What is the value of R in PV nRT if the pressure temperature and volume are measured in atmosphere Kelvin and liter respectively?

The temperature, T

The temperature has to be in kelvin. Don't forget to add 273 if you are given a temperature in degrees Celsius.

Using the ideal gas equation

Calculations using the ideal gas equation are included in my calculations book (see the link at the very bottom of the page), and I can't repeat them here. There are, however, a couple of calculations that I haven't done in the book which give a reasonable idea of how the ideal gas equation works.

The molar volume at stp

If you have done simple calculations from equations, you have probably used the molar volume of a gas.

1 mole of any gas occupies 22.4 dm3 at stp (standard temperature and pressure, taken as 0°C and 1 atmosphere pressure). You may also have used a value of 24.0 dm3 at room temperature and pressure (taken as about 20°C and 1 atmosphere).

These figures are actually only true for an ideal gas, and we'll have a look at where they come from.

We can use the ideal gas equation to calculate the volume of 1 mole of an ideal gas at 0°C and 1 atmosphere pressure.

First, we have to get the units right.

0°C is 273 K. T = 273 K

1 atmosphere = 101325 Pa. p = 101325 Pa

We know that n = 1, because we are trying to calculate the volume of 1 mole of gas.

And, finally, R = 8.31441 J K-1 mol-1.

Slotting all of this into the ideal gas equation and then rearranging it gives:

And finally, because we are interested in the volume in cubic decimetres, you have to remember to multiply this by 1000 to convert from cubic metres into cubic decimetres.

The molar volume of an ideal gas is therefore 22.4 dm3 at stp.

And, of course, you could redo this calculation to find the volume of 1 mole of an ideal gas at room temperature and pressure - or any other temperature and pressure.

Finding the relative formula mass of a gas from its density

This is about as tricky as it gets using the ideal gas equation.

The density of ethane is 1.264 g dm-3 at 20°C and 1 atmosphere. Calculate the relative formula mass of ethane.

The density value means that 1 dm3 of ethane weighs 1.264 g.

Again, before we do anything else, get the awkward units sorted out.

A pressure of 1 atmosphere is 101325 Pa.

The volume of 1 dm3 has to be converted to cubic metres, by dividing by 1000. We have a volume of 0.001 m3.

The temperature is 293 K.

Now put all the numbers into the form of the ideal gas equation which lets you work with masses, and rearrange it to work out the mass of 1 mole.

The mass of 1 mole of anything is simply the relative formula mass in grams.

So the relative formula mass of ethane is 30.4, to 3 sig figs.

Now, if you add up the relative formula mass of ethane, C2H6 using accurate values of relative atomic masses, you get an answer of 30.07 to 4 significant figures. Which is different from our answer - so what's wrong?

There are two possibilities.

The density value I have used may not be correct. I did the sum again using a slightly different value quoted at a different temperature from another source. This time I got an answer of 30.3. So the density values may not be entirely accurate, but they are both giving much the same sort of answer.
Ethane isn't an ideal gas. Well, of course it isn't an ideal gas - there's no such thing! However, assuming that the density values are close to correct, the error is within 1% of what you would expect. So although ethane isn't exactly behaving like an ideal gas, it isn't far off.

If you need to know about real gases, now is a good time to read about them.

An Ideal Gas is also known as a Perfect Gas.

An Ideal Gas is one which obeys Boyle's Law and Charles' Law exactly.

