PV = nRT: The Ideal Gas Law

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The Ideal Gas Law was first written in 1834 by Emil Clapeyron.

This is just one way to "derive" the Ideal Gas Law:

For a static sample of gas, we can write each of the six gas laws as follows:

PV = k1
V / T = k2
P / T = k3
V / n = k4
P / n = k5
1 / nT = 1 / k6

Note that the last law is written in reciprocal form. The subscripts on k indicate that six different values would be obtained.

When you multiply them all together, you get:

P3V3 / n3T3 = k1k2k3k4k5 / k6

Let the cube root of k1k2k3k4k5 / k6 be called R.

The units work out:

k1 = L-atm
k2 = L / K
k3 = atm / K
k4 = L / mol
k5 = atm / mol
1 / k6 = 1 / mol-K

Each unit occurs three times and the cube root yields L-atm / mol-K, the correct units for R when used in a gas law context.

Consequently, we have:

PV / nT = R

or, more commonly:

PV = nRT

R is called the gas constant. Sometimes it is referred to as the universal gas constant. If you wind up taking enough chemistry, you will see it showing up over and over and over.


The Numerical Value for R

R's value can be determined many ways. This is just one way:

We will assume we have 1.000 mol of a gas at STP. The volume of this amount of gas under the conditions of STP is known to a high degree of precision. We will use the value of 22.414 L.

By the way, 22.414 L at STP has a name. It is called molar volume. It is the volume of ANY ideal gas at standard temperature and pressure.

Let's plug our numbers into the equation:

(1.000 atm) (22.414 L) = (1.000 mol) (R) (273.15 K)

Notice how atmospheres were used as well as the exact value for standard temperature.

Solving for R gives 0.08206 L atm / mol K, when rounded to four significant figures. This is usually enough. Remember the value. You'll need it for problem solving.

Notice the weird unit on R: say out loud "liter atmospheres per mole Kelvin."

This is not the only value of R that can exist. (Here's a whole bunch.) It depends on which units you select. Those of you that take more chemistry than high school level will meet up with 8.31447 Joules per mole Kelvin, but that's for another time. The ChemTeam will use the 0.08206 value in gas-related problems almost every time.


Example #1: A sample of gas at 25.0 °C has a volume of 11.0 L and exerts a pressure of 660.0 mmHg. How many moles of gas are in the sample?

Solution:

1) PV = nRT:

(660.0 mmHg / 760.0 mmHg/1.00 atm) (11.0 L) = (n) (0.08206 L atm / mol K) (298 K)

Note the conversion from mmHg to atm and from Celsius to Kelvin.

2) Solve:

    (0.868421 atm) (11.0 L)
n = –––––––––––––––––––––––––––
    (0.08206 L atm / mol K) (298 K)

n = 0.391 mol


Example #2: A sample of gas at 28.0 °C has a volume of 6.20 L and exerts a pressure of 720.0 mmHg. How many moles of gas are in the sample?

Solution:

1) PV = nRT:

(720.0 mmHg / 760.0 mmHg/1.00 atm) (6.20 L) = (n) (0.08206 L atm / mol K) (301 K)

Note the conversion from mmHg to atm and from Celsius to Kelvin.

2) Calculate:

n = 0.238 mol

Example #3: Calculate the approximate volume of a 0.400 mol sample of gas at 11.0 °C and a pressure of 2.43 atm.

Solution:

1) PV = nRT:

(2.43 atm) (V) = (0.400 mol) (0.08206 L atm / mol K) (284 K)

2) Calculate:

    (0.400 mol) (0.08206 L atm / mol K) (284 K)
V = –––––––––––––––––––––––––––––––––––––
    (2.43 atm)

V = 3.84 L


Example #4: Calculate the approximate temperature of a 0.30 mol sample of gas at 780 mmHg and a volume of 6.0 L.

