States and Properties of Matter  The Gas Laws
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Shortcut to Questions
Q: 1 2 3
4 5
6 7 8 9 10 11

1

What will be the volume of a sample of carbon dioxide at 98.0 kPa if its volume at 96.4 kPa is 223 mL , the temperature being unchanged? 

2 
What volume will be occupied by a sample of hydrogen at 153 kPa
if the volume at 98.6 kPa is 325 mL, the temperature being held
constant?


3 
A sample of hydrogen sulfide was collected
in a 250 mL flask at a pressure of 98.6 kPa and 310 K. What volume
would the gas occupy at 369 kPa and 313 K?


4 
Calculate the molar weight of a gas a sample of which (0.638 g)
occupies a volume of 223 mL at 300 K and 750 mmHg.


5 
A mixture of Ar (3.28 x 10^{3} mole) and N_{2} (1.92 x 10^{2} mole) was
collected in a vessel whose capacity was 62.5 mL. What would be
the total gas pressure at 298 K?


6 
A spark is passed through a mixture formed by adding hydrogen (3.00 litre), oxygen (1.00 litre), and neon (2.00 litre), all the volumes being measured at 273 K and 101 kPa. What is the final volume of the reaction mixture at these conditions?


7 
(a) If air is 80% nitrogen by mass and 20% oxygen by mass, how
many molecules would be present in a sample (2.17 g) of air?
(b) What volume would this occupy at 273 K and 101 kPa?


8 
(Advanced question) The mass of an evacuated
glass bulb is 35.194 g. When filled with O_{2} at 101 kPa
pressure and 298 K its mass is 36.406 g. When filled with a mixture
of CH_{4} and C_{2}H_{6} under the same
physical conditions its mass is 36.033 g. What is the % by volume
of CH_{4} in the mixture?


9 
(Advanced question) An equilibrium mixture of
gaseous NO_{2} and N_{2}O_{4} exerts a total
pressure of 101 kPa in a container at 298 K, and the mole fraction
of NO_{2} in the mixture is 0.200. What is the total pressure
in the container at 473 K if all the N_{2}O_{4}
has been converted into NO_{2}?


10 
(Advanced question) Pure phosphine gas (PH_{3})
was prepared by complete decomposition of Ca_{3}P_{2} (1.00 g) with excess
acid. The (dry) PH_{3} gas was heated to 673 K to decompose it into
its gaseous elements, P_{z} and H_{2} . The final volume was 1.060 litre
at 673 K and 101 kPa pressure. What is the molecular formula of
gaseous phosphorus under these conditions?


11 
(Advanced question) Chlorine gas(Cl_{2})
can be prepared by adding concentrated hydrochloric acid to MnO_{2}.
The equation for the reaction is
MnO_{2} + 4HCl
® MnCl_{2} + Cl_{2} + 2H_{2}O Calculate (a) the mass of MnO_{2} and, (b) the
volume of hydrochloric acid of density 1.12 g ml^{1} and containing
40.0% HCl by mass needed to produce 2.50 litre of Cl_{2} gas at 293
K and 103 kPa.


The Gas Laws (Answers)


1

As temperature is constant, Boyle's Law can
be used. PV = constant or P_{1}V_{1} = P_{2}V_{2}
P_{1} = 96.4 kPa and V_{1}
= 223 mL
P2 = 98.0 kPa and V2 = ?
So 96.4 x 223 = 98.0 x V_{2}
V_{2} = (96.4 x 223) / 98.0 = 219 mL
[Note: the units of P and V must be the same
on both sides of the equation.]


2 
As temperature is constant, Boyle's Law can
be used. PV = constant or P_{1}V_{1} = P_{2}V_{2}
P_{1} = 98.6 kPa and V_{1}
= 325 mL
P_{2} = 153 kPa and V_{2} = ?
So 98.6 x 325 = 153 x V_{2}
V_{2} = (98.6 x 325) / 153 = 209 mL


3 
The temperature and pressure both change so the combined Boyle's
and Charles's Law expression must be used. P_{1}V_{1
}/ T_{1} = P_{2}V_{2 }/ T_{2}
P_{1} = 98.6 kPa, V_{1} = 250 mL and T_{1} = 310 K (Note that temperature
must be expressed in Kelvin)
P_{2} = 369 kPa, V_{2} = ? and T_{2} = 313 K
So (98.6 x 250) / 310 = (369 x V_{2}) / 313
V_{2} = (98.6 x 250 x 313) / (310 x 369) = 67.4 mL


