**1. A fixed mass of gas at 102300Pa pressure has a volume of 25cm3.Calculate its volume if the pressure is doubled.**

Working

P_{1} V_{1} = P_{2} V_{2 }Substituting :102300 x 25 = (102300 x **2**) x V_{2}

V_{2} = 102300 x 25 = **12.5cm3**

(102300 x 2)

**2. Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm ^{-2}.**

P_{1} V_{1} = P_{2} V_{2 }Substituting :100000 x 100 = P_{2} x (100 x **3**)

V_{2} = 100000 x 100 = **33333.3333 Nm ^{-2}**

(100 x 3)

**3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa?**

P_{1} V_{1} = P_{2} V_{2 }Substituting :101325 x 60 = 90000 x V_{2}

V_{2} = 101325 x 60 = **67.55 cm3**

90000

The new volume at 67.55 cm3 **exceed** ballon capacity of 60.00 cm3.It will **burst** before reaching destination.

7.Charles law states that“**the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant/fixed pressure **”

Mathematically:

Volume α Pressure **(Fixed /constant pressure)**

V α T ** (Fixed /constant P)** ie **V = Constant**(k)

** **T

From Charles law , an **increase** in temperature of a gas cause an **increase** in volume. i.e **doubling** the temperature cause the volume to be **doubled**.

Gases expand/increase by ** ^{1}/_{273} **by volume on heating.Gases contact/decrease by

**by volume on cooling at constant/fixed pressure.**

^{1}/_{273}The volume of a gas continue decreasing with decrease in temperature until at **-273 ^{o}C /0 K** the volume is

**zero**. i.e. there is no gas.

This temperature is called **absolute zero. **It is the **lowest** temperature at which a gas **can** exist.

Graphically therefore a plot of volume(**V**)** against** Temperature(**T**) in:

(**i**)^{o}C produces a **straight line **that is **extrapolated** to the absolute zero of -273** ^{o}C **.

(**ii**)Kelvin/K produces a **straight line **from absolute zero of **O Kelvin**

4. **500cm3 of carbon(IV)oxide at 0 ^{o}C was transfered into a cylinder at -4^{o}C. If the capacity of the cylinder is 450 cm3,explain what happened.**

V_{1 } = V_{2} substituting 500 = V_{2}

T_{1 } T_{2} (0 +273) (-4 +273)

= 500 x (-4 x 273) = **492.674cm3**

(0 + 273)

The capacity of cylinder (500cm3) is **less** than new volume(492.674cm3).

**7.326cm3**(500-492.674cm3)of carbon(IV)oxide gas did not fit into the cylinder.

**5. A mechanic was filling a deflated tyre with air in his closed garage using a hand pump. The capacity of the tyre was 40,000cm3 at room temperature. He rolled the tyre into the car outside. The temperature outside was 30 ^{o}C.Explain what happens.**

V_{1 } = V_{2} substituting 40000 = V_{2}

T_{1 } T_{2} (**25** +273) (**30** +273)

= 40000 x (30 x 273) = **40671.1409cm3**

(25 + 273)

The capacity of a tyre (40000cm3) is **less** than new volume(40671.1409cm3).

The tyre thus bursts.

**6. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30 ^{o}C , explain what happened.**

V_{1 } = V_{2} substituting 80 = V_{2}

T_{1 } T_{2} (**25** +273) (-**30** +273)

= 80 x (-30 x 273) = **65.2349cm3**

(25 + 273)

The capacity of balloon (80cm3) is **more** than new volume (65.2349cm3).

The balloon thus remained intact.

7. The continuous random motion of gases differ from gas to the other.The movement of molecules (of a gas) from region of high concentration to a region of low concentration is called **diffusion.**

The rate of diffusion of a gas depends on its density. i.e. **The higher the rate of diffusion, the less dense the gas**.

The density of a gas depends on its molar mass/relative molecular mass. i.e. **The **higher** the density the **higher** the molar mass/relative atomic mass and thus the **lower** the rate of diffusion.**

Examples

1.Carbon (IV)oxide(CO_{2}) has a molar mass of 44g.Nitrogen(N_{2})has a molar mass of 28g. (N_{2})is thus lighter/less dense than Carbon (IV)oxide(CO_{2}). N_{2 }diffuses faster than CO_{2.}

2.Ammonia(NH_{3}) has a molar mass of 17g.Nitrogen(N_{2})has a molar mass of 28g. (N_{2})is thus about **twice **lighter/less dense than Ammonia(NH_{3}). Ammonia(NH_{3})diffuses twice faster than N_{2.} 3. Ammonia(NH_{3}) has a molar mass of 17g.Hydrogen chloride gas has a molar mass of 36.5g.Both gases on contact react to form **white fumes** of ammonium chloride .When a glass/cotton wool dipped in ammonia and another glass/cotton wool dipped in hydrochloric acid are placed at opposite ends of a glass tube, both gases diffuse towards each other. A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower

The rate of diffusion of a gas is in accordance to **Grahams law of diffusion**.Grahams law states that:

**“the rate of diffusion of a gas is inversely proportional to the square root of its density, at the same/constant/fixed temperature and pressure”**

Mathematically

**R α 1 ** and since density is proportional to mass then **R α 1 **

** √ p √ m**

For two gases then:

**R _{1 } = R_{2}** where: R

**and R**

_{1}**is the**

_{2}**rate**of diffusion of

**1**

^{st}and

**2**

^{nd}gas.

