Chapter 3, Problem 3B.29E

(a)

Interpretation Introduction

Interpretation:

Volume of air at $52\text{kPa}$ and $-5°\text{C}$ 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.29E

Volume of air at $52\text{kPa,}-5°\text{C}$ 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 air 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 $21.1\text{°C}$ to $\text{K}$ is as follows:

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =21.1\text{°C}+273.15\\ =294.25\text{K}\end{array}$

The conversion factor to convert $-5\text{°C}$ to $\text{K}$ is as follows:

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =-5\text{°C}+273.15\\ =268.15\text{K}\end{array}$

Substitute $2{\text{m}}^3$ for $V_1$, $104\text{kPa}$ for $P_1$, $294.25\text{K}$ for $T_1$, $52\text{kPa}$ for $P_2$ and $268.15\text{K}$ for $T_2$ in equation (3).to calculate $V_2$.

  $\begin{array}{c}{V}_2=\frac{\left(268.15\text{K}\right)\left(104\text{kPa}\right)\left(2{\text{m}}^3\right)}{\left(52\text{kPa}\right)\left(294.25\text{K}\right)}\\ =3.6451{\text{m}}^3\end{array}$

Volume of air at $52\text{kPa,}-5°\text{C}$ is $3.6451{\text{m}}^3$.

(b)

Interpretation Introduction

Interpretation:

Volume of air at $880\text{Pa}$ and $-52°\text{C}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.29E

Volume of air at $880\text{Pa,}-52°\text{C}$ 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 air 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 $21.1\text{°C}$ to $\text{K}$ is as follows:

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =21.1\text{°C}+273.15\\ =294.25\text{K}\end{array}$

The conversion factor to convert $-52\text{°C}$ to $\text{K}$ is as follows:

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =-52\text{°C}+273.15\\ =221.15\text{K}\end{array}$

Substitute $2{\text{m}}^3$ for $V_1$, $104\text{kPa}$ for $P_1$, $294.25\text{K}$ for $T_1$, $880\text{Pa}$ for $P_2$ and $221.15\text{K}$ for $T_2$ in equation (3) to calculate $V_2$.

  $\begin{array}{c}{V}_2=\frac{\left(221.15\text{K}\right)\left(104\text{kPa}\right)\left(2{\text{m}}^3\right)}{\left(880\text{Pa}\right)\left(294.25\text{K}\right)}\\ =0.1777{\text{m}}^3\end{array}$

Volume of air at $52\text{kPa,}-5°\text{C}$ is $0.1777{\text{m}}^3$.