Boyle’s Law and Charles’ Law - · PDF fileA gas has a pressure of 1000 Pa at a...

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Boyle’s Law and Charles’ Law

© www.cgrahamphysics.com 2015

© www.cgrahamphysics.com 2015

kTP=P

T

Temperature (K)

Pressure (torr)

k (torr/K)

248 691.6 2.79 273 760.0 2.78 298 828.4 2.78 373 1,041.2 2.79

© www.cgrahamphysics.com 2015

The higher the temperature of a gas, the greater the kinetic energy of the particles. They move faster and collide with the walls of the container more frequently and with greater force.

This increases the pressure of the gas when volume is constant.

Pressure and temperature are directly proportional.

higher temperature higher pressure

pressure (Pa) temperature (K)

= constant

This means that for a gas at constant volume:

P1 T1

= P2 T2

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A gas has a pressure of 1000 Pa at a temperature of 295 K. What is the pressure when it is heated to 395 K?

P1 T1

P2 =

P2 =

P2 = 1339 Pa

The formula linking temperature and pressure in a gas at constant volume is called the Pressure-Temperature Law:

P2 T2

=

× T2 P1 T1

1000 295 × 395

𝑃/𝑇  = constant

© www.cgrahamphysics.com 2015

kTP=P

T

} The pressure and absolute temperature (K) of a gas are directly related ◦ at constant mass & volume

© www.cgrahamphysics.com 2015

© www.cgrahamphysics.com 2015

P

V PV = k

Volume (mL)

Pressure (torr)

k (mL·torr)

10.0 760.0 7.60 x 103 20.0 379.6 7.59 x 103 30.0 253.2 7.60 x 103 40.0 191.0 7.64 x 103

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P1 × V1 = P2 × V2

When the temperature of a gas is kept constant, a change in volume causes a change in pressure.

Particles collide more frequently with the container wall when the volume is smaller. This increases the pressure.

larger volume lower pressure

Pressure and volume are inversely proportional.

pressure (Pa) × volume (m3) = constant

This means that for a gas at constant temperature:

© www.cgrahamphysics.com 2015

P

V

PV = k

} The pressure and volume of a gas are inversely related ◦ at constant mass & temp

© www.cgrahamphysics.com 2015

© www.cgrahamphysics.com 2015

The formula linking volume and pressure in a gas at constant temperature is called Boyle’s Law:

Example: A gas syringe contains 0.00005 m3 of CO2 at a pressure of 100 kPa. If the volume is reduced to 0.000035 m3, what is the new pressure?

P1 × V1 = P2 × V2

P1 × V1 V2

100,000 × 0.00005 0.000035

P2 = 143 kPa

P2 =

P2 =

P × V = constant

© www.cgrahamphysics.com 2015

kTV=V

T

Volume (mL)

Temperature (K)

k (mL/K)

40.0 273.2 0.146 44.0 298.2 0.148 47.7 323.2 0.148 51.3 348.2 0.147

© www.cgrahamphysics.com 2015

kTV=V

T

} The volume and absolute temperature (K) of a gas are directly related ◦ at constant mass &

pressure

© www.cgrahamphysics.com 2015

© www.cgrahamphysics.com 2015

If a gas is in a container that does not have a fixed volume, then it will expand as it is heated. The pressure on the walls of the container will cause it to expand, so pressure stays constant.

higher temperature larger volume

Temperature and volume are directly proportional.

V1 T1

volume (m3) temperature (K)

= constant

This means that for a gas at constant pressure:

V2 T2

=

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The formula linking volume and temperature for a gas at a constant pressure is called Charles’ law:

Example: A gas has an initial temperature of 300 K and initial volume of 0.2 m3. What is the volume when it is heated to 330 K?

V1 T1

V2 =

V2 =

V2 = 0.22 m3

= V2 T2

× T2 V1 T1

× 330 0.2 300

𝑉/𝑇 = constant

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The Combined Gas Law brings together the Pressure-Temperature Law, Boyle’s Law and Charles’ Law.

It is generally written as:

pressure × volume temperature

To compare the behaviour of a gas in different conditions, the following form of the equation is used:

P1V1 T1

It describes the relationship between volume, pressure and temperature, whether or not any variable is kept fixed.

= constant

P2V2 T2

=

© www.cgrahamphysics.com 2015

}  A gas has an initial temperature of 3 oC and occupies an initial volume of 100 mL at 150kPa. Find its new volume when its pressure changes to 200 kPa and the temperature is changed to 8 oC.

Solution

}  𝑃↓1 = 150kPa and 𝑃↓2  = 200 kPa

}  𝑇↓1  = 3 oC and 𝑇↓2  = 8 oC

}  𝑉↓1  = 100 mL and 𝑉↓2  = ?

}  𝑃↓1 𝑥𝑉↓1 /𝑇↓1  = 𝑃↓2 𝑥  𝑉↓2 /𝑇↓2   solve for 𝑉↓2 

𝑉↓2 = 𝑃↓1 𝑥𝑉↓1 /𝑇↓1   x 𝑇↓2 /𝑃↓2  

Convert ^↑0↓𝐶   𝑡𝑜  𝐾𝑒𝑙𝑣𝑖𝑛: 𝑇↓1  = 273 + 3 = 276K 𝑇↓2  = 273 + 8 = 281K Plug in numbers: 𝑉↓2 = 150𝑥100𝑥281/276𝑥200 =76.35  𝑚𝐿

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}  A gas pressure is 765 torr at 296 K. At what temperature will the pressure be 560 torr if the volume is held constant?

Solution

}  𝑃↓1 /𝑇↓1  = 𝑃↓2 /𝑇↓2  

}  𝑇↓2  = 𝑃↓2 𝑥𝑇↓1 /𝑃↓1  

𝑇↓2 = 560𝑥296/765 =216.68𝐾 Given: 𝑃↓1  = 765 torr 𝑃↓2 = 560 torr 𝑇↓1 = 296 K 𝑇↓2  = ?

© www.cgrahamphysics.com 2015

© www.cgrahamphysics.com 2015

© www.cgrahamphysics.com 2015

© www.cgrahamphysics.com 2015

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