All Solutions
Page 348: Section Review
textbf{(a)} \
The atmospheric pressure outside of the two boxes is equal.
textbf{(b)} \
To solve for the magnitude of the total force exerted by the air, we use the equation below
$$
P = frac{F}{A}
$$
Isolating $F$ on one side of the equation, we have
$$
F = PA
$$
We can infer from the equation above that, since the box with dimensions $20 text{cm} times 20 text{cm} times 40 text{cm}$ have larger surface area, then the magnitude of the force of air exerted on it will also be larger.
The pressure on the boxes is independent of the dimensions of the boxes and the pressures are the same.
According to the equation $F=PA$, the force on the boxes is proportional to the area of the boxes. Therefore the force on the box with dimensions 20x20x40 is greater.
textbf{Given:} \
$P_1 = 1.01325 times 10^5 text{Pa}$ (Mean sea-level atmospheric pressure) \
$V_1 = 25.0 text{m}^3$ \
$P_2 = 0.82 times 10^5 text{Pa}$ \
textbf{Find:} \
$V_2 = ?$ \
textbf{Calculation:}\
To solve for the final volume $V_2$ of the gas, we use the Boyle’s Law
$$
P_1V_1 = P_2V_2
$$
Isolating $V_2$ on one side of the equation
$$
V_2 = frac{P_1V_1}{P_2}
$$
Plugging in the given values, we have
$$
V_2 = frac{(1.01325 times 10^5) cdot (25.0)}{(0.82 times 10^5)}
$$
$$
boxed{V_2 = 30.9 text{m}^3}
$$
V_2 = 30.9 text{m}^3
$$
Temperature is unchanged, therefore:
$P_1 V_1 = P_2 V_2$
Sole for $V_2$:
$V_2 = dfrac{P_1 V_1}{P_2}$
$V_2 = dfrac{(1.01e5)*(25.0)}{(0.82e5)}$
$$
V_2 = 30.8 m^3
$$
30.8 m^3
$$
textbf{Given:} \
$V_1 = 0.0021 text{m}^3$ \
$P_1 = 1.01325 times 10^5 text{Pa}$ (Atmospheric Pressure) \
$T_1 = 303 text{K}$ \
$V_2 = 0.0003 text{m}^3$ \
$P_2 = 20.1 times 10^5 text{Pa}$ \
textbf{Find:} \
$T_2 = ?$ \
textbf{Calculation:}\
To solve for the final temperature $T_2$ of air, we use the Combined Gas Law
$$
frac{P_1V_1}{T_1} = frac{P_2V_2}{T_2}
$$
Isolating $T_2$ on the left side of the equation
$$
T_2 = frac{P_2V_2T_1}{P_1V_1}
$$
Plugging in the given values, we have
$$
T_2 = frac{(20.1 times 10^5) cdot (0.0003) cdot (303)}{(1.0325 times 10^5) cdot (0.0021)}
$$
$$
boxed{T_2 = 859 text{K}}
$$
T_2 = 859 text{K}
$$
Solve for $T_2$:
$T_2 = dfrac{T_1 P_2 V_2}{P_1 V_1}$
$T_2 = dfrac{(303)*(20.1e5)*(0.0003)}{(1.01e5)*(0.0021)}$
$$
T_2 = 861 K
$$
861 K
$$
textbf{Given:} \
$T_1 = 0 ^circ text{C} = 273 text{K}$ \
$T_2 = 4 ^circ text{C} = 277 text{K}$ (For the first question) \
$T_2 = 8 ^circ text{C} = 281 text{K}$ (For the second question)\
$d_1 = 999.82 frac{text{kg}}{text{m}^3}$ (Density of water at $0 ^circ text{C}$) \
textbf{Find:} \
$d_2 = ?$ at $4 ^circ text{C}$ and $8 ^circ text{C}$\
textbf{Calculation:}\
To get an expression for density, we use the Ideal Gas Law
begin{equation}
PV = nRT
end{equation}
To express the equation above in terms of mass, we use the equation below
begin{equation} tag{2}
n = frac{m}{M}
end{equation}
