5.1 Ideal gas law and general gas equation (Page 5/4)

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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

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?

Aislinn Reply

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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance

Jude Reply

Can you compute that for me. Ty

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what is viscosity?

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what is inorganic

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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes

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chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.

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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity

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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.

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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s

Samuel Reply

can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?

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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time

Joseph

Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing

Joseph

"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true

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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?

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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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