Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

Solved Examples - Gas Laws

For all questions:
(1.) Identify the gas law(s) applicable to the question
(2.) Show all work

(1.) $-13k^2 - 52$


$ -13k^2 - 52 \\[3ex] Factor\: by\: GCF \\[3ex] GCF = -13 \\[3ex] = -13(k^2 + 4) $
(2.) JAMB $50cm^3$ of gas was collected over water at $10^oC$ and $765mmHg$.
Calculate the volume of the gas at s.t.p. if the saturated vapour pressure of water at $10^oC$ is $5mmHg$


Dalton's Law of Partial Pressures
Let:
$P_{total}$ = pressure of the gas saturated with water vapor = $765mmHg$
$P_{water\:vapour}$ = pressure of the water vapour = $15mmHg$
$P_{dry\:gas}$ = pressure of the dry gas

$ P_{total} = P_{dry\:gas} + P_{water\:vapour}...Dalton's\:\:Law \\[3ex] P_{dry\:gas} = P_{total} - P_{water\:vapour} \\[3ex] P_{dry\:gas} = 765 - 5 = 760mmHg \\[3ex] $ $P_1$ = $P_{dry\:gas}$ = $760mmHg$

$V_1$ = volume of the carbon(IV)oxide gas = $50cm^3$

$T_1$ = temperature of the carbon(IV)oxide gas = $18^oC = 10 + 273 = 283K$

$P_2$ = standard pressure = $760mmHg$

$V_2$ = ?

$T_2$ = standard temperature = $273K$

$ \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\[5ex] V_2 = \dfrac{P_1V_1T_2}{P_2T_1} \\[5ex] V_2 = \dfrac{760 * 50 * 273}{760 * 283} \\[5ex] V_2 = \dfrac{50 * 273}{283} \\[5ex] V_2 = 48.23321555cm^3 $
(3.) $9p^2 + 27p^3 - 18p^4$


$ 9p^2 + 27p^3 - 18p^4 \\[3ex] Arrange\: in\: descending\: order\: of\: exponents \\[3ex] -18p^4 + 27p^3 + 9p^2 \\[3ex] Factor\: by\: GCF \\[3ex] GCF = -9p^2 \\[3ex] = -9p^2(2p^2 - 3p - 1) $
(4.) WASSCE If $200cm^3$ of carbon(IV)oxide were collected over water at $18^oC$ and $700mmHg$, determine the volume of the dry gas at s.t.p.
[Standard vapour pressure of water at $18^oC$ = 15mmHg]


Dalton's Law of Partial Pressures
Let:
$P_{total}$ = pressure of the carbon(IV)oxide gas saturated with water vapor = $700mmHg$
$P_{water\:vapour}$ = pressure of the water vapour = $15mmHg$
$P_{dry\:gas}$ = pressure of the dry carbon(IV)oxide gas

$ P_{total} = P_{dry\:gas} + P_{water\:vapour}...Dalton's\:\:Law \\[3ex] P_{dry\:gas} = P_{total} - P_{water\:vapour} \\[3ex] P_{dry\:gas} = 700 - 15 = 685mmHg \\[3ex] $ $P_1$ = $P_{dry\:gas}$ = $685mmHg$

$V_1$ = volume of the carbon(IV)oxide gas = $200cm^3$

$T_1$ = temperature of the carbon(IV)oxide gas = $18^oC = 18 + 273 = 291K$

$P_2$ = standard pressure = $760mmHg$

$V_2$ = ?

$T_2$ = standard temperature = $273K$

$ \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\[5ex] V_2 = \dfrac{P_1V_1T_2}{P_2T_1} \\[5ex] V_2 = \dfrac{685 * 200 * 273}{760 * 291} \\[5ex] V_2 = \dfrac{37401000}{221160} \\[5ex] V_2 = 169.1128595cm^3 $
(5.) $3c(c^2 + 5) + 7(c^2 + 5)$


$ 3c(c^2 + 5) + 7(c^2 + 5) \\[3ex] Factor\: by\: GCF \\[3ex] GCF = c^2 + 5 \\[3ex] = (c^2 + 5)[3c + 7] \\[3ex] = (c^2 + 5)(3c + 7) $
(6.) WASSCE A mixture of gases with total pressure of $120kNm^{-2}$ consists of $0.175$ moles of hydrogen, $0.067$ moles of nitrogen and $0.025$ moles of oxygen at $25^oC$.
Calculate the:
$(\alpha.)$ total volume of the gaseous mixture;
$(\beta.)$ partial pressure of hydrogen in the mixture.
$[R = 8.314JK^{-1}mol^{-1}]$


Dalton's Law of Partial Pressures
All the units should be applicable to the unit of $R$

