Chapter 16, Problem 30P

(a)

To determine

The mass of the gas.

Answer to Problem 30P

The mass of the gas is $1.18\times {10}^{-3}\text{kg}$.

Explanation of Solution

Given information:Temperature of the air is $300\text{K}$, atmospheric pressure of the air is $1.013\times {10}^5\text{Pa}$, equivalent molar mass of the air is $28.9\text{g}/\text{mol}$, edge of the cube is $10.0\text{cm}$.

Calculate the volume of the air.

$V=x^3$

Here,

$V$ is the volume of the air.

$x$ is the edge of the cube.

Formula to calculate number of moles of gas.

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$ (I)

Here,

$n$ is the number of moles of gas.

$P$ is the atmospheric pressure of the air.

$R$ is the ideal gas constant.

$T$ is the temperature of the air

Substitute $x^3$ for $V$ in equation (I) to find $n$,

$n=\frac{P{x}^3}{RT}$ (II)

The value of ideal gas constant is $8.314\text{J}/\text{molK}$.

Substitute $8.314\text{J}/\text{molK}$ for $R$, $\left(1.013\times {10}^5\text{Pa}\right)$ for $P$, $10.0\text{cm}$ for $x$, $\left(300\text{K}\right)$ for $T$ in equation (II) to find $n$.

$\begin{array}{c}n=\frac{\left(1.013\times {10}^5\text{Pa}\right)\times \left({\left(10.0\text{cm}\right)}^3\times {\left(\frac{1\text{m}}{100\text{cm}}\right)}^3\right)}{\left(8.314\text{J}/\text{molK}\right)\times \left(300\text{K}\right)}\\ =0.040614\text{moles}\\ \approx 4.1\times {10}^{-2}\text{moles}\end{array}$

Thus, the number of moles of the gas is $4.1\times {10}^{-2}\text{moles}$.

Formula to calculate mass of the gas.

$n=\frac{m}{M}$

$m=n\times M$ (III)

Here,

$n$ is the number of moles of gas.

$m$ is the mass of the gas.

$M$ is the equivalent molar mass.

Substitute $4.1\times {10}^{-2}\text{moles}$ for $n$, $28.9\text{g}/\text{mol}$ for $M$ in equation (III) to find $m$.

$\begin{array}{c}m=4.1\times {10}^{-2}\text{moles}\times 28.9\text{g}/\text{mole}\\ =1.18\text{g}\\ =1.18\text{g}\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\\ =1.18\times {10}^{-3}\text{kg}\end{array}$

Conclusion:

Therefore, the mass of the gas is $1.18\times {10}^{-3}\text{kg}$.

(b)

To determine

The gravitational force exerted on the container.

Answer to Problem 30P

The gravitational force exerted on the container is $0.0115\text{N}$.

Explanation of Solution

Given information:Temperature of the air is $300\text{K}$, atmospheric pressure of the air is $1.013\times {10}^5\text{Pa}$, equivalent molar mass of the air is $28.9\text{g}/\text{mol}$, edge of the cube is $10.0\text{cm}$.

Formula to calculate gravitational force exerted on the container.

$F_g=mg$ (IV)

Here,

$F_g$ is the gravitational force exerted on the container.

$m$ is the mass of the gas.

$g$ is the acceleration due to gravity.

The value of acceleration due to gravity is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Substitute $1.18\times {10}^{-3}\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (IV) to find $F_g$.

$\begin{array}{c}{F}_g=mg\\ =\left(1.15\times {10}^{-3}\text{kg}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =0.0115\text{N}\end{array}$

Conclusion:

Therefore, the gravitational force exerted on the container is $0.0115\text{N}$.

(c)

To determine

The force exerted on each face of the cube.

Answer to Problem 30P

The force exerted on each face of the cube $1.92\times {10}^{-3}\text{N}$.

Explanation of Solution

Given information:Temperature of the air is $300\text{K}$, atmospheric pressure of the air is $1.013\times {10}^5\text{Pa}$, equivalent molar mass of the air is $28.9\text{g}/\text{mol}$, edge of the cube is $10.0\text{cm}$.

Formula to calculate force exerted on each face of the cube.

$F=\frac{F_g}{a}$ (V)

Here,

$F$ is the force exerted on each face of the cube.

$a$ is the number of faces of the cube.

The number of faces of the cube are $6$.

Substitute $0.0115\text{N}$ for $F_g$, $6$ for $a$ in equation (V) to find $F$,

$\begin{array}{c}F=\frac{0.0115\text{N}}{6}\\ =1.92\times {10}^{-3}\text{N}\end{array}$

Conclusion:

Therefore, the force exerted on each face of the cube $1.92\times {10}^{-3}\text{N}$.

(d)

To determine

To Explain:The reason why such a small sample exert such a great force.

Answer to Problem 30P

The small sample exert such a great force because this force is exerted due to the collision of the gas molecules.

Explanation of Solution

A small sample exerts such a great force because this force is exerted due to the collision of the gas molecules with the wall of the container that’s why it experiences such a great force.

Conclusion:

Therefore, a small sample exert such a great force this force is exerted due to the collision of the gas molecules.