Solution:

1) PV = nRT:

(780 mmHg) (6.0 L) = (0.30 mol) (62.3638 L mmHg / mol K) (284 K)

Note the different value and unit for R, to be in agreement with using mmHg for the volume unit.

2) T = PV / nR

T = [(780) (6.0)] / [(0.30) (62.3638)]

T = 250 K

The problem does not specify the final unit, but Celsius is most often requested.

250 - 273 = -23 °C


Example #5: What is the pressure exerted by 2.3 mol of a gas with a temperature of 40. °C and a volume of 3.5 L?

Solution:

PV = nRT

(P) (3.5 L) = (2.3 mol) (0.08206 L atm / mol K) (313 K)

P = [(2.3) (0.08206) (313)] / 3.5

P = 16.9 atm


Example #6: A sample of dry gas weighing 2.1025 grams is found to occupy 2.850 L at 22.00 °C and 740.0 mmHg. How many moles of the gas are present?

Solution:

1) I will change the units for pressure to atm., so as to keep with my preferred value for R:

740.0 mm Hg ÷ 760.0 mm Hg/atm = 0.973684 atm

2) Now, plug into the equation and solve for n:

(0.973684 atm) (2.850 L) = (n) (0.08206 L atm / mol K) (295.0 K)

To four sig figs, the answer is 0.1146 mol

I will use 0.1146329 mol in the next example and round off at the end.


Example #7: Using the problem above, what is the molar mass of the gas?

This is a very common use of this law and the odds are very good you will see this type of question on a test.

The key is to remember the units on molar mass: grams per mole.

We know from the problem statement that 2.1025 grams of the gas is involved and we also know how many moles that is.

We know that from doing the calculation above and getting 0.1146329 mol.

So all we have to do is divide the grams of gas by how many moles it is:

2.1025 g ÷ 0.1146329 mol = 18.34 g/mol

Let's go over those steps for using the Ideal Gas Law to calculate the molar mass of the gas:

1) You have to know the grams of gas involved. Usually the problem will just give you the value, but not always. You might have to calculate it.
2) You are going to have to calculate the moles of gas. Use PV = nRT and solve for n. Make sure to use L, atm and K.
3) Divide grams by moles and there's your answer.

Example #8: At STP, a 5.00 L flask filled with air has a mass of 543.251 g. The air in the flask is replaced with another gas and the mass of the flask is 566.107 g. The density of air is 1.29 g/L. What is the gas that replaced the air?

Solution:

1) Calculate mass of air in flask:

1.29 g/L times 5.00 L = 6.45 g

2) Calculate mass of flask:

543.251 g minus 6.45 g = 536.801 g

3) Calculate mass of unknown gas:

566.107 g minus 536.801 g = 29.306 g

4) Calculate moles of unknown gas:

PV = nRT

(1.00 atm) (5.00 L) = (n) (0.08206) (273 K)

n = 0.22319 mol

5) Calculate molar mass of unknown gas:

29.306 g / 0.22319 mol = 131.3 g/mol

The unknown gas could be xenon.


Example #9: (a) A gas has a temperature of 300. K and a pressure of 104 kPa. Find the volume occupied by 1.05 mol of this gas, assuming it is ideal. (b) Assuming the gas molecules can be approximated as small spheres of diameter 3.0 x 10¯10 m , determine the fraction of the volume found in part (a) that is occupied by the molecules.

Solution:

1) PV = nRT:

(104 kPa / 101.325 kPa/atm) (V) = (1.05 mol) (0.08206 L atm / mol K) (300. K)

V = 25.184 L

To three sig figs, the answer is 25.2 L

2) PV = nRT with slightly different numbers:

PV = nRT where

P = 104000 Pa, n = 1.05 mol, R = 8.314 Pa m3/mol K, T = 300. K.

V = nRT/p = (315) (8.31447 m3) / 104000 = 0.025183 m3 <--- note the 315, which comes from 300. x 1.05

Note how only the final unit that survives in the answer is shown.

