Calculations on Gas Laws

Solved Examples on Gas Laws

Samuel Dominic Chukwuemeka (SamDom For Peace)

Prerequisites:
(1.) Balance Chemical Reactions
(2.) Measurements and Units
For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve all questions.
Show all work.

(1.) WASSCE (a.) State Graham's law of diffusion.
(b.) Arrange the following gases,

$He$ ,

$CH_4$ , and

$N_2$ in order of increasing rates of diffusion.
Give a reason for the order.

$(H = 1,\:\: He = 4,\:\: C = 12,\:\: N = 14)$

(a.)
Graham's law of diffusion states that at a constant temperature and pressure, the rate of the diffusion of a gas is inversely proprotional to the square root of it's vapour density.

(b.)
Based on Graham's law,
At a constant temperature and pressure:
This means that the lower the square root of the vapour density of the gas, the faster the rate of diffusion of the gas
Similarly, the higher the square root of the vapour density of the gas, the slower the rate of diffusion of the gas

In other words:
At a constant temperature and pressure:
Lighter gases diffuse faster
Heavier gases diffuse slower
The heavier the gas, the slower the diffusion.

Increasing rates of diffusion means from least to greatest
It means from slowest to highest.

Let us compare this using a table. It is much better to use a table.
Relative Molecular Mass = RMM
Vapour Density = VD

Gas	$RMM$	$VD = \dfrac{RMM}{2}$	$\sqrt{VD}$
$He$	$4$	$\dfrac{4}{2}$ $= 2$	$\sqrt{2}$
$CH_4$	$12 + (1 * 4)$ $= 12 + 4$ $= 16$	$\dfrac{16}{2}$ $= 8$	$\sqrt{8}$
$N_2$	$14 * 2$ $= 28$	$\dfrac{28}{2}$ $= 14$	$\sqrt{14}$
$\sqrt{14} \gt \sqrt{8} \gt \sqrt{2}$
Decreasing Order of Vapour Density: $N_2$ , $CH_4$ , $He$
Increasing Rate of Diffusion: $N_2$ , $CH_4$ , $He$

(2.) If a gas sample has a volume of

$15.0$ liters at

$1.5$ atmospheres and

$45^\circ C$ , what will be it's volume at standard temperature and pressure (STP)?
(At STP, temperature =

$0^\circ C$ , pressure =

$1$ atmosphere)

Ensure that all corresponding quantities have the same units.
Covert all units to the S.I unit (Internation System of Units)

$P_1 = 1.5\:atm \\[3ex] P_2 = 1\: atm \\[3ex] V_1 = 15\:L \\[3ex] V_2 = ? \\[3ex] T_1 = 45^\circ = 45 + 273 = 318\:K \\[3ex] T_2 = 0^\circ = 0 + 273 = 273\:K \\[3ex] \underline{General\:\:Gas\:\:Equation} \\[3ex] \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\[5ex] Cross\:\:Multiply \\[3ex] T_1 * P_2 * V_2 = P_1 * V_1 * T_2 \\[3ex] V_2 = \dfrac{P_1 * V_1 * T_2}{T_1 * P_2} \\[5ex] V_2 = \dfrac{1.5 * 15 * 273}{318 * 1} \\[5ex] V_2 = \dfrac{6142.5}{318} \\[5ex] V_2 = 19.31603774\:L \\[3ex] V_2 \approx 19\:L$

(3.) WASSCE Consider the following reaction equation:

$2{H_2}_{(g)} + {O_2}_{(g)} \rightarrow 2{H_2O}_{(g)} \\[3ex]$ Calculate the volume of unused oxygen gas when

$40\;cm^3$ of hydrogen gas is sparked with

$30\;cm^3$ of oxygen gas

Based on the question, we already know that oxygen is the excess reactant because it was not used completely
This implies that we shall use hydrogen: the limiting reactant

$2{H_2}_{(g)} + {O_2}_{(g)} \rightarrow 2{H_2O}_{(g)} \\[3ex] 2\;mol\;H_2 \rightarrow 1\;mol\;O_2 \\[3ex] 2\;cm^3\;H_2 \rightarrow 1\;cm^3\;O_2 ...Gay\;Lussac's\;\;Law\;\;of\;\;Combining\;\;Volumes \\[3ex]$

