# Jamb Chemistry Tutorials S02

by Philip Obhenimen · Published · Updated

**INTRODUCTION**

Welcome to the second series of this **Jamb chemistry tutoria**l this **Jamb Chemistry tutorial** series is all about solving gas law problems as it will be seen in your exams.Questions from gas laws are inevitable. Check out jamb chemistry hot topics, these are topics jamb set almost every year by** clicking here.**

Grab your writing material (Biro and jotter). Also, make sure you get a glass of chilled fruit drink or water to refresh yourself as we start up with this **Jamb chemistry tutorial** series.

**AIM OF THIS JAMB CHEMISTRY TUTORIALS**

In this **Jamb Chemistry Tutorial** series one we will achieve the following objectives:

- Solve Jamb chemistry problems concerning Boyle’s law.
- Solve Jamb chemistry problems concerning Charles law.
- Solve Jamb chemistry problems concerning Pressure law.
- Solve Jamb problems concerning chemistry General gas law.
- Solve Jamb chemistry problems concerning Ideal gas equation.
- Solve Jamb chemistry problems concerning Avogadro’s law.
- Solve Jamb chemistry problems concerning Gay lussac law of combining volume.
- Solve Jamb chemistry problems concerning Dalton’s law of partial pressure.
- Solve Jamb chemistry problems concerning Graham’s law of diffusion.

Do you find it hard to concentrate while reading ** click here** to see the steps you can take to discipline yourself to study well.

Click here to open my * Jamb online calculator * in another tab on your browser. When you are practising for jamb calculations like physics,chemistry and mathematics, please use a very simple calculator just like the one here.

##### Solving Questions involving Boyle’s Law

Please note this: When solving gas law problems generally, please ensure unit consistency.

As we dive into this Jamb Chemistry Tutorial i will explain the consistency in unit for each gas law.

###### Example 1

The pressure of a fixed mass of gas in a container of 200cm^{3} is 150kpa. What will be the volume of the gas if the pressure is increased to 350kpa?

Solution:

First lets list out the parameters and the unknown in the equation;

Initial pressure, P1 = 150kpa

Initial volume, V1 = 200cm^{3}

Final pressure, P2 = 350kpa

Final volume, V2 = ?

Secondly, check for unit consistency. Here, we can see that both pressure, P1 and P2 have the same unit of pressure in Kpa (Kilopascals)

Thirdly, apply the Boyle’s law formula;

P_{1}V_{1} = P_{2}V_{2}

150*200 = 350*V_{2}

V_{2} = \frac{150*200}{350}

Using my calculator,

V_{2} = 85.71cm^{3}

###### Example 2

If the volume of a gas is tripled what will be the pressure if the gas was initially at a pressure of 560 mmHg with the temperature constant?

Solution;

Questions like this should be solved without your pen or pencil.

If the volume of the gas is tripled it simply means the pressure should should be reduced by a factor of 3 (That is, the pressure should be one-third the initial pressure). This is because pressure and volumes are inversely related at constant temperature so that as pressure increases volume decreases or vice versa;

Therefore, P2 = 560/3 = 186.67 mmHg

##### Example 3

The volume of a fixed mass of gas is increased 10 times its initial value. By what factor will the pressure be increased?

solution:

If the volume is increased by 10 then according to Boyle’s law, the pressure is to be decreased by a factor of 10.

Therefore, since the question asked what factor will the pressure be increased;

The pressure will increase by a factor of one-tenth

###### Example 4:

Helium gas stored in a vessel of volume 500cm³ has a pressure of 3atm what will be the volume of the gas if the system was further compressed to 6080 mmHg.

Solution:

Writing out our parameters;

P1 = 3 atm

V1 = 500cm³

P2 = 6080 mmHg

V2 = ?

You will notice that the pressure are not consistent in unit

P1 is in atmospheres while P2 is in mmHg

Let’s convert in mmHg to atm you can decide to convert the atm to mmHg

Conversion factor: 1 atm = 760mmHg

Therefore, 6080mmHg will be \frac{6080}{760} = 8 atm

We can now use the formula:

P_{1}V_{1} = P_{2}V_{2}

3*500 = 8*V2

V2 =\frac{3*500}{8} = 187.5cm³

##### Solving Questions involving Charle’s law

When solving problems using the equation postulated by Sir Charles

The important point to note is that your temperature must be in Kelvin

NB: Convert all temperature to Kelvin before using the Charles law equation

###### Example 1:

A given mass of gas occupies 2 litres at a temperature of 50°c. What will be the

Volume if the gas is heated to 120°C at constant pressure?

