Chapter 5, Problem 5.137P

(a)

Interpretation Introduction

Interpretation:

The mass of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CO}}_{\text{2}}$ gas exhaled in $1\text{ h}$ of sleep are to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.137P

The mass of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ exhaling in $1\text{ h}$ of sleep is $8.3\text{ g}$ and the mass of ${\text{CO}}_{\text{2}}\left(g\right)$ exhaling in $1\text{ h}$ of sleep is $20.5\text{ g}$.

Explanation of Solution

The equation for the reaction of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)$ with ${\text{O}}_{\text{2}}\left(g\right)$ is as follows:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)+{\text{6O}}_{\text{2}}\left(g\right)\to 6{\text{CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$        (1)

The formula to convert $°\text{C}$ to Kelvin is:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$        (2)

Substitute $\text{37 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (2).

$\begin{array}{c}T\left(\text{K}\right)=\text{37 }°\text{C}+273.15\\ =310.15\text{ K}\end{array}$

Convert pressure from $\text{torr}$ into $\text{atm}$ is as follows:

$\begin{array}{c}P=30\text{ torr}\left(\frac{1\text{ atm}}{\text{760 torr}}\right)\\ =0.0394737\text{ atm}\end{array}$

The expression to calculate the moles of the ${\text{CO}}_{\text{2}}\left(g\right)$ is as follows,

$PV=nRT$        (3)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of ${\text{CO}}_{\text{2}}$ and $R$ is the gas constant.

Rearrange equation (3) for $n$ as follows:

$n=\frac{PV}{RT}$        (4)

Substitute $0.0394737\text{ atm}$ for $P$, $310.15\text{ K}$ for $T$, $\text{300 L}$ for $V$ and $0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}$ for $R$ in the equation (4).

$\begin{array}{c}n=\frac{\left(0.0394737\text{ atm}\right)\left(\text{300 L}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)\left(\text{310}\text{.15 K}\right)}\\ =0.4650653\text{ mol}\end{array}$

From the equation (1), six moles of ${\text{CO}}_{\text{2}}\left(g\right)$ and six moles of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ are formed in the reaction. Thus, the moles of ${\text{H}}_{\text{2}}\text{O}$ is calculated from ${\text{CO}}_{\text{2}}\left(g\right)$ as follows:

${\text{Moles of H}}_{\text{2}}\text{O}\left(l\right)=\left[\left({\text{Moles of CO}}_{\text{2}}\left(g\right)\right)\left(\frac{{\text{6 mol of H}}_{\text{2}}\text{O}\left(l\right)}{{\text{6 mol of CO}}_{\text{2}}\left(g\right)}\right)\right]$        (5)

Substitute the value $0.4650653\text{ mol}$ for moles of ${\text{CO}}_{\text{2}}\left(g\right)$ in the equation (5).

$\begin{array}{c}{\text{Moles of H}}_{\text{2}}\text{O}\left(l\right)=\left[\left(0.4650653\text{ mol}\right)\left(\frac{{\text{6 mol of H}}_{\text{2}}\text{O}\left(l\right)}{{\text{6 mol of CO}}_{\text{2}}\left(g\right)}\right)\right]\\ =0.4650653\text{ mol}\end{array}$

The expression to calculate the mass of ${\text{CO}}_{\text{2}}\left(g\right)$ exhaled in $1\text{ h}$ of sleep is as follows:

${\text{Moles of CO}}_{\text{2}}\left(g\right)=\frac{{\text{Mass of CO}}_{\text{2}}\left(g\right)}{{\text{Molar mass of CO}}_{\text{2}}\left(g\right)}$        (6)

Rearrange the equation (6) to calculate the mass of ${\text{CO}}_{\text{2}}\left(g\right)$ is as follows:

${\text{Mass of CO}}_{\text{2}}\left(g\right)=\left({\text{Moles of CO}}_{\text{2}}\left(g\right)\right)\left({\text{Molar mass of CO}}_{\text{2}}\left(g\right)\right)$        (7)

Substitute the value $0.4650653\text{ mol}$ for moles of ${\text{CO}}_{\text{2}}\left(g\right)$ and $44.01\text{ g/mol}$ for molar mass of ${\text{CO}}_{\text{2}}\left(g\right)$ in the equation (7).

$\begin{array}{c}{\text{Mass of CO}}_{\text{2}}\left(g\right)=\left(0.4650653\text{ mol}\right)\left(44.01\text{ g/mol}\right)\\ =20.4675\text{ g}\\ \approx 20.5\text{ g}\end{array}$

Hence, the mass of ${\text{CO}}_{\text{2}}\left(g\right)$ exhaled in $1\text{ h}$ of sleep is $20.5\text{ g}$.