Proportional Reasoning Method
$H_2\;(cm^3)$	$O_2\;(cm^3)$
$2$	$1$
$40$	$p$

$\dfrac{p}{1} = \dfrac{40}{2} \\[5ex] p = 20\;cm^3 ...used\;\;amount\;\;of\;\;O_2 \\[3ex] Given\;\;amount\;\;of\;\;O_2 = 30\;cm^3 \\[3ex] Unused\;\;amount\;\;of\;\;O_2 = 30 - 20 = 10\;cm^3$

(4.) If a gas sample has a volume of

$6.0$ liters at

$1.5$ atmospheres and

$45^\circ C$ , what will be it's volume at standard temperature and pressure (STP)?
(At STP, temperature =

$0^\circ C$ , pressure =

$1$ atmosphere)

Ensure that all corresponding quantities have the same units.
Covert all units to the S.I unit (Internation System of Units)

$P_1 = 1.5\:atm \\[3ex] P_2 = 1\: atm \\[3ex] V_1 = 6\:L \\[3ex] V_2 = ? \\[3ex] T_1 = 35^\circ = 35 + 273 = 308\:K \\[3ex] T_2 = 0^\circ = 0 + 273 = 273\:K \\[3ex] \underline{General\:\:Gas\:\:Equation} \\[3ex] \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\[5ex] Cross\:\:Multiply \\[3ex] T_1 * P_2 * V_2 = P_1 * V_1 * T_2 \\[3ex] V_2 = \dfrac{P_1 * V_1 * T_2}{T_1 * P_2} \\[5ex] V_2 = \dfrac{1.5 * 6 * 273}{308 * 1} \\[5ex] V_2 = \dfrac{2457}{308} \\[5ex] V_2 = 7.977272727\:L \\[3ex] V_2 \approx 8\:L$

(5.) WASSCE

$t = 4\:years \\[3ex] Semiannual\:\:bond \rightarrow m = 2 \\[3ex] FV = \$1000 \\[3ex] BCR = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] YTM = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] BP = ? \\[3ex] BP = \dfrac{FV * BCR}{YTM} * \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] + \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[10ex] Too\:\:long...let\:\:us\:\:solve\:\:in\:\:parts \\[3ex] \underline{First\:\:Part} \\[3ex] \dfrac{FV * BCR}{YTM} \\[5ex] = \dfrac{1000 * 0.06}{0.07} \\[5ex] = \dfrac{60}{0.07} \\[5ex] = 857.142857 \\[3ex] \underline{Second\:\:Part} \\[3ex] \left[1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}}\right] \\[10ex] mt = 2(30) = 60 \\[3ex] 1 + \dfrac{YTM}{m} = 1 + \dfrac{0.07}{2} = 1 + 0.035 = 1.035 \\[5ex] \left(1 + \dfrac{YTM}{m}\right)^{mt} = (1.035)^{60} = 7.8780909 \\[7ex] \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = \dfrac{1}{7.8780909} = 0.126934306 \\[10ex] 1 - \dfrac{1}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} = 1 - 0.126934306 = 0.873065694 \\[10ex] \underline{Third\:\:Part} \\[3ex] \dfrac{FV}{\left(1 + \dfrac{YTM}{m}\right)^{mt}} \\[7ex] = \dfrac{1000}{7.8780909} \\[5ex] = 126.934306 \\[3ex] \underline{Bond\:\:Price} \\[3ex] BP = 857.142857 * 0.873065694 + 126.934306 \\[3ex] BP = 748.342023 + 126.934306 \\[3ex] BP = 875.276329 \\[3ex] BP \approx \$875.28$

(6.) WASSCE A mole of an ideal gas occupies

$22.4\;dm^3$ at

$273.15\;K$ and a pressure of

$1.0132 * 10^5\;Nm^{-2}$
Determine the gas constant

$R$

$P = 1.0132 * 10^5\;Nm^{-2} \\[3ex] V = 22.4\;dm^3 = 0.0224m^3 \\[3ex] n = 1\;mol \\[3ex] T = 273.15\;K \\[3ex] R = ? \\[3ex] \underline{Ideal\;\;Gas\;\;Equation} \\[3ex] PV = nRT \\[3ex] nRT = PV \\[3ex] R = \dfrac{PV}{nT} \;\; \left(\dfrac{Nm^{-2} * m^3}{mol * K}\right) \\[5ex] R = \dfrac{1.0132 * 10^5 * 0.0224}{1 * 273.15} \\[5ex] R = \dfrac{2269.568}{273.15} \\[5ex] R = 8.308870584\;Nm\;mol^{-1}K^{-1}$