4

The Ideal Gas Equation is required to solve this problem. PV =
nRT
The constant R has a value which depends on the units for P and
V. (Units for n are always moles).
If P is expressed in kPa and V in litres, R = 8.314 J K^{1}
mol^{1}. Temperature is always in Kelvin.
Let the molar mass of the gas = m
P = 750 mmHg
Converting this to kPa: P = (750 / 760) x 101.3 kPa = 100.0 kPa
V 223 mL = 0.223 L
n = mass / molar mass = 0.638 / m mol
R = 8.314 J K^{1} mol.
T = 300 K (Note: T must be in Kelvin)
PV = nRT
So 100.0 x 0.223 = (0.638 / m) x 8.314 x 300
m = (0.638 x 8.314 x 300) / (100.0 x 0.223)
= 71.4 g mol^{1}


5 
The Ideal Gas Equation is required to solve this problem. PV =
nRT
The constant R has a value which depends on the units for P and
V. (Units for n are always moles).
If P is expressed in kPa and V in litres, R = 8.314 J K^{1}
mol^{1}. Temperature is in Kelvin.
P = ? kPa, V 62.5 mL = 0.0625 L
and n = total moles of gas present = 3.28 x 10^{3} + 1.92
x 10^{2}
= 2.248 x 10^{2} mol
R = 8.314 J K^{1} mol^{1} and T = 298 K (Note: T must be in
Kelvin)
P x 0.0625 = 2.248 x 10^{2} x 8.314 x 298
P = (2.248 x 10^{2} x 8.314 x 298) / 0.0625
= 891 kPa


6 
The hydrogen and oxygen gases combine to give water which is a
liquid at the prevailing conditions of temperature and pressure.
The neon gas does not react.
The equation for the reaction is 2H_{2}(g) + O_{2}(g)
® 2H_{2}O(l)
From the stoichiometry of the equation, 2.00 mole of hydrogen reacts
exactly with 1.00 mole of oxygen.
Because one mole of any gas occupies the same volume as one mole
of any other gas at the same conditions of temperature and pressure,
then these gases also combine in the same ratio by volume as they
do by moles
i.e. 2.00 volume of hydrogen gas reacts exactly with 1.00 volume
of oxygen gas.
If all the oxygen (1.00 L) is used in the reaction, it would require
2.00 L of hydrogen, leaving 1.00 L of hydrogen gas unreacted.
The volume of the water liquid formed is negligible, so the final
volume is just 1.00 L of hydrogen gas + the nonreacting neon gas
(2.00 L) = 3.00 L
Note: Gases combine in simple ratio by volume, the same ratio
as for moles, because gases expand to uniformly fill their container.
However, the same is not true for liquids or solids which have a
specified volume for a given mass.


7 
(a) Mass of nitrogen, N_{2}, = 0.80 x 2.17 = 1.74 g
Mass of oxygen, O_{2}, = 0.20 x 2.17 = 0.434 g
Moles of N_{2 }= mass / molar mass = 1.736 / (2 x 14.01)
= 0.0620 mol
Moles of O_{2} = mass / molar mass = 0.434 / (2 x 16.00)
= 0.0136 mol
Total moles of gas present = 0.0620 + 0.0136 = 0.0756 mol
1 mole of any gas contains an Avogadro number of molecules, 6.022
x 10^{23}.
Therefore 0.0756 mole contains 0.0756 x 6.022 x 10^{23}
molecules
= 4.6 x 10^{22} molecules.
(b) Two methods can be used.
Either: from the ideal gas equation PV = nRT, V = nRT / P
For V in L and P in kPa, R = 8.324 J K^{1} mol^{1}
V = (0.0756 x 8.314 x 273) / 101 = 1.7 L
or: one mole of any gas at 273 K and 101 kPa occupies a volume
of 22.4 L
0.0756 mole of gas occupies 0.0756 x 22.4 L = 1.7 L


8 
The essential concept required in this question is that at a given
temperature and pressure, a fixed volume of any gas or mixture of
gases will always have the same number of moles present regardless
of which gases are present. This follows from the property of all
gases that they expand to completely and uniformly occupy any fixedvolume
container because the attractive forces between gas phase molecules
are very small.
From the data given for the system when oxygen is present, the moles
of O_{2}(g) can be deduced:
Mass of oxygen = 36.406  35.194 = 1.212 g
So, moles of O_{2} = mass/molar mass = 1.212 / 32.00 = 0.03788
mol
Thus moles of (methane + ethane) = 0.03788 mol as the temperature,
pressure and volume are unchanged.
Let moles of methane present = m and moles of ethane present = y.
As total moles = 0.03788 mol,
0.03788 = m + y .............(1)
A second equation in m and y can be obtained from the mass of the
mixture of methane + ethane.
Mass of methane + ethane = 36.033  35.194 = 0.839 g
This mass is made up of m moles of methane + y moles of ethane.
Molar mass of CH_{4} = 16.04 g mol1 and molar mass of
C_{2}H_{6}
= 30.08 g mol^{1}
Therefore m x 16.04 + y x 30.08 = 0.839 ....................(2)
From (1), y = (0.03788  m)
Substituting in (2)
m x 16.04 + (0.03788  m) x 30.08 = 0.839
whence 14.04 m = 0.300
m = 0.0214 mol
The fraction of methane by volume will be the same as its fraction
by moles as these are gases.
Fraction of CH_{4} in the mixture = m / (m + y)
= 0.0214 / 0.03788 = 0.565
and % methane = 100 x 0.565 = 56.6 %