An Ideal Gas obeys the Ideal Gas Law (also known as the General gas equation):

PV = nRT

where P = pressure
V = volume

n = moles of gas
T = temperature
R = gas constant (dependent on the units of pressure, temperature and volume)

R = 8.314 J K-1 mol-1 if (a) Pressure is in kilopascals (kPa) (b) Volume is in litres (L)
(c) Temperature is in kelvin (K)

R = 0.0821 L atm K-1 mol-1 if (a) Pressure is in atmospheres (atm) (b) Volume is in litres (L)
(c) Temperature is in kelvin (K)

An Ideal Gas is modelled on the Kinetic Theory of Gases which has 4 basic postulates:

Gases consist of small particles (molecules) which are in continuous random motion

The volume of the molecules present is negligible compared to the total volume occupied by the gas

Intermolecular forces are negligible

Pressure is due to the gas molecules colliding with the walls of the container

Real Gases deviate from Ideal Gas Behaviour because:

at low temperatures the gas molecules have less kinetic energy (move around less) so they do attract each other

at high pressures the gas molecules are forced closer together so that the volume of the gas molecules becomes significant compared to the volume the gas occupies

Under ordinary conditions, deviations from Ideal Gas behaviour are so slight that they can be neglected

A gas which deviates from Ideal Gas behaviour is called a non-ideal gas.

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Deriving the Ideal Gas Equation

For a constant amount of gas:

Boyle's Law tells us that the volume of the gas (V) is inversely proportional to its pressure (P):

Charles' Law tells us that the volume of the gas (V) is directly proportional to its temperature in Kelvin (T):

and the Combined Gas Equation tells us that the volume of the gas (V) is directly proportional to its temperature in Kelvin (T) and inversely proportional to its pressure (P):

But what if we do not have a constant amount of gas? What if we were to add more gas, or, remove some gas? What would happen to the volume of the gas then?

Avogadro's Principle tells us that, for a gas at constant temperature and pressure, the volume of gas (V) is directly proportional to the number of gas molecules (N):

V ∝ N

and since a mole is equivalent to Avogadro's Number of molecules, we can say that the volume of a gas (V) is directly proportional to the moles of gas (n)

V ∝ n

Remember that we combined Boyle's law and Charles' Law to give the Combined Gas Equation:

So now we can also include V ∝ n to give a new relationship between the amount of gas (n), its volume (V), pressure (P), and temperature (T):

We can turn this relationship into an equation by using a constant of proportionality, R:

If we rearrange this equation, we find that

If we know the values P, V, n and T then we can find the value of the constant, R.

The volume (V) of different amounts of gas (n) was measured at a constant pressure of 101.3 kPa (1 atm) and 298 K (25° C).
These values were then used to calculate the value of the constant, R, as shown in the table below:

Experiment

Pressure
(kPa)

Volume
(L)

Moles
(mol)

Temperature
(K)

101.3

24.5

1.00

298

101.3 × 24.5
1 × 298

R = 8.3

101.3

48.9

2.00

298

101.3 × 48.9
2 × 298

R = 8.3

101.3

73.4

3.00

298

101.3 × 73.4
3 × 298

R = 8.3

101.3

97.8

4.00

298

101.3 × 97.8
4 × 298

R = 8.3

101.3

122

5.00

298

101.3 × 122
5 × 298

R = 8.3

If we measure pressure in kilopascals (kPa), volume in litres (L), temperature in Kelvin (K) and the amount of gas in moles (mol), then we find that R = 8.314 and it has the units kPa L K-1 mol-1.
The pressure exerted by the gas in this volume is actually a measure of the energy of the gas particles, so the units of this Gas Constant, R, are most often expressed as Joules per Kelvin per mole, J K-1 mol-1.
R = 8.314 J K-1 mol-1

The gas constant R is very special because its value does not depend on the nature of the gas used.
If the experiment described above is done using:

carbon dioxide gas, CO2(g), then R = 8.314 J K-1 mol-1

nitrogen gas, N2(g), then R = 8.314 J K-1 mol-1

helium gas, He(g), then R = 8.314 J K-1 mol-1

So R is called the Gas Constant, and the equation PV = nRT is known as the Ideal Gas Equation, or, as the Ideal Gas Law.

Since R depends only on the amount of gas present, and does not depend on what the gas molecules are made up of, then the molecules of gas must not interact with each other, for, if they did, the value of R would be expected to change and reflect the impact of these interactions on the value of pressure (for instance).