Solution:

Writing out the parameters;

Initial volume ,V1 = 2 litres

Initial temperature = 50°c converting to kelvin

T1 = 50 + 273 = 323K

Final temperature = 120°C, converting to kelvin

T2 = 120 + 273 = 393K

Final volume, V2 = ?

Using the formula;

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

\frac{2}{393} = \frac{V_{2}}{393}

V2 =\frac{393*2}{323}

V2 = 2.43 litres

###### EXAMPLE 2:

The temperature of hydrogen gas in a container of volume 22.4dm³ is

Reduced to one-eighth of its initial temperature, what will be the volume of the gas?

Solution:

Initial Volume,V_{1} = 22.4dm³

Initial temperature, ,T_{1} = T (we are assigning it T because its value wasn’t given)

Final temperature, T_{2} =\frac{T}{8} (According to the question)

Final volume, V2 = ?

Using the formula: \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

\frac{22.4}{T} = \frac{V_{2}}{(1/8)T}

T is common to both side of the equation above, therefore, we can remove or cancel them out.

22.4 = 8*V_{2}

solving for V_{2} gives;

V_{2} = \frac{22.4}{8}

V_{2} = 2.8dm^{3}

###### Example 3

At 25ºC, a sample of hydrogen occupies 120 cm^{3}. What will the volume be at 200ºC, if the pressure remains constant?

Solution

writing out the parameters;

T_{1} = (25 + 273)K = 298 K

V_{1} = 120 cm^{3}

T_{2} = (200 + 273)K = 473 K

V_{2} = ?

using the formula; \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

\frac{120}{298} = \frac{V_{2}}{473}

V_{2} = \frac{120*473}{298}

V_{2} = 190.47cm^{3}

##### Solving Questions involving Pressure law

When solving problems using the equation postulated by gay lussac

The important point to note is that your temperature must be in Kelvin

NB: Convert all temperature to Kelvin before using the Pressure law equation

###### Example 1

The pressure of a gas in a container of temperature 450 K is 700 mmHg. What will be the new pressure if the container is heated to 400ºC?

Solution;

writing the parameters out:

P_{1} = 700 mmHg

T_{1} = 450K

T_{2} = (400 + 273)K = 673K

P_{2} = ?

Using the formula:\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}

Inputing the parameters;

\frac{700}{450} = \frac{P_{2}}{673}

P_{2} = \frac{700*673}{450}

P_{2}= 1046.89 mmHg

###### Example 2

The Pressure of a fixed mass of gas is reduced to one-fifths its initial pressure by what factor does the temperature increase or decrease?

solution;

It should be noted that since in pressure law, as the temperature increases, the pressure also decreases by the same factor.

Reading the question carefully, we can clearly see that the pressure of the gas was reduced by one-fifths. It therefore means that the temperature must also reduce by the same factor.

Ans: The temperature reduce to one-fifths its initial value.

###### Example 3

It was noticed that the temperature of a gas decreases to -25ºC when the pressure was decreased to one-third its initially pressure. Find the temperature of the gas before the pressure was reduce.(The volume remain constant during the process)

solution;

In this question, you are asked to compute for the initial pressure which is unlike the questions we have been solving.

P_{1} = P since its values wasn’t given

T_{1} = ?

P_{2} = \frac{P}{3}

T_{2} = (-25 + 273)K = 248K

Using the formula:\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}

Substituting the values;

\frac{P}{T_{1}} = \frac{(1/3)*P}{248}

P is common to both sides of the equation, so we can cancel out the P’s

Simplifying further becomes

\frac{1}{T_{1}} = \frac{1}{3*248}

cross-multiplying gives:

T_{1} = 3*248

T_{1} = 744K

##### Solving Questions involving General gas law

The general gas equation as its name implies, includes all Pressure,Temperature and Volume in its equation.

As a result of this, it is import for you to know what S.T.P means;

S.T.P is an acronym for Standard Temperature and Standard pressure.

The Standard temperature has a value of 0.15ºC or 273.15K.

The Standard pressure has a value of 1 atm or 760 mmHg or 101325N/m^{2}

For most calculation in jamb you will be using 273K as the value for Standard Temperature.

###### Example 1

A certain mass of gas occupies 200cm^{3} at a pressure of 500mmHg and temperature of 370K. What will be the temperature of the gas in other to occupy a volume of 350cm^{3} at a pressure of 800mmHg.

solution:

Writing out the parameters;

P_{1} = 500 mmHg

V_{1} = 200cm^{3}

T_{1} = 370K

P_{2} = 800 mmHg

V_{2} = 350 cm^{3}

T_{2} = ?

Using the general gas equation formula;

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

From the above general gas equation;

T_{2} = \frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

Inputing the values into the equation;

T_{2} = \frac{800*350*370}{500*200}

Using our online calculator

T_{2} = 196K

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