5.1 Ideal gas law and general gas equation

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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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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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