as follows:

  $\frac{P_1{V}_1}{n_1{T}_1}=\frac{P_2{V}_2}{n_2{T}_2}$        (1)

Here,

$P_1$ is initial pressure of gas.

$P_2$ is final pressure of gas.

$V_1$ is initial volume of gas.

$V_2$ is final volume of gas.

$n_1$ denotes the initial number of moles of gas.

$n_2$ denotes the initial number of moles of gas.

$T_1$ is initial temperature of gas.

$T_2$ is final temperature of gas.

Moles of $\text{Xe}$ remain constant so $n_1$ is equal to $n_2$.

So the modified ideal gas equation is as follows:

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$        (2)

Rearrange equation (2) to calculate $V_2$.

  $V_2=\frac{T_2{P}_1{V}_1}{P_2{T}_1}$        (3)

The conversion factor to convert $-45\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =-45\text{°C}+273.15\\ =228.15\text{K}\end{array}$

Substitute $35.5\text{mL}$ for $V_1$, $0.255\text{atm}$ for $P_1$, $228.15\text{K}$ for $T_1$, $1\text{atm}$ for $P_2$ and $298\text{K}$ for $T_2$ in equation (3).to calculate $V_2$.

  $\begin{array}{c}{V}_2=\frac{\left(298\text{K}\right)\left(0.255\text{atm}\right)\left(35.5\text{mL}\right)}{\left(1\text{atm}\right)\left(228.15\text{K}\right)}\\ =11.82399\text{mL}\end{array}$

Hence, volume of sample at $1\text{atm}$ and $298\text{K}$ is $11.82399\text{mL}$.

(b)

Interpretation Introduction

Interpretation:

Pressure of sample at $20°\text{C}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.17E

Pressure of sample at $20°\text{C}$ is $1.03167\text{atm}$.

Explanation of Solution

The conversion factor to convert $20\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =20\text{°C}+273.15\\ =293.15\text{K}\end{array}$

Rearrange equation (2) to calculate $P_2$.

  $P_2=\frac{P_1{V}_1{T}_2}{T_1{V}_2}$        (4)

Substitute $35.5\text{mL}$ for $V_1$, $0.255\text{atm}$ for $P_1$, $228.15\text{K}$ for $T_1$, $12\text{mL}$ for $V_2$ and $293.15\text{K}$ for $T_2$ in equation (4).to calculate $P_2$.

  $\begin{array}{c}{P}_2=\frac{\left(0.255\text{atm}\right)\left(35.5\text{mL}\right)\left(293.15\text{K}\right)}{\left(228.15\text{K}\right)\left(12\text{mL}\right)}\\ =0.9692\text{atm}\end{array}$

Hence, pressure of sample at $20°\text{C}$ is $0.9692\text{atm}$.

(c)

Interpretation Introduction

Interpretation:

Temperature for xenon at $5\times {10}^2\text{Torr}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.17E

Temperature for xenon at $5\times {10}^2\text{Torr}$ is $588.622\text{K}$.

Explanation of Solution

Since volume remains constant so $V_1$ is equal to $V_2$.

So the modified ideal gas equation is as follows:

  $\frac{P_1}{T_1}=\frac{P_2}{T_2}$        (5)

Rearrange equation (5) to calculate $T_2$.

  $T_2=\frac{T_1{P}_2}{P_1}$        (6)

Substitute $0.255\text{atm}$ for $P_1$, $228.15\text{K}$ for $T_1$ and $5\times {10}^2\text{Torr}$ for $P_2$ in equation (6).to calculate $T_2$.

  $\begin{array}{c}{T}_2=\frac{\left(228.15\text{K}\right)\left(5\times {10}^2\text{Torr}\right)}{\left(0.255\text{atm}\right)\left(\frac{760\text{Torr}}{\text{1}\text{atm}}\right)}\\ =588.622\text{K}\end{array}$

Hence, temperature for xenon at $5\times {10}^2\text{Torr}$ is $588.622\text{K}$.