All Solutions
Page 423: Lesson Check
That being said, when looking at the pressure exerted by a gas on a certain surface, the direction of the force it experiences will always be at right angles with that surface.
$$
P = dfrac{N cdot R cdot T}{V}
$$
Where $N$ is the number of molecules of the gas,
$V$ is the volume of the gas,
$T$ is the temperature of the gas,
$k$ is the Boltzmann constant whose value is $k = 1.38 cdot 10^{-23} text{ }dfrac{text{J}}{text{K}}$ and will never change.
So, the three ways to increase the pressure of an enclosed gas are:
1) Increasing the temperature of the gas.
2) Decreasing the volume of the gas.
3) Increasing the number of moles of the gas.
1) Increasing the temperature of the gas.
2) Decreasing the volume of the gas.
3) Increasing the number of moles of the gas.
2. By decreasing the volume.
3. By increasing the number of gas molecule.
This allows us to calculate the number of moles contained in the flask as follows:
$$
begin{equation*}
n = frac{m}{M}
end{equation*}
$$
After that the number of particles is routinely found using Avogadro’s constant:
$$
begin{equation*}
N = n cdot Na
end{equation*}
$$
– The number of moles of each gas
$$
n_{text{A}} = 10text{ mol} quad,quad n_{text{B}} = 20text{ mol} quad,quad n_{text{C}} = 50text{ mol} quad,quad n_{text{D}} = 5text{ mol}
$$
– The pressure of each gas
$$
P_{text{A}} = 100text{ kPa} quad,quad P_{text{B}} = 200text{ kPa} quad,quad P_{text{C}} = 50text{ kPa} quad,quad P_{text{D}} = 50text{ kPa}
$$
– The volume of each gas
$$
V_{text{A}} = 1text{ m}^3 quad,quad V_{text{B}} = 2text{ m}^3 quad,quad V_{text{C}} = 1text{ m}^3 quad,quad V_{text{D}} = 4text{ m}^3
$$
To find the temperature of each gas we write the Ideal Gas Law:
$$
begin{equation*}
PV = n , R , T
end{equation*}
$$
rearranging for temperature we have:
$$
begin{equation*}
T = frac{PV}{n , R}
end{equation*}
$$
Plugging in the numbers for each gas we get:
$$
begin{align*}
T_{text{A}} &= frac{P_{text{A}} , V_{text{A}} }{n_{text{A}} , R } = frac{100text{ kPa} cdot 1text{ m}^3 }{10text{ mol} , R} = frac{10}{R} , frac{text{KJ}}{mol} \
T_{text{B}} &= frac{P_{text{B}} , V_{text{B}} }{n_{text{B}} , R } = frac{200text{ kPa} cdot 2text{ m}^3}{20text{ mol} , R} = frac{20}{R} , frac{text{KJ}}{mol} \
T_{text{C}} &= frac{P_{text{C}} , V_{text{C}} }{n_{text{C}} , R } = frac{50text{ kPa} cdot 1text{ m}^3}{50text{ mol} , R} = frac{1}{R} , frac{text{KJ}}{mol} \
T_{text{D}} &= frac{P_{text{D}} , V_{text{D}} }{n_{text{D}} , R } = frac{50text{ kPa} cdot 4text{ m}^3 }{5text{ mol} , R} = frac{40}{R} , frac{text{KJ}}{mol} \
end{align*}
$$
Ranking in order of increasing temperature:
$$
begin{align*}
T_{text{C}} < T_{text{A}} < T_{text{B}} < T_{text{D}}
end{align*}
$$
$$
begin{align*}
T_{text{C}} < T_{text{A}} < T_{text{B}} < T_{text{D}}
end{align*}
$$
$PV=NkT$,
or
$P=frac{NkT}{V}$
So the ratio
$frac{P_A}{P_B}=frac{(N_AkT)/V_A}{(N_BkT)/V_B}=frac{N_AV_B}{N_BV_A}$
Given $N_A=N_B/2$ and $V_A=2V_B$, usiing this values we get
$$
frac{P_A}{P_B}=frac{(N_B/2)V_B}{N_B(2V_B)}=frac{1}{4}
$$
frac{1}{4}
$$
$$
P = dfrac{N cdot R cdot T}{V}
$$
Where $N$ is the number of molecules of the gas,
$V$ is the volume of the gas,
$T$ is the temperature of the gas,
$k$ is the Boltzmann constant whose value is $k = 1.38 cdot 10^{-23} text{ }dfrac{text{J}}{text{K}}$ and will never change.