By rearranging the Ideal Gas Law (Ideal Gas Equation) PV=nRT, it can be used to calculate the pressure (P), volume (V), temperature (T) or amount (n) of gas:

To calculate gas pressure:	P =	nRT V
To calculate gas volume:	V =	nRT P
To calculate gas temperature:	T =	PV nR
To calculate amount of gas:	n =	PV RT

Note that the Ideal Gas Law is supported by the Kinetic Theory of Gases:

Ideal Gas Law says that at constant temperature (T) and volume (V), the pressure of a gas (P) is directly proportional to the amount of gas (n)
P ∝ n

That is, increase the moles of gas molecules and the pressure will increase.
Reduce the moles of gas molecules and the pressure will decrease.

Kinetic Theory of Gases tells us that pressure is caused by gas molecules colliding with the walls of the container, in other words, for a given volume of gas (at constant pressure):

P ∝ number of gas molecules

So, since the number of gas molecules is related to the amount of gas in moles, the Ideal Gas Law is supported by the Kinetic Theory of Gases.

Ideal Gas Law says that the volume of a gas (V) is dependent on the amount of gas (n), its temperature (T) and its volume (V). Note that the volume of a gas is not dependent on the size, or volume, of the gas molecules.
The volume of a gas depends on the amount of gas and NOT on any other properties of the gas.

Kinetic Theory of Gases tells us that a gas is made up of a huge number of gas particles which are so small that their sizes are negligible compared to the average distances between them, that is, the volume of gas molecules is negligible compared to the volume of space occupied by the gas.
In other words, the volume of space occupied by a gas depends on the number of the gas particles, but does not depend on the nature of the gas particles.

V ∝ number of gas molecules

So, since the number of gas molecules is related to the amount of gas in moles, the Ideal Gas Law is supported by the Kinetic Theory of Gases.

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Question : What volume is needed to store 0.050 moles of helium gas at 202.6 kPa and 400 K?

Solution:

(Based on the StoPGoPS approach to problem solving.)

What is the question asking you to do?
Calculate gas volume
V = ? L

What data (information) have you been given in the question?
Extract the data from the question:

P = gas pressure = 202.6 kPa T = gas temperature = 400 K
n = moles of gas = 0.050 mol

What is the relationship between what you know and what you need to find out?
Assume Ideal Gas behaviour
R = 8.314 J K-1 mol-1 (from Data Sheet)

PV = nRT

Rearrange this equation by dividing both sides by P:

Substitute in the values and solve for V:

V = nRT
P

= 0.050 × 8.314 × 400
202.6

= 0.82 L

Is your answer plausible?
Consider the molar volume of a gas at room temperature (298 K) and pressure (100 kPa): that is 1 mole of gas under these conditions occupies a volume of 24.79 L (say 25 L) So, under the same conditions, 0.050 moles of gas would have a volume of 0.05 × 25 = 1.25 L The pressure is 2 times atmospheric pressure, so this will reduce the volume by half, ½ × 1.25 L = 0.625 L
BUT, if room temperature is about 300K (298 K ≈ 300 K) then the temperature of the experiment is also higher by a factor of about 400/300 = 4/3 which means that the reduction in volume due to pressure will not be so great, that is,

gas volume = 4/3 × 0.625 L = 0.83 L
Since this approximate calculation for volume is about the same as that we carefully calculated, we are reasonably confident that our answer is plausible.

State your solution to the problem "gas volume":
V = 0.82 L

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Question : What pressure will be exerted by 20.16 g hydrogen gas in a 7.5 L cylinder at 20oC?

Solution:

(Based on the StoPGoPS approach to problem solving.)