**√M _{2 }√M_{1}**

_{ }M

**and M**

_{1}**is the**

_{2}**molar mass**of

**1**

^{st}and

**2**

^{nd}gas.

Since rate is inverse of time. i.e. the higher the rate the less the time:

For two gases then:

T_{1}_{ }= T_{2} where: T** _{1}** and T

**is the**

_{2}**time taken**for

**1**

^{st}and

**2**

^{nd}gas to diffuse.

**√**M_{1 }**√**M_{2 } M** _{1}** and M

**is the**

_{2}**molar mass**of

**1**

^{st}and

**2**

^{nd}gas.

8.

**It takes 30 seconds for 100cm3 of carbon(IV)oxide to diffuse across a porous plate. How long will it take 150cm3 of nitrogen(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,N=14.0=16.0)**

Molar mass CO_{2}=44.0 Molar mass NO_{2}=46.0

Method 1

100cm3 CO_{2 } takes 30seconds

150cm3 takes 150 x30 = 45seconds

100

T CO_{2 }= √ molar mass CO_{2} => 45seconds = √ 44.0

T NO_{2 } √ molar mass NO_{2 }T NO_{2 }√ 46.0

T NO_{2} =45seconds x √ 46.0 = **46.0114 **seconds

_{ }√ 44.0

Method 2

100cm3 CO_{2 } takes 30seconds

1cm3 takes 100 x1 = **3.3333cm3sec ^{-1}**

30

R CO_{2 }= √ molar mass NO_{2} => 3.3333cm3sec^{-1} = √ 46.0

R NO_{2 } √ molar mass CO_{2 }R NO_{2 }√ 44.0

R NO_{2} = 3.3333cm3sec^{-1} x √ 44.0 = **3.2601**cm3sec^{-1}

_{ }√ 46.0

3.2601cm3 takes 1seconds

150cm3 take 150cm3 = **46.0109seconds**

_{ } 3.2601cm3

**9. How long would 200cm3 of Hydrogen chloride take to diffuse through a porous plug if carbon(IV)oxide takes 200seconds to diffuse through.**

Molar mass CO_{2} = 44g Molar mass HCl = 36.5g

T CO_{2 }= √ molar mass CO_{2} => 200 seconds = √ 44.0

T HCl_{ } √ molar mass HCl_{ }T HCl_{ }√ 36.5

T HCl = 200seconds x √ 36.5 = **182.1588 **seconds

_{ }√ 44.0

**10. Oxygen gas takes 250 seconds to diffuse through a porous diaphragm. Calculate the molar mass of gas Z which takes 227 second to diffuse.**

Molar mass O_{2} = 32g Molar mass Z = x g

T O_{2 }= √ molar mass O_{2} => 250 seconds = √ 32.0

T Z_{ } √ molar mass Z_{ }227seconds_{ }√ x

√ x = 227seconds x √ 32 = **26.3828 **grams

_{ } 250

11**. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,0=16.0)**

Molar mass CO_{2 }= 44.0 Molar mass CO = 28.0

Method 1

25cm3 CO takes 25seconds

75cm3 takes 75 x25 = 75seconds

25

T CO_{2 }= √ molar mass CO_{2} => T CO_{2}seconds = √ 44.0

T CO_{ } √ molar mass CO_{ }75_{ }√ 28.0

T CO_{2} =75seconds x √ 44.0 = **94.0175 **seconds

_{ }√ 28.0

Method 2

25cm3 CO_{2 } takes 25seconds

1cm3 takes 25 x1 = **1.0cm3sec ^{-1}**

25

R CO_{2 }= √ molar mass CO => x cm3sec^{-1} = √ 28.0

R CO_{ } √ molar mass CO_{2 }1.0cm3sec^{-1}_{ }√ 44.0

R CO_{2} = 1.0cm3sec^{-1} x √ 28.0 = **0.7977**cm3sec^{-1}

_{ }√ 44.0

0.7977cm3 takes 1 seconds

75cm3 takes 75cm3 = **94.0203seconds**

_{ } 0.7977cm3