The expression to calculate the mass of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ exhaled in $1\text{ h}$ of sleep is as follows:

${\text{Moles of H}}_{\text{2}}\text{O}\left(l\right)=\frac{{\text{Mass of H}}_{\text{2}}\text{O}\left(l\right)}{{\text{Molar mass of H}}_{\text{2}}\text{O}\left(l\right)}$        (8)

Rearrange the equation (8) to calculate the mass of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ is as follows:

${\text{Mass of H}}_{\text{2}}\text{O}\left(l\right)=\left({\text{Moles of H}}_{\text{2}}\text{O}\left(l\right)\right)\left({\text{Molar mass of H}}_{\text{2}}\text{O}\left(l\right)\right)$        (9)

Substitute the value $0.4650653\text{ mol}$ for moles of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ and $18.01\text{ g/mol}$ for molar mass of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ in the equation (9).

$\begin{array}{c}{\text{Mass of H}}_{\text{2}}\text{O}\left(l\right)=\left(0.4650653\text{ mol}\right)\left(18.02\text{ g/mol}\right)\\ =8.3805\text{ g}\\ \approx 8.38\text{ g}\end{array}$

Hence, the mass of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ exhaled in $1\text{ h}$ of sleep is $8.3\text{ g}$.

Conclusion

The mass of ${\text{H}}_{\text{2}}\text{O}$ exhaled in $1\text{ h}$ of sleep is $8.3\text{ g}$ and the mass of ${\text{CO}}_{\text{2}}$ exhaled in $1\text{ h}$ of sleep is $20.5\text{ g}$.

(b)

Interpretation Introduction

Interpretation:

The grams of the body mass lost in $8\text{ h}$ of sleep is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

Answer to Problem 5.137P

The grams of the body mass lost in $8\text{ h}$ of sleep is $1.0\times {10}^2\text{ g}$.

Explanation of Solution

The expression to calculate the moles of ${\text{CO}}_{\text{2}}$ exhaled in $8\text{ h}$ is as follows:

${\text{Moles of CO}}_{\text{2}}=\left({\text{Moles of CO}}_{\text{2}}\text{ exhaled in 1hr}\right)\left(\text{time}\right)$        (10)

Substitute the value $0.4650653\text{ mol}$ for moles of ${\text{CO}}_{\text{2}}$ exhaled in $1\text{ hr}$ and $8\text{ h}$ for time in the equation (10).

$\begin{array}{c}{\text{Moles of CO}}_{\text{2}}=\left(0.4650653\text{ mol}\right)\left(8\text{ h}\right)\\ =3.720523\text{ mol}\end{array}$

From the equation (1), one mole of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)$ will be formed six moles of ${\text{CO}}_2$. Thus, the moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)$ is calculated from ${\text{CO}}_2$ as follows:

${\text{Moles of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)=\left[\left({\text{Moles of CO}}_2\right)\left(\frac{{\text{1 mol of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)}{{\text{6 mol of CO}}_2}\right)\right]$        (11)

Substitute the value $0.4650653\text{ mol}$ for moles of ${\text{CO}}_{\text{2}}\left(g\right)$ in the equation (11).

$\begin{array}{c}{\text{Moles of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)=\left[\left(0.4650653\text{ mol}\right)\left(\frac{{\text{1 mol of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)}{{\text{6 mol of CO}}_2}\right)\right]\\ =0.0775108\text{ mol}\end{array}$

The expression to calculate the mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)$ is as follows:

${\text{Moles of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)=\frac{{\text{Mass of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)}{{\text{Molar mass of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)}$        (12)

Rearrange the equation (12) to calculate the mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)$ is as follows:

${\text{Mass of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)=\left({\text{Moles of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)\right)\left({\text{Molar mass of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)\right)$        (13)

Substitute the value $0.0775108\text{ mol}$ for moles of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ and $180.16\text{ g/mol}$ for the molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in the equation (13).

$\begin{array}{c}{\text{Mass of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}=\left(0.0775108\text{ mol}\right)\left(180.16\text{ g/mol}\right)\\ =111.7149\text{ g}\\ \approx 1.0\times {10}^2\text{ g}\end{array}$

Hence, the grams of the body mass lost in $8\text{ h}$ of sleep is $1.0\times {10}^2\text{ g}$.

Conclusion

The grams of the body mass lose in $8\text{ h}$ of sleep is $1.0\times {10}^2\text{ g}$.