What is the question asking you to do?
Calculate gas pressure
P = ? kPa

What data (information) have you been given in the question?
Extract the data from the question:

gas is hydrogen gas V = gas volume = 7.5 L T = gas temperature = 20°C
Convert temperature in °C to K
T = 273 + 20 = 293 K
m(H2(g)) = mass of hydrogen gas = 20.16 g

What is the relationship between what you know and what you need to find out?
Assume Ideal Gas behaviour
R = 8.314 J K-1 mol-1 (from Data Sheet)

PV = nRT

Rearrange this equation by dividing both sides by V:

Calculate moles of hydrogen gas, n,
Hydrogen gas is a diatomic molecule with the molecular formula H2(g)

moles = mass ÷ molar mass mass = 20.16 g (given in question)
molar mass of a hydrogen atom = 1.008 g mol-1 (from Periodic Table)

molar mass of hydrogen gas molecule = 2 × 1.008 = 2.016 g mol-1
n(H2(g)) = 20.16 ÷ 2.016 = 10.00 mol

Substitute in the values and solve for P:

P = n × R × T
V

= 10 × 8.314 × 293
7.5

= 3248 kPa

Is your answer plausible?
Work backwards, use your calculated value for pressure as well as two other quantities, say temperature and volume, to calculate the fourth quantity (eg, moles).

n = P × V
R × T

= 3248 × 7.5
8.314 × 293

= 10 mol

1 mole H2 has a mass of about 2 g mol-1
So 10 mole H2 has a mass of about 10 mol × 2 g mol-1 = 20 g
Since this "rough calculation" is in good agreement with the value given in the question, we are reasonably confident that our answer for gas pressure is plausible.

State your solution to the problem "pressure of gas":
P = 3248 kPa

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Question : A 50.00 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 30.00°C.
How many moles of argon gas are in the cylinder?

Solution:

(Based on the StoPGoPS approach to problem solving.)

What is the question asking you to do?
Calculate gas moles of argon gas
n = ? mol

What data (information) have you been given in the question?
Extract the data from the question:

V = gas volume = 50.00 L P = gas pressure = 10130.0 kPa T = gas temperature = 30.00°C
Convert temperature in °C to K

T = 273 + 30.00 = 303 K

What is the relationship between what you know and what you need to find out?
Assume Ideal Gas behaviour
R = 8.314 J K-1 mol-1 (from Data Sheet)

PV = nRT

Rearrange this equation by dividing both sides by RT:

Substitute in the values and solve for n:

n = PV
RT

= 10130.0 × 50.00
8.314 × 303

= 506500
2519.142

= 201 mol

Is your answer plausible?
Perform a "rough" calculation to see if your answer is the right order of magnitude. Let V = 50 L P = 10000 kPa T = 300 K
R = 10
Then n = PV ÷ RT n = (10000 × 50) ÷ (300 × 10) n = 500000 ÷ 3000
n = 500 ÷ 3 ≈ 170 mol

170 mol is the same order of magnitude as 201 mol so we are reasonably confident that our answer is plausible.

State your solution to the problem "moles of argon gas":
n = 201 mol

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Question : To what temperature in Kelvin does a 250.0 mL cylinder containing 0.4000 g helium gas need to be cooled in order for the pressure to be 253.25 kPa?

Solution:

(Based on the StoPGoPS approach to problem solving.)

What is the question asking you to do?
Calculate gas temperature in Kelvin
T = ? K

What data (information) have you been given in the question?
Extract the data from the question:

Gas is helium gas m = mass of helium gas = 0.4000 g V = gas volume = 250.0 mL
Convert volume in mL to L
V = 250.0 mL ÷ 1000 mL/L = 0.2500 L
P = gas pressure = 253.25 kPa

What is the relationship between what you know and what you need to find out?
Assume Ideal Gas behaviour
R = 8.314 J K-1 mol-1 (from Data Sheet)

PV = nRT

Calculate moles of helium gas, n(He(g)),
Helium gas is a Group 18 element (Noble Gas) therefore a monatomic molecule with the molecular formula He(g)
moles(He(g)) = mass(He(g)) ÷ molar mass(He(g))
mass(He(g)) = 0.4000 g (from question)
molar mass(He(g)) = 4.003 g mol-1 (from Periodic Table)
n(