9 
The volume is fixed but not P or T. Therefore any convenient volume
can be taken as long as the same volume is used throughout the calculation.
Take V = 1.00 L
Given P = 101 kPa and T = 298 K, then:
At 298 K, total moles of gas present in 1.00 L = PV / RT
= (101 x 1.00) / (8.314 x 298) = 0.04077 mol
0.200 of this total number of moles is NO_{2}(g) and 0.800
of the total moles is N_{2}O_{4}(g).
Therefore, moles of NO_{2} = 0.200 x 0.04077 = 0.00815 mol
and moles of N_{2}O_{4} = 0.800 x 0.04077 = 0.03261
mol
At 473 K, all the N_{2}O_{4} has been converted
to NO_{2},
N_{2}O_{4}(g) ®
2NO_{2}(g)
1 mole ® 2 moles
Thus the total moles of gas present = moles of NO_{2} originally
present
at 298 K + 2 x moles of N_{2}O_{4} originally present
at 298 K.
i.e. moles of gas at 473 K = 0.00815 + 2 x 0.03261
= 0.0734 mol
Substituting in the ideal gas equation,
P = nRT / V = (0.0734 x 8.314 x 473) / 1.00
= 289 kPa


10 
Each mole of Ca_{3}P_{2} would produce 2 moles
of PH_{3}(g) as there are 2 phosphorous atoms in each formula
unit of Ca_{3}P_{2}.
The PH_{3} then undergoes decomposition according to the
equation
PH_{3} ® (1/z)P_{z}
+ (3/2)H_{2}.
From this equation, 1 mole of PH_{3} results in the formation
of 1/z moles of gaseous P_{z} molecules and 3/2 moles of
hydrogen gas, H_{2}. Therefore from 2 moles of PH_{3}
there would be formed 2 x 1/z moles of P_{z} and 2 x 3/2
moles of H_{2}.
i.e. 1 mole of Ca_{3}P_{2} would result in the
formation of 2/z moles of P_{z} and 3 moles of H_{2}.
Moles of Ca_{3}P_{2} = mass / molar mass = 1.00
/ 182.2 = 0.005488 mol
Therefore moles of P_{z} resulting from the decomposition
of the PH_{3} formed
= (2/z) x 0.005488 mol = 0.01098 / z mol.
Moles of H_{2} resulting from the decomposition of the
PH_{3} = 3 x 0.005488 mol
= 0.01646 mol.
Total moles of gas present = moles of P_{z} + moles of
H_{2}
= (0.01098 / z + 0.01646) mol
In the Ideal Gas Equation,
n = PV / RT
P = 101 kPa
V = 1.060 L
T = 673 K
R = 8.314 J K^{1} mol^{1}
i.e. 0.01098 / z + 0.01646 = (101 x 1.060) / (8.314 x 673)
0.01913 = 0.01098 / z + 0.01646
z = 0.01098 / 0.00267 = 4.1
As z must be an integer, the formula P_{z} is P_{4}


11 
MnO_{2} + 4HCl ® MnCl_{2} + Cl_{2} + 2H_{2}O
From the equation,
1 mole of Cl_{2} requires 1 mole of MnO_{2} and
4 moles of HCl.
Moles of Cl_{2} formed can be calculated from the Ideal Gas Equation.
n = PV / RT = (103 x 2.50) / (8.314 x 293) mol
= 0.1057 mol.
(a) Moles of MnO_{2} = moles of Cl_{2} = 0.1057 mol.
Therefore mass of MnO_{2} = 0.1057 x molar mass
= 0.1057 x 86.94 = 9.19 g
(b) Moles of HCl = 4 x moles of Cl_{2} = 4 x 0.1057 = 0.4228 mol.
[Note: while only 2 moles of HCl are needed to form 1 mole of Cl_{2},
the additional 2 moles of HCl are needed to provide the required
total moles of H^{+} to form the 2 moles of H_{2}O.]
So, mass of HCl = mol HCl x molar mass HCl
= 0.4228 x 36.46 = 15.42 g
The hydrochloric acid solution contains 40.0 % HCl by mass and has
a
density = 1.12 g mL.
i.e 1.00 mL of solution contains 0.400 x 1.12 g of HCl
Therefore 1.00 / (0.400 x 1.12) mL contains 1.00 g of HCl
and (15.42 x 1.00) / (0.400 x 1.12) mL contains 15.42 g of HCl
= 34.4 mL
 