# The Ideal Gas Equation

####
**The
Ideal Gas Equation**

**❒**Let us summarize the gas laws we have discussed so far:

**❒**We can combine all three expressions to form a single master equation for the behavior

of gases:.

where R , the
proportionality constant, is called the gas constant.

**❒**The ideal gas equation, describes the relationship among the four variables P, V, T, and n.

####
**Important
Notes of Ideal gas**

**(1)**An ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation.

**(2)**The molecules of an ideal gas do not attract or repel one another, and their volume is negligible compared with the volume of the container.

**(3)**Although there is no such thing in nature as an ideal gas, the ideal gas approximation works rather well for most reasonable temperature and pressure ranges.

**(4)**we can safely use the ideal gas equation to solve many gas problems.

####
**The
gas constant**

**❒**Before we can apply the ideal gas equation to a real system, we must evaluate the gas constant R.

**❒**At 0

^{o}C (273.15 K) and 1 atm pressure, many real gases behave like an ideal gas.

**❒**Experiments show that under these conditions, 1 mole of an ideal gas occupies 22.414 L, which is somewhat greater than the volume of a basketball, as shown in The following Figure:

**❒**The conditions 0

^{o}C and 1 atm are called standard temperature and pressure, often abbreviated STP.

**❒**From the ideal gas Equation we can write:

**❒**The dots between L and atm and between K and mol remind us that both L and atm are in the numerator and both K and mol are in the denominator. For most calculations, we will round off the value of R to three significant figures (0.0821 L.atm/K.mol) and use 22.41 L for the molar volume of a gas at STP.

**❒**Example (1) shows that if we know the quantity, volume, and temperature of a gas, we can calculate its pressure using the ideal gas equation. Unless otherwise stated, we assume that the temperatures given in °C in calculations are exact so that they do not affect the number of significant figures.

**Example (1): Sulfur hexafluoride (SF**

_{6}) is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5^{o}C.**Strategy:**

The problem gives the amount of the gas and
its volume and temperature. Is the gas undergoing a change in any of its
properties? What equation should we use to solve for the pressure? What
temperature unit should we use?

**Solution:**

Because no changes in gas properties occur, we
can use the ideal gas equation to calculate the pressure. Rearranging the ideal
gas Equation, we write:

**❒**By using the fact that the molar volume of a gas occupies 22.41 L at STP, we can calculate the volume of a gas at STP without using the ideal gas equation.

**Example (2): Calculate the volume (in liters) occupied by 7.40 g of NH**

_{3}at STP.**Strategy:**

What is the volume of one mole of an ideal gas
at STP? How many moles are there in 7.40 g of NH

_{3}?**Solution:**

Recognizing
that 1 mole of an ideal gas occupies 22.41 L at STP and using the molar mass of
NH

_{3}(17.03 g), we write the sequence of conversions as
so the volume
of NH

_{3}is given by:
It is often true in chemistry, particularly
in gas-law calculations, that a problem can be solved in more than one way.
Here the problem can also be solved by first converting 7.40 g of NH

_{3}to number of moles of NH_{3}, and then applying the ideal gas equation ( V = nRT/P ). Try it.**Check:**

Because 7.40 g of NH

_{3}is smaller than its molar mass, its volume at STP should be smaller than 22.41 L. Therefore, the answer is reasonable.####
**Combined
gas law equation**

**❒**The ideal gas equation is useful for problems that do not involve changes in P , V , T , and n for a gas sample. Thus, if we know any three of the variables we can calculate the fourth one using the equation. At times, however, we need to deal with changes in pressure, volume, and temperature, or even in the amount of gas. When conditions change, we must employ a modified form of the ideal gas equation that takes into account the initial and final conditions.

We derive the
modified equation as follows.

From Ideal gas
equation:

**❒**It is interesting to note that all the gas laws can be derived from the previous Equation.

**❒**If n

_{1}= n

_{2}, as is usually the case because the amount of gas normally does not change, the equation then becomes:

**Example (3): An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon?**

**Strategy**:

The amount of gas inside the balloon and its
temperature remain constant, but both the pressure
and the volume change. What gas law do you need?

**solution:**

Because n

_{1}= n_{2}and T_{1}= T_{2},
which is
Boyle’s law. The given information is tabulated:

**Check:**

When pressure applied on the balloon is
reduced (at constant temperature), the helium gas expands and the balloon’s
volume increases. The final volume is greater than the initial volume, so the
answer is reasonable.

**Example (4): Argon is an inert gas used in light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.20 atm and 18**

^{o}C is heated to 85^{o}C at constant volume. Calculate its final pressure (in atm).**Strategy:**

The temperature and pressure of argon change
but the amount and volume of gas remain the same. What equation would you use
to solve for the final pressure? What temperature unit should you use?

**Solution:**

from combined
gas law:

Because n

_{1}= n_{2}and V_{1}= V_{2}, the Equation becomes:
which is Charles’s
law. Next we write

**Check:**

At constant volume, the pressure of a given
amount of gas is directly proportional to its absolute temperature. Therefore
the increase in pressure is reasonable.

**Example (5): A small bubble rises from the bottom of a lake, where the temperature and pressure are 8**

^{o}C and 6.4 atm, to the water’s surface, where the temperature is 25^{o}C and the pressure is 1.0 atm. Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL.**Strategy:**

In solving this kind of problem, where a lot
of information is given, it is sometimes helpful to make a sketch of the
situation, as shown here:

What
temperature unit should be used in the calculation?

**Solution**:

According to combined gas law Equation:

The given
information is summarized:

**Check:**

We see that the fi nal volume involves
multiplying the initial volume by a ratio of pressures (P

_{1}/P_{2}) and a ratio of temperatures (T_{2}/T_{1}). Recall that volume is inversely proportional to pressure, and volume is directly proportional to temperature. Because the pressure decreases and temperature increases as the bubble rises, we expect the bubble’s volume to increase. In fact, here the change in pressure plays a greater role in the volume change.####
**Density
of gas **

If we rearrange
the ideal gas equation, we can calculate the density of a gas:

where m is the
mass of the gas in grams and µ is its molar mass. Therefore

Because
density, d, is mass per unit volume, we can write:

Unlike
molecules in condensed matter (that is, in liquids and solids), gaseous
molecules are separated
by distances that are large compared with their size. Consequently, the density of
gases is very low under atmospheric conditions. For this reason, gas densities are
usually expressed in grams per liter (g/L) rather than grams per milliliter (g/mL).

**Example (6):**

**Calculate the density of carbon dioxide (CO**

_{2}) in grams per liter (g/L) at 0.990 atm and 55^{o}C.**Strategy:**

We need the combined gas Equation to calculate
gas density. Is sufficient information provided in the problem? What
temperature unit should be used?

**Solution:**

To use the combined gas Equation, we convert
temperature to kelvins ( T = 273 + 55 = 328 K) and use 44.01 g for the molar
mass of CO

_{2}:
Alternatively,
we can solve for the density by writing:

Assuming that
we have 1 mole of CO

_{2}, the mass is 44.01 g. The volume of the gas can be obtained from the ideal gas equation
Therefore, the
density of CO

_{2}is given by:**Comment:**

ln
units of grams per milliliter, the gas density is 1.62 × 10

^{-3}g/mL, which is a very small number. In comparison, the density of water is 1.0 g/mL and that of gold is 19.3 g/cm^{3}.

*Reference:*

*Chemistry*

*/ Raymond Chang ,*

*Williams College*

*/*

*(10th edition).*

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