Chapter 12, Problem 12.123P

(a)

Interpretation Introduction

Interpretation:

The mass of liquid water that condenses inside the bottle is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$        (1)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

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

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273$        (2)

Answer to Problem 12.123P

The mass of liquid water that condenses inside the bottle is $0.0027\text{g}$.

Explanation of Solution

Substitute $22°\text{C}$ in equation (2) to calculate the temperature $T_{\text{initial}}$ in Kelvin as follows:

$\begin{array}{c}{T}_{\text{initial}}\left(\text{K}\right)=22°\text{C}+273\\ =295\text{ K}\end{array}$

Substitute $0°\text{C}$ in equation (2) to calculate the temperature $T_{\text{final}}$ in Kelvin as follows:

$\begin{array}{c}{T}_{\text{final}}\left(\text{K}\right)=0°\text{C}+273\\ =273\text{ K}\end{array}$

The relative humidity is $44\%$ and therefore the initial pressure is calculated as follows:

$\begin{array}{c}\text{Pressure}=44\%\left(19.8\text{torr}\right)\\ =8.712\text{torr}\end{array}$

Rearrange the equation (1) to calculate the initial number of moles.

$n_{\text{initial}}=\frac{PV}{RT}$        (3)

Substitute the value $8.712\text{torr}$ for $P$, $295\text{ K}$ for $T$, $0.75\text{L}$ for $V$ and $0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}$ for $R$ in the equation (3).

$\begin{array}{c}{n}_{\text{initial}}=\frac{\left(8.712\text{torr}\right)\left(0.75\text{L}\right)}{\left(0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(295\text{ K}\right)}\left(\frac{1\text{atm}}{\text{760}\text{torr}}\right)\\ =0.000354977\text{mol}\end{array}$

The formula to calculate initial mass is as follows:

$m_{\text{initial}}=\left(n_{\text{initial}}\right)\left(M_{{\text{H}}_{\text{2}}\text{O}}\right)$        (4)

Substitute the value $0.000354977\text{mol}$ for $n_{\text{initial}}$ and $18.02\text{ g/mol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$ in the equation (4).

$\begin{array}{c}{m}_{\text{initial}}=\left(0.000354977\text{mol}\right)\left(18.02\text{ g/mol}\right)\\ =0.0063966877\text{ g}\end{array}$

Rearrange the equation (1) to calculate the final number of moles.

$n_{\text{final}}=\frac{PV}{RT}$        (5)

Substitute the value $4.6\text{torr}$ for $P$, $273\text{ K}$ for $T$, $0.75\text{L}$ for $V$ and $0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}$ for $R$ in the equation (5).

$\begin{array}{c}{n}_{\text{initial}}=\frac{\left(4.6\text{torr}\right)\left(0.75\text{L}\right)}{\left(0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(273\text{ K}\right)}\left(\frac{1\text{atm}}{\text{760}\text{torr}}\right)\\ =0.000202534\text{mol}\end{array}$

The formula to calculate final mass is as follows:

$m_{\text{final}}=\left(n_{\text{final}}\right)\left(M_{{\text{H}}_{\text{2}}\text{O}}\right)$        (6)

Substitute the value $0.000202534\text{mol}$ for $n_{\text{final}}$ and $18.02\text{ g/mol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$ in the equation (6).

$\begin{array}{c}{m}_{\text{final}}=\left(0.000202534\text{mol}\right)\left(18.02\text{ g/mol}\right)\\ =0.0036470057\text{ g }\end{array}$

The formula to calculate the mass of ${\text{H}}_{\text{2}}\text{O}$ condensed is as follows:

${\text{Mass of H}}_{\text{2}}\text{O condensed}=m_{\text{initial}}-m_{\text{final}}$        (7)

Substitute the value $0.0063966877\text{ g}$ for $m_{\text{initial}}$ and $0.0036470057\text{ g }$ for $m_{\text{final}}$ in the equation (7).

$\begin{array}{c}{\text{Mass of H}}_{\text{2}}\text{O condensed}=0.0063966877\text{ g}-0.0036470057\text{ g }\\ =0.002749682\text{g}\\ \approx 0.0027\text{g}\end{array}$

Conclusion

The mass of liquid water that condenses inside the bottle is $0.0027\text{g}$.

(b)

Interpretation Introduction

Interpretation:

Whether the liquid water condenses at $10°\text{C}$ is to be determined.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$        (1)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

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

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273$        (2)

Answer to Problem 12.123P

The liquid water cannot condense at $10°\text{C}$.

Explanation of Solution

Substitute $10°\text{C}$ in equation (2) to calculate the temperature $T$ in Kelvin as follows:

$\begin{array}{c}T\left(\text{K}\right)=10°\text{C}+273\\ =283\text{ K}\end{array}$

Rearrange the equation (1) to calculate the number of moles at $10°\text{C}$.

$n=\frac{PV}{RT}$        (3)

Substitute the value $9.2\text{torr}$ for $P$, $283\text{ K}$ for $T$, $0.75\text{L}$ for $V$ and $0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}$ for $R$ in the equation (3).

$\begin{array}{c}n=\frac{\left(9.2\text{torr}\right)\left(0.75\text{L}\right)}{\left(0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(283\text{ K}\right)}\left(\frac{1\text{atm}}{\text{760}\text{torr}}\right)\\ =0.00040506964\text{mol}\end{array}$

The formula to calculate mass at $10°\text{C}$ is as follows:

$m=\left(n\right)\left(M_{{\text{H}}_{\text{2}}\text{O}}\right)$        (4)

Substitute the value $0.00040506964\text{mol}$ for $n$ and $18.02\text{ g/mol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$ in the equation (4).

$\begin{array}{c}m=\left(0.00040506964\text{mol}\right)\left(18.02\text{ g/mol}\right)\\ =0.007041427\text{ g}\end{array}$

Mass of water in equilibrium with vapor state is $0.007041427\text{ g}$ and the initial mass is $0.0063\text{ g}$. Therefore the liquid water cannot condense at $10°\text{C}$.

Conclusion

The liquid water cannot condense at $10°\text{C}$.