5.1 Ideal gas law and general gas equation (Page 7/4)

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The general gas equation

All the gas laws we have described so far rely on the fact that at least one variable (T, p or V) remains constant. Since this is unlikely to be the case most times, it is useful to combine the relationships into one equation. These relationships are as follows:

Boyle's law: p $\propto$ $\frac{1}{V}$ (constant T)

Relationship between p and T: p $\propto$ T (constant V)

If we combine these relationships, we get p $\propto$ $\frac{T}{V}$

If we introduce the proportionality constant k, we get p = k $\frac{T}{V}$

or, rearranging the equation...

p V = k T

We can also rewrite this relationship as follows:

\frac{p V}{T} = k

Provided the mass of the gas stays the same, we can also say that:

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

In the above equation, the subscripts 1 and 2 refer to two pressure and volume readings for the same mass of gas under different conditions. This is known as the general gas equation . Temperature is always in kelvin and the units used for pressure and volume must be the same on both sides of the equation.

Remember that the general gas equation only applies if the mass of the gas is fixed.

At the beginning of a journey, a truck tyre has a volume of 30 dm $^{3}$ and an internal pressure of 170 kPa. The temperature of the tyre is 16 $^{\circ}$ C. By the end of the trip, the volume of the tyre has increased to 32 dm $^{3}$ and the temperature of the air inside the tyre is 35 $^{\circ}$ C. What is the tyre pressure at the end of the journey?

Write down all the information that you know about the gas.
p $_{1}$ = 170 kPa and p $_{2}$ = ?

V $_{1}$ = 30 dm $^{3}$ and V $_{2}$ = 32 dm $^{3}$

T $_{1}$ = 16 $^{\circ}$ C and T $_{2}$ = 40 $^{\circ}$ C
Convert the known values to SI units if necessary.
Here, temperature must be converted into Kelvin, therefore:

T $_{1}$ = 16 + 273 = 289 K

T $_{2}$ = 40 + 273 = 313 K
Choose a relevant gas law equation that will allow you to calculate the unknown variable.
Use the general gas equation to solve this problem:

$\frac{p_{1} \times V_{1}}{T_{1}} = \frac{p_{2} \times V_{2}}{T_{2}}$

Therefore,

$p_{2} = \frac{p_{1} \times V_{1} \times T_{2}}{T_{1} \times V_{2}}$
Substitute the known values into the equation. Calculate the unknown variable.
$p_{2} = \frac{170 \times 30 \times 313}{289 \times 32} = 173 k P a$

The pressure of the tyre at the end of the journey is 173 kPa.

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A cylinder that contains methane gas is kept at a temperature of 15 $^{\circ}$ C and exerts a pressure of 7 atm. If the temperature of the cylinder increases to 25 $^{\circ}$ C, what pressure does the gas now exert? (Refer to [link] to see what an 'atm' is.

Write down all the information that you know about the gas.
p $_{1}$ = 7 atm and p $_{2}$ = ?

T $_{1}$ = 15 $^{\circ}$ C and T $_{2}$ = 25 $^{\circ}$ C
Convert the known values to SI units if necessary.
Here, temperature must be converted into Kelvin, therefore:

T $_{1}$ = 15 + 273 = 288 K

T $_{2}$ = 25 + 273 = 298 K
Choose a relevant gas law equation that will allow you to calculate the unknown variable.
Since the volume of the cylinder is constant, we can write:

$\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}$

Therefore,

$p_{2} = \frac{p_{1} \times T_{2}}{T_{1}}$
Substitute the known values into the equation. Calculate the unknown variable.
$p_{2} = \frac{7 \times 298}{288} = 7.24 a t m$

The pressure of the gas is 7.24 atm.

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A gas container can withstand a pressure of 130 kPa before it will start to leak. Assuming that the volume of the gas in the container stays the same, at what temperature will the container start to leak if the gas exerts a pressure of 100 kPa at 15 $^{\circ}$ C?

Write down all the information that you know about the gas.
p $_{1}$ = 100 kPa and p $_{2}$ = 130 kPa

T $_{1}$ = 15 $^{\circ}$ C and T $_{2}$ = ?
Convert the known values to SI units if necessary.
Here, temperature must be converted into Kelvin, therefore:

T $_{1}$ = 15 + 273 = 288 K
Choose a relevant gas law equation that will allow you to calculate the unknown variable.
Since the volume of the container is constant, we can write:

$\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}$

Therefore,

$\frac{1}{T_{2}} = \frac{p_{1}}{T_{1} \times p_{2}}$

Therefore,

$T_{2} = \frac{T_{1} \times p_{2}}{p_{1}}$
Substitute the known values into the equation. Calculate the unknown variable.
$T_{2} = \frac{288 \times 130}{100} = 374.4 K = 101 . 4^{\circ} C$

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Questions & Answers

what is microbiology

Agebe Reply

What is a cell

Odelana Reply

what is cell

Mohammed

how does Neisseria cause meningitis

Nyibol Reply

what is microbiologist

Muhammad Reply

what is errata

Muhammad

is the branch of biology that deals with the study of microorganisms.

Ntefuni Reply

What is microbiology

Mercy Reply

studies of microbes

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when we takee the specimen which lumbar,spin,

Ziyad Reply

How bacteria create energy to survive?

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Bacteria doesn't produce energy they are dependent upon their substrate in case of lack of nutrients they are able to make spores which helps them to sustain in harsh environments

_Adnan

But not all bacteria make spores, l mean Eukaryotic cells have Mitochondria which acts as powerhouse for them, since bacteria don't have it, what is the substitution for it?

Muhamad

they make spores

Louisiaste

what is sporadic nd endemic, epidemic

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the significance of food webs for disease transmission

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food webs brings about an infection as an individual depends on number of diseased foods or carriers dully.

Mark

explain assimilatory nitrate reduction

Esinniobiwa Reply

Assimilatory nitrate reduction is a process that occurs in some microorganisms, such as bacteria and archaea, in which nitrate (NO3-) is reduced to nitrite (NO2-), and then further reduced to ammonia (NH3).

Elkana

This process is called assimilatory nitrate reduction because the nitrogen that is produced is incorporated in the cells of microorganisms where it can be used in the synthesis of amino acids and other nitrogen products

Elkana

Examples of thermophilic organisms

Shu Reply

Give Examples of thermophilic organisms

Shu

advantages of normal Flora to the host

Micheal Reply

Prevent foreign microbes to the host

Abubakar

they provide healthier benefits to their hosts

ayesha

They are friends to host only when Host immune system is strong and become enemies when the host immune system is weakened . very bad relationship!

Mark

what is cell

faisal Reply

cell is the smallest unit of life

Fauziya

cell is the smallest unit of life

Akanni

Innocent

cell is the structural and functional unit of life

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is the fundamental units of Life

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Micheal Reply

There are nothing like emergency disease but there are some common medical emergency which can occur simultaneously like Bleeding,heart attack,Breathing difficulties,severe pain heart stock.Hope you will get my point .Have a nice day ❣️

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I think infection prevention and control is the avoidance of all things we do that gives out break of infections and promotion of health practices that promote life

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Heyy Lubega hussein where are u from?

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en français

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Many sites of the body have it Skin Nasal cavity Oral cavity Gastro intestinal tract

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skin

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skin,Oral,Nasal,GIt

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all

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by fussion

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what are the advantages of normal Flora to the host

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what are the ways of control and prevention of nosocomial infection in the hospital

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Source: OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2

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