V = 0.0252 m3

3) The number of molecules in 1.05 moles:

(1.05 mol) (6.022 x 1023 molecules mol¯1) = 6.3231 x 1023 molecules

4) The volume of an individual molecule is:

(4/3) (3.14159) (1.5 x 10¯10 m)3 = 1.4137 x 10¯29 m3

5) The total volume of the molecules in m3:

(6.3231 x 1023 molecules) (1.4137 x 10¯29 m3 / molecule) = 0.00000894 m3

6) The fraction of the gas volume that is occupied by matter is:

0.00000894 m3 / 0.025183 m3 = 0.000355 or about 1/2817 of the total volume

Example #10: Of the following gases, which has density of 0.906 g/L at 315 K and 1.16 atm.

A) Ne; B) Ar; C) Kr; D) Xe; E) He

Solution #1:

PV = nRT

(1.16 atm) (1.00 L) = (n) (0.08206 L atm / mol K) (315 K)

n = 0.044876 mol

0.906 g / 0.044876 mol = 20.19 g/mol <--- the molar mass of neon

Solution #2:

PV = nRT

PV = (m/M)RT <--- where m = mass, M = molar mass

Since density = m/V, we have this (after substituting and rearranging):

PVM = mRT

M = mRT / PV

M = [(0.906 g) (0.08206 L atm / mol K) (315 K)] / [(1.00 L) (1.16 atm)] = 20.19 g/mol

Neon.

Solution #3 (without PV = nRT):

At STP , 1 mol of gas has a volume of 22.414 L

Using the Combined Gas Law, convert the volume to 315 K and 1.16 atm:

22.414 * (315/273) * (1.00/1.16) = 22.295 L

Now, you need to calculate densities until you get the correct answer:

Ne---> 20.180 g/22.295 L = 0.905 g/L <--- the correct answer

For comparison's sake, here is argon:

Ar ---> 39.95 g/22.295 L = 1.79 g/L <--- this is a wrong answer


Example #11: A 0.105 g sample of an unknown diatomic gas contained in a 125 mL vessel has a pressure of 560 torr at 23 &175;C. What is the molar mass of the gas? What is the identity of the gas?

Solution:

PV = nRT

(560 torr / 760 torr/atm) (0.125 L) = (n) (0.08206 L atm / mol K) (296 K)

n = 0.003792 mol

0.105 g / 0.003792 mol = 27.7 g/mol

Nitrogen has a molar mass of 28.0 g/mol


Example #12: If 1.0 g of each of the following gases is taken at STP, which one would occupy the greatest volume?

(a) CO
(b) H2O
(c) CH4
(d) NO
(e) They would all occupy the same volume.

Solution:

1) Modify PV = nRT as follows:

V = nRT / P

Since RT/P are the same for each gas, the greatest volume will be for the gas with the greatest number of moles.

2) Consequently, the greatest number of moles in 1.0 g of a gas will be for the compound with the lowest molar mass:

CO ---> 28.0 g/mol
H2O ---> 18.0 g/mol
CH4 ---> 16.0 g/mol
NO ---> 30.0 g/mol

1.0 g of methane, CH4, will have the greatest volume at STP.


Example #13: If 12.8 g of liquid helium at 1.7 K is completely vaporized, what volume does the helium occupy at STP?

Solution:

We don't care about the 1.7 K. One reason is that, at 1.7 K, the helium is not yet a gas, since helium vaporizes at about 4.25 K. Another reason is that we only care about helium's situation at STP, we don't care what the helium did at lower temperatures.

1) Determine moles of helium:

12.8 g / 4.0026 g/mol = 3.1979 mol

2) Use PV = nRT:

(1.00 atm) (x) = (3.1979 mol) (0.08206 L atm / mol K) (273.15 K)

x = 71.6799 L

To three sig figs, 71.7 L

Using the more common 273 K for standard temperature results in 71.6 L. Make sure to use the value for standard temperature that your teacher uses. That is the value he/she uses to calculate the answers on the test.


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