Plugging this into the Ideal Gas Law, we have
$$
PV = frac{mRT}{M}
$$
Isolating $R$ on the right side of the equation, we have
begin{equation} tag{3}
frac{PVM}{MT} = R
end{equation}
We know that density is given by the equation
$$
d = frac{m}{V}
$$
Then, we can express Equation (3) as follows
begin{equation} tag{4}
frac{PM}{dT} = R
end{equation}
Since we can treat $R$ as a proportionality constant, we can express Equation (4) as follows
$$
frac{P_1M_1}{d_1T_1} = frac{P_2M_2}{d_2T_2}
$$
Assuming that the $P$ and $M$ are constant, we now have
$$
d_1T_1 = d_2T_2
$$
Isolating $d_2$ on one side of the equation
begin{equation} tag{5}
d_2 = frac{d_1T_1}{T_2}
end{equation}
The density at $T_2 = 277 text{K}$ can now be calculated using the Equation (5)
$$
d_2 = frac{(999.82) cdot (273)}{(277)}
$$
$$
boxed{d_2 = 986 frac{text{kg}}{text{m}^3}}
$$
The density at $T_2 = 281 text{K}$ can now be calculated using the Equation (5)
$$
d_2 = frac{(999.82) cdot (273)}{(281)}
$$
$$
boxed{d_2 = 971 frac{text{kg}}{text{m}^3}}
$$
$d_2 = 971 frac{text{kg}}{text{m}^3}$ at $8 ^circ text{C}$
textbf{Given:} \
$n = 1 text{mol}$ \
$P = 1.01325 times 10^5 text{Pa}$ \
$R = 8.31 frac{text{Pa} cdot text{m}^3}{text{mol} cdot text{K}}$ \
$T = 273 text{K}$ \
textbf{Find:} \
$V = ?$ \
textbf{Calculation:}\
To solve for the volume of the gas, we use the Ideal Gas Law
$$
PV = nRT
$$
Isolating $V$ on one side of the equation
$$
V = frac{nRT}{P}
$$
Plugging in the given values, we have
$$
V = frac{(1) cdot (8.31) cdot (273)}{(1.01325 times 10^5)}
$$
$$
boxed{V = 0.0225 text{m}^3}
$$
V = 0.0225 text{m}^3
$$
Solve for V:
$V = dfrac{n R T}{P}$
$V = dfrac{(1)*(8.31)*(273)}{(1.01e5)}$
$$
V = 0.0225 m^3
$$
0.0225 m^3
$$
textbf{Given:} \
$V = 0.635 text{m}^3$ \
$T = 2.00 ^circ text{C} = 275 text{K}$ \
$P = 1.01325 times 10^5 text{Pa}$ (Atmospheric pressure) \
$R = 8.31 frac{text{Pa} cdot text{m}^3}{text{K} cdot text{mol}}$ \
$M_text{air} = 29 frac{text{g}}{text{mol}}$ \
textbf{Find:} \
$n = ?$ \
$m = ?$ \
textbf{Calculation:}\
To solve for the number of moles of air in the refrigerator, we use the Ideal Gas Law
$$
PV = nRT
$$
Isolating $n$ on the left side of the equation
$$
n = frac{PV}{RT}
$$
Plugging in the given values, we have
$$
n = frac{(1.01325 times 10^5) cdot (0.635)}{(8.31) cdot (275)}
$$
$$
boxed{n = 28.2 text{mol}}
$$
To solve for the mass of the air, we use the equation
$$
m = M_text{air}n
$$
Plugging in the given values, we have
$$
m = (29) cdot (28.2)
$$
$$
boxed{m = 818 text{g}}
$$
$$
m = 818 text{g}
$$
$P = 1 atm = 1.01e5 Pa$
We have:
$n = dfrac{P V}{R T} = dfrac{(1.01e5)*(0.635)}{(8.31)*(2+273)} = 28.1 moles$
The mass of the air:
$$
m = M n = (29)*(28.1) = 814 g
$$
$$
814 g
$$
$n = dfrac{P V}{R T}$
The number of particles only depends on the volume, temperature and pressure and it is independent of size of the particles. Therefore the number of the particles are the same.
