– In this subject, we will discuss the Ideal Gas Equation (Definition, Formula, Notes)

**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:

– This Equation is called the ideal gas equation 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.

**Solved problem on Ideal Gas Equation**

**Problem (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.

**Problem (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 the 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:

**Solved Problems On Combined gas law Equation**

**Problem (1)**

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:

– We start with Equation:

– Because n_{1}= n_{2} and T_{1} = T_{2} ,

– which is Boyle’s law.

– The given information is tabulated:

Check:

– When the 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.

**Problem (2)**

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.

**Problem (3)**

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 the combined gas law Equation:

– We assume that the amount of air in the bubble remains constant, that is, n_{1} = n_{2}so that

– The given information is summarized:

Check:

– We see that the final 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).

**Solved problems on the Density of gas **

**Problem (1)**

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).*