All Solutions
Page 344: Practice Problems
textbf{Given:} \
$P = 1.0 times 10^5 text{Pa}$ \
$l = 152 text{cm} = 1.52 text{m}$ \
$w = 76 text{cm} = 0.76 text{m}$ \
textbf{Calculation:}\
To solve for the force exerted by the air on the top of a desk at sea level, we start with the equation below
$$
P = frac{F}{A}
$$
Isolating $F$ on one side of the equation, we have
begin{equation}
F = PA
end{equation}
To solve for the area at the top of the desk, we use the equation below
begin{equation} tag{2}
A = lw
end{equation}
Combining equations (1) and (2), we have
$$
F = Plw
$$
Plugging in the given values
$$
F = (1.0 times 10^5) cdot (1.52) cdot (0.76)
$$
$$
boxed{F = 1.2 times 10^5 text{N}}
$$
F = 1.2 times 10^5 text{N}
$$
F = PA = Plw
$$
F = left( 1.0 times 10^{5} Paright)left( 1.52 mright)left( 0.76 mright)
$$
textbf{Given:} \
$m = 925 text{kg}$ \
$A_text{tire} = 12 text{cm} times 18 text{cm} = 0.12 text{m} times 0.18 text{m}$ \
$g = 9.8 frac{text{m}}{text{s}^2}$ \
textbf{Calculation:}\
To solve for the area of the four tires, we uese the equation below
$$
A = 4 cdot A_text{tire}
$$
Plugging in the given values, we have
$$
A = 4 cdot (0.12 times 0.18)
$$
$$
A = 0.0864 text{m}^2
$$
To solve for the weight of the car, we use the equation below
$$
F = mg
$$
Plugging in the given values, we have
$$
F = (925) cdot (9.8)
$$
$$
F = 9065 text{N}
$$
To solve for the pressure the car tires exert on the ground, we use the equation below
$$
P = frac{F}{A}
$$
Plugging in the calculated $F$ and $A$, we have
$$
P = frac{9065}{0.0864}
$$
$$
boxed{P = 1.0 times 10^5 text{Pa}}
$$
P = 1.0 times 10^5 text{Pa}
$$
$a = (0.12)*(0.18) = 0.0216 m^2$
The area of the four tires:
$A = 4 a = (4) *(0.0216) = 0.0864 m^2$
The weight of the car is:
$F = m g = (925)*(9.8) = 9065 N$
The pressure:
$$
P = dfrac{F}{A} = dfrac{9065}{0.0864} = 1.05 times 10^5 Pa
$$
1.05 times 10^5 Pa
$$
textbf{Given:} \
$V = 5.0 text{cm} times 10.0 text{cm} times 20.0 text{cm}$ \
$rho = 11.8 frac{text{g}}{text{cm}^3}$ \
$g = 9.8 frac{text{m}}{text{s}^2}$ \
textbf{Calculation:}\
To solve for the smallest face of the brick, we calculate the surface area formed by the two smallest sides, $5.0 text{cm} = 0.05 text{m}$ and $10.0 text{cm} = 0.10 text{m}$, as follows
$$
A = 0.05 times 0.10
$$
$$
A = 5 times 10^{-3} text{m}^2
$$
To solve for the mass of the lead brick, we use the definition of density below
$$
rho = frac{m}{V}
$$
Isolate $m$ on one side of the equation, we have
$$
m = rho V
$$
Plugging in the given values
$$
m = (11.8) cdot (5.0 cdot 10.0 cdot 20.0)
$$
$$
m = 11.8 text{kg}
$$
To solve for the weight of the lead brick, we use the equation below
$$
F = mg
$$
Plugging in the calculated $m$, we have
$$
F = (11.8) cdot (9.8)
$$
$$
F = 115.64 text{N}
$$
To solve for the pressure exerted by the brick on the ground, we use the equation below
$$
P = frac{F}{A}
$$
Plugging in the calculated $F$ and $A$, we have
$$
P = frac{115.64}{5 times 10^{-3}}
$$
$$
boxed{P = 2.3 times 10^4 text{Pa} = 23 text{kPa}}
$$
P = 2.3 times 10^4 text{Pa} = 23 text{kPa}
$$
$V = (5.0)*(10.0)*(20.0) = 1000 cm^3$