$ P = 120kNm^{-2} = 120kPa = 120000Pa \\[3ex] n_{hydrogen} = 0.175moles \\[3ex] n_{nitrogen} = 0.067moles \\[3ex] n_{oxygen} = 0.025moles \\[3ex] n_{total} = 0.175 + 0.067 + 0.025 = 0.267moles \\[3ex] T = 25^oC = 25 + 273 = 298K \\[3ex] R = 8.314JK^{-1}mol^{-1} \\[3ex] V = ? \\[3ex] PV = nRT ...Ideal\:\:Gas\:\:Equation \\[3ex] V = \dfrac{n_{total}RT}{P} \\[5ex] V = \dfrac{0.267 * 8.314 * 298}{120000} \\[5ex] V = \dfrac{661.511724}{120000} \\[5ex] (\alpha.)\:\: V = 0.0055125977m^3 \\[3ex] $ To calculate the partial pressure of hydrogen, we can do it in two ways

$ 1st\:\:Method...More\:\:Work \\[3ex] PV = nRT ...Ideal\:\:Gas\:\:Equation \\[3ex] V_{hydrogen} = V = 0.0055125977m^3 \\[3ex] n_{hydrogen} = 0.175moles \\[3ex] T_{hydrogen} = T = 25^oC = 25 + 273 = 298K \\[3ex] R = 8.314JK^{-1}mol^{-1} \\[3ex] P_{hydrogen} = ? \\[3ex] P_{hydrogen} = \dfrac{n_{hydrogen}RT}{V} \\[5ex] P_{hydrogen} = \dfrac{0.175 * 8.314 * 298}{0.0055125977} \\[5ex] P_{hydrogen} = \dfrac{433.5751}{0.0055125977} \\[5ex] P_{hydrogen} = 78651.6854Pa \\[5ex] 2nd\:\:Method...Less\:\:Work \\[3ex] P_{hydrogen} = \dfrac{n_{hydrogen}}{n_{total}} * P \\[5ex] P_{hydrogen} = \dfrac{0.175}{0.267} * 120000 \\[5ex] P_{hydrogen} = \dfrac{21000}{0.267} \\[5ex] P_{hydrogen} = 78651.6854Pa $
(7.) $6a^3 - 9a^2 - 2a + 3$


$ 6a^3 - 9a^2 - 2a + 3 \\[3ex] Factor\: by\: Grouping \\[3ex] Group\: in\: twos \\[3ex] Use\: Vertical\: Method \\[3ex] 6a^3 - 9a^2 = 3a^2(2a - 3) \\[3ex] -2a + 3 = -1(2a - 3) \\[3ex] = (2a - 3)(3a^2 - 1) $
(8.) $3y^2 - xy - 6y + 2x$


$$ 3y^2 - xy - 6y + 2x \\[3ex] Factor\: by\: Grouping \\[3ex] Group\: in\: twos \\[3ex] Use\: Vertical\: Method \\[3ex] 3y^2 - xy = y(3y - x) \\[3ex] -6y + 2x = -2(3y - x) \\[3ex] = (3y - x)(y - 2) $$
(9.) AP Chemistry A sample of a pure, gaseous hydrocarbon is introduced into a previously evacuated rigid $1.00$L vessel. The pressure of the gas is $0.200$ atm at a temperature of $127^oC$.
$(a.)$ Calculate the number of moles of the hydrocarbon in the vessel.
$(b.)$ ${O_2}_{(g)}$ is introduced into the same vessel containing the hydrocarbon. After the addition of the ${O_2}_{(g)}$, the total pressure of the gas mixture in the vessel is $1.40$ atm at $127^oC$. Calculate the partial pressure of ${O_2}_{(g)}$ in the vessel.
The mixture of the hydrocarbon and oxygen is sparked so that a complete combustion reaction occurs, producing ${CO_2}_{(g)}$ and ${H_2O}_{(g)}$. The partial pressures of these gases at $127^oC$ are $0.600$ atm for ${CO_2}_{(g)}$ and $0.800$ atm for ${H_2O}_{(g)}$. There is ${O_2}_{(g)}$ remaining in the container after the reaction is complete.
$(c.)$ Use the partial pressures of ${CO_2}_{(g)}$ and ${H_2O}_{(g)}$ to calculate the partial pressure of the ${O_2}_{(g)}$ consumed in the combustion.


$ 4d^3 + 6d^2 - 2d - 3 \\[3ex] Factor\: by\: Grouping \\[3ex] Group\: in\: twos \\[3ex] Use\: Vertical\: Method \\[3ex] 4d^3 + 6d^2 = 2d^2(2d + 3) \\[3ex] -2d - 3 = -1(2d + 3) \\[3ex] = (2d + 3)(2d^2 - 1) $
(10.) $-12p - 45d$


$ -12p -45d \\[3ex] Factor\: by\: GCF \\[3ex] GCF = -3 \\[3ex] = -3(4p + 15d) $
(11.) $-k^2 + 13k - 42$


$ -k^2 + 13k - 42 \\[3ex] Factor\: Quadratic\: Trinomials \\[3ex] = -1(k^2 - 13k + 42) \\[3ex] k^2 - 13k + 42 \\[3ex] Compare\: to\: ak^2 + bk + c \\[3ex] a = 1, b = -13, c = 42 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-13)^2 - 4(1)(42) \\[3ex] = 169 - 168 = 1 \\[3ex] 1 \:\:is\:\:a\:\: Perfect\: Square \\[3ex] (k^2)(42) = 42k^2 \\[3ex] Factors\: are\: -6k, -7k \\[3ex] k^2 - 6k - 7k + 42 \\[3ex] k^2 - 6k = k(k - 6) \\[3ex] -7k + 42 = -7(k - 6) \\[3ex] (k - 6)(k -7) \\[3ex] = -1(k - 6)(k - 7) \\[3ex] $ Check your solution