Pressure in container $A$ will be:
$$
P_A = dfrac{dfrac{N}{2} cdot R cdot T}{2V}
$$
Pressure in container $B$ will be:
$$
P_A = dfrac{N cdot R cdot T}{V}
$$
The ratio $dfrac{P_A}{P_B}$ will be:
$$
begin{align*}
dfrac{P_A}{P_B} &= dfrac{dfrac{N cdot R cdot T}{4V}}{dfrac{N cdot R cdot T}{V}}\
\
dfrac{P_A}{P_B} &= boxed{dfrac{1}{4}}\
end{align*}
$$
– The number of particles $N = 2.2 cdot 10^{23}$
– The volume $V = 0.026text{ m}^3$
– The temperature $T = 290text{ K}$
To solve the problem we use the Ideal Gas Law:
$$
begin{equation*}
PV = N , k , T
end{equation*}
$$
rearranging for pressure we have:
$$
begin{equation*}
P = frac{N , k , T}{V}
end{equation*}
$$
Plugging in the numbers:
$$
begin{align*}
P &= frac{2.2 cdot 10^{23} cdot 1.38 cdot 10^{-23} , frac{text{J}}{text{K}} cdot 290text{ K} }{0.026text{ m}^3 } \
P &= 33.86text{ kPa}
end{align*}
$$
– The number of particles $N = 4.1 cdot 10^{23}$
– The pressure $P = 195text{ kPa}$
– The temperature $T = 262 text{ K}$
To solve the problem we use the Ideal Gas Law:
$$
begin{equation*}
PV = N , k , T
end{equation*}
$$
rearranging for volume we have:
$$
begin{equation*}
V = frac{N , k , T}{P}
end{equation*}
$$
Plugging in the numbers:
$$
begin{align*}
V &= frac{4.1 cdot 10^{23} cdot 1.38 cdot 10^{-23} , frac{text{J}}{text{K}} cdot 262text{ K} }{192 cdot 10^{3}text{ Pa} } \
V &= 0.0076text{ m}^3
end{align*}
$$
– The number of moles of air in the tire $n = 1.5text{ mol}$
– The gauge pressure in the tire $P = 205text{ kPa}$
– The volume of the tire $V = 0.012text{ m}^3$
To calculate the temperature of the tire we use the Ideal Gas Equation:
$$
begin{equation*}
PV = n , R , T
end{equation*}
$$
rearranging for temperature:
$$
begin{equation*}
T = frac{pV}{n , R}
end{equation*}
$$
Where the pressure in the tire is the sum of the gauge pressure and atmospheric pressure $P = P_{text{gauge}} + P_{text{atmosphere}}$
Plugging in the values we get:
$$
begin{align*}
T = frac{(205 + 101)text{ kPa} cdot 0.012text{ m}^3 }{1.5text{ mol} cdot 8.31 , frac{text{J}}{text{mol K}} } = 302.23text{ K}
end{align*}
$$
– The gauge pressure of the tire in the morning $P_{text{gauge1}} = 401text{ kPa}$
– The gauge pressure of the tire in the afternoon $P_{text{gauge2}} = 419 text{ kPa}$
– The morning temperature $T_1 = 286text{ K}$
The total pressure in the tire is the sum of the gauge and atmosphere pressures. We use the Ideal Gas Equation for the gas in the tire in the morning and afternoon:
$$
begin{align*}
P_1 V &= n , R , T_1 \
P_2 V &= n , R , T_2
end{align*}
$$
Where $P_1 = P_{text{gauge1}} + P_{text{atmosphere}}$ and $P_2 = P_{text{gauge2}} + P_{text{atmosphere}}$
Dividing we get:
$$
begin{align*}
frac{P_1}{P_2} = frac{ P_{text{gauge1}} + P_{text{atmosphere}} }{P_{text{gauge2}} + P_{text{atmosphere}} } = frac{T_1}{T_2}
end{align*}
$$
rearranging for $T_2$, and plugging in the values we get:
$$
begin{align*}
T_2 &= 286text{ K} cdot frac{(419 + 101)text{ kPa}}{(401 + 101)text{ kPa} } \
T_2 &= 296.25text{ K}
end{align*}
$$