The mass of the brick is:
$m = rho V = (11.8)*(1000) = 11800 grams = 11.8 kg$
The weight of the brick is:
$F = m g = (11.8)*(9.8) = 115.64 N$
The area of the smallest face is:
$$
A = (0.05)*(0.10) = 0.005 cm^2
$$
The pressure is:
$$
P = dfrac{F}{A} = dfrac{115.64}{0.005} = 2.3 times 10^4 Pa
$$
2.3 times 10^4 Pa
$$
textbf{Given:} \
$P = 15% text{of} P_text{atm}$ \
$A = 195 text{cm} times 91 text{cm} = 1.95 text{m} times 0.91 text{m}$ \
textbf{Calculation:}\
To solve for the pressure, we use the equation below
$$
P = 0.15P_text{atm}
$$
Plugging in the atmospheric pressure $P_text{atm} = 1.0$
$$
P = 0.15 cdot (1.0)
$$
$$
P = 0.15 text{atm}
$$
Converting the unit of $P$ from $text{atm}$ to $text{Pa}$, we use the confersion factor below
$$
P = 0.15 text{atm} cdot left( frac{1.01325 times 10^5 text{Pa}}{1 text{atm}} right)
$$
$$
P = 15198.75 text{Pa}
$$
To solve for the force exerted on the door, we use the equation below
$$
P = frac{F}{A}
$$
Isolating $F$ on one side of the equation
$$
F = PA
$$
Plugging in the given values, we have
$$
F = (15198.75) cdot (1.95 times 0.91)
$$
$$
boxed{F = 2.7 times 10^4 text{N}}
$$
Since the pressure outside the house dropped by $15%$, the pressure inside the house will be much larger than the pressure outside thus the force exerted on the door is directed $boxed{text{outward}}$
textbf{Given:} \
$m = 454 text{kg}$ \
$P = 5.0 times 10^4 text{Pa}$ \
textbf{Calculation:}\
To solve for the weight of the device, we use the equation below
$$
F = mg
$$
Plugging in the given values, we have
$$
F = (454) cdot (9.8)
$$
$$
F = 4449.2 text{N}
$$
To solve for the area required for the steel support plate to withstand the device, we use the equation below
$$
P = frac{F}{A}
$$
Isolating $A$ on one side of the equation, we have
$$
A = frac{F}{P}
$$
Plugging in the value of $F$ we calculated earlier and the given $P$
$$
A = frac{4449.2}{5.0 times 10^4}
$$
$$
boxed{A = 0.089 text{m}^2}
$$
A = 0.089 text{m}^2
$$
$A = dfrac{(454)*(9.8)}{(5.0 times 10^4)}$
$$
A = 0.089 m^2
$$
0.089 m^2
$$
textbf{Given:} \
$P_1 = 15.5 times 10^6 text{Pa}$ \
$T_1 = 293 text{K}$ \
$V_1 = 0.020 text{m}^3$ \
$P_2 = 1.00 text{atm}$ \
$T_2 = 323 text{K}$
textbf{Calculation:}\
To convert the unit of $P_2$ from $text{atm}$ to text{Pa}, we use the conversion factor below
$$
P_2 = 1.00 text{atm} cdot left( frac{1.01325 times 10^5 text{Pa}}{1 text{atm}} right)
$$
$$
P_2 = 1.01325 times 10^5 text{Pa}
$$
To solve for the $V_2$, we use the Combined Gas Law
$$
frac{P_1V_1}{T_1} = frac{P_2V_2}{T_2}
$$
Isolating $V_2$ on one side of the equation
$$
V_2 = frac{P_1V_1T_2}{P_2T_1}
$$
Plugging in the given values, we have
$$
V_2 = frac{(15.5 times 10^6) cdot (0.020) cdot (323)}{(1.01325 times 10^5) cdot (293)}
$$
$$
boxed{V_2 = 3.37 text{m}^3}
$$
V_2 = 3.37 text{m}^3
$$
$V_1 = 0.020 m^3$
$T_1 = 293 K$
$P_2 = 1.00 atm = 1.01e5 Pa$
$V_2 = ?$
$T_2 = 323 K$
We have:
$dfrac{P_2 V_2}{T_2} = dfrac{P_1 V_1}{T_1}$
Solve for $V_2$:
$V_2 = dfrac{T_2 P_1 V_1}{P_2 T_1}$
$V_2 = dfrac{(323)*(15.5e6)*(0.020)}{(1.01e5)*(293)}$
$$
V_2 = 3.38 m^3
$$
3.38 m^3
$$