$ -1(k - 6)(k - 7) \\[3ex] = -1(k^2 - 7k - 6k + 42) \\[3ex] = -1(k^2 - 13k + 42) \\[3ex] = -k^2 + 13k - 42 $
(12.) $x^2 - 144$


$ x^2 - 144 \\[3ex] = x^2 - 12^2 \\[3ex] Compare\: to\: x^2 - y^2 \\[3ex] x = x \\[3ex] y = 12 \\[3ex] = (x + 12)(x - 12) $
(13.) WASSCE The volume of a sample of methane collected over water at a temperature of $12^oC$ and a pressure of $700mmHg$ was $30cm^3$.
Calculate the volume of the dry gas at s.t.p.
[Saturated vapour pressure of water at $12^oC$ is $10mmHg$]


Dalton's Law of Partial Pressures
Let:
$P_{total}$ = pressure of the methane gas saturated with water vapor = $700mmHg$
$P_{water\:vapour}$ = pressure of the water vapour = $10mmHg$
$P_{dry\:gas}$ = pressure of the dry methane gas

$ P_{total} = P_{dry\:gas} + P_{water\:vapour}...Dalton's\:\:Law \\[3ex] P_{dry\:gas} = P_{total} - P_{water\:vapour} \\[3ex] P_{dry\:gas} = 700 - 10 = 690mmHg \\[3ex] $ $P_1$ = $P_{dry\:gas}$ = $690mmHg$

$V_1$ = volume of the methane gas = $30cm^3$

$T_1$ = temperature of the methane gas = $12^oC = 12 + 273 = 285K$

$P_2$ = standard pressure = $760mmHg$

$V_2$ = ?

$T_2$ = standard temperature = $273K$

$ \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\[5ex] V_2 = \dfrac{P_1V_1T_2}{P_2T_1} \\[5ex] V_2 = \dfrac{690 * 30 * 273}{760 * 285} \\[5ex] V_2 = \dfrac{5651100}{216600} \\[5ex] V_2 = 26.0900277cm^3 $
(14.)


(15.) $q^2 + 12 + 13q$


$ q^2 + 12 + 13q \\[3ex] = q^2 + 13q + 12 \\[3ex] Compare\: to\: aq^2 + bq + c \\[3ex] a = 1, b = 13, c = 12 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = 13^2 - 4(1)(12) \\[3ex] = 169 - 48 = 125 \\[3ex] 125 \:\:is\:\:a\:\: Perfect\: Square \\[3ex] Factor\: Quadratic\: Trinomials \\[3ex] (q^2)(12) = 12q^2 \\[3ex] Factors\: are\: q, 12q \\[3ex] q^2 + q + 12q + 12 \\[3ex] q^2 + q = q(q + 1) \\[3ex] 12q + 12 = 12(q + 1) \\[3ex] (q + 1)(q + 12) \\[3ex] $ OR use the method explained in *Step $3$* on "Factor Quadratic Trinomials"

$ q^2 + 12 + 13q \\[3ex] (q + 1)(q + 12) \\[3ex] $ Check your solution

$ (q + 1)(q + 12) \\[3ex] = q^2 + 12q + q + 12 \\[3ex] = q^2 + 13q + 12 \\[3ex] = q^2 + 12 + 13q $
(16.) $d^4 - 17d^3 + 60d^2$


$ d^4 - 17d^3 + 60d^2 \\[3ex] Factor\:\:by\:\:GCF \\[3ex] GCF = d^2 \\[3ex] = d^2(d^2 - 17d + 60) \\[3ex] $ Let us deal with $d^2 - 17d + 60$
We shall include the $d^2$ in our final answer

$ d^2 - 17d + 60 \\[3ex] Compare\: to\: ad^2 + bd + c \\[3ex] a = 1, b = -17, c = 60 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] Discriminant = (-17)^2 - 4(1)(60) \\[3ex] Discriminant = 289 - 240 = 49 \\[3ex] 49 \:\:is\:\:a\:\: Perfect\: Square \\[3ex] Factor\: Quadratic\: Trinomials \\[3ex] (d^2)(60) = 60d^2 \\[3ex] Factors\: are\: -5d, -12d \\[3ex] (d - 5)(d - 12) \\[3ex] \rightarrow d^4 - 17d^3 + 60d^2 \\[3ex] = d^2(d - 5)(d - 12) \\[3ex] $ Check your solution

$ d^2(d - 5)(d - 12) \\[3ex] = d^2(d^2 - 12d - 5d + 60) \\[3ex] = d^2(d^2 - 17d + 60) \\[3ex] = d^4 - 17d^3 + 60d^2 $