textbf{Given:} \
$M_text{Helium} = 4.00 frac{text{g}}{text{mol}}$ \
$P = 15.5 times 10^6 text{Pa}$ \
$T = 293 text{K}$ \
$V = 0.020 text{m}^2$ \
$R = 8.31 frac{text{J}}{text{K} cdot text{mol}}$
textbf{Calculation:}\
To solve for the number of moles of the Helium gas, we first use the Ideal Gas Law
$$
PV = nRT
$$
Isolating $n$ on one side of the equation
begin{equation}
n = frac{PV}{RT}
end{equation}
To solve for the mass of the Helium gas, we use the equation below
begin{equation} tag{2}
m = M_text{Helium}n
end{equation}
Combining equations (1) and (2)
$$
m = frac{M_text{Helium}PV}{RT}
$$
Plugging in the given values, we have
$$
m = frac{(4.00) cdot (15.5 times 10^6) cdot (0.020)}{(8.31) cdot (293)}
$$
$$
boxed{m = 509 text{g}}
$$
m = 509 text{g}
$$
$n = dfrac{P V}{R T}$
$n = dfrac{(15.5e6)*(0.020)}{(8.31)*(293)}$
$n = 127.3 mol$
mass:
$m = M n$
$$
m = (4.00)*(127.3) = 509 g
$$
509 g
$$
textbf{Given:} \
$V_1 = 200.0 text{L}$ \
$T_1 = 0.0 ^circ text{C} = 273 text{K}$ \
$P_1 = 156 text{kPa} = 1.56 times 10^5 text{Pa}$ \
$T_2 = 95 ^circ text{C} = 368 text{K}$ \
$V_2 = 175 text{L}$ \
Make sure to convert the unit of temperatures from $^circ text{C}$ to $K$ to prevent the denominator from being 0
textbf{Calculation:}\
To solve for the new pressure, we first use the Combined Gas Law
$$
frac{P_1V_1}{T_1} = frac{P_2V_2}{T_2}
$$
Isolating $P_2$ on one side of the equation
$$
P_2 = frac{P_1V_1T_2}{V_2T_1}
$$
Plugging in the given values, we have
$$
P_2 = frac{(156) cdot (200.0) cdot (368)}{(175) cdot (273)}
$$
$$
boxed{P_2 = 2.4 times 10^5 text{Pa} = 240 text{kPa}}
$$
P_2 = 2.4 times 10^5 text{Pa} = 240 text{kPa}
$$
$V_1 = 200 L = 0.200 m^3$
$T_1 = 0 C = 273 K$
$P_2 = ?$
$V_2 = 175 L = 0.175 m^3$
$T_2 = 95 C = 368 K$
We have:
$dfrac{P_2 V_2}{T_2} = dfrac{P_1 V_1}{T_1}$
Solve for $P_2$:
$P_2 = dfrac{T_2 P_1 V_1}{V_2 T_1}$
$P_2 = dfrac{(368)*(156e3)*(0.200)}{(0.175)*(273)}$
$$
P_2 = 2.40 times 10^5 Pa
$$
2.40 times 10^5 Pa
$$
$M_text{air} = 29 frac{text{g}}{text{mol}}$
$m = 1.0 text{kg} = 1000 text{g}$
$P = 1.01 times 10^5 text{Pa}$
$T = 20.0 ^circ text{C} = 293 text{K}$
$R = 8.31 frac{text{m}^3 cdot text{Pa}}{text{mol} cdot text{K}}$
**Unknown:**
$V = ?$
**Approach:**
To solve for the volume, we will use the Ideal Gas Law. Since the $n$ is not given, we will use the definition of molar mass $M = frac{m}{n}$.
To solve for the volume of air, we use Ideal Gas Law
$$
PV = nRT
$$
Isolating $V$ on one side of the equation
$$begin{aligned} tag{1}
V = frac{nRT}{P}
end{aligned}$$
$$begin{aligned}
M = frac{m}{n}
end{aligned}$$
Isolating $n$ on on side of the equation
$$begin{aligned} tag{2}
n = frac{m}{M}
end{aligned}$$
$$begin{aligned}
V = frac{mRT}{MP}
end{aligned}$$
Plugging in the given values, we have
$$begin{aligned}
V &= frac{(1000) cdot (8.31) cdot (293)}{(29) cdot (1.01 times 10^5)} \ \
V &= 0.84 text{m}^3
end{aligned}$$
The volume of the gas is $boxed{V = 0.84 text{m}^3}$
$n = dfrac{m}{M} = dfrac{1000 kg}{29} = 34.48 moles$
find the volume:
$V = dfrac{n R T}{P}$
$V = dfrac{(34.48)*(8.31)*(293)}{(1.01e5)}$
$$
V = 0.83 m^3
$$
0.83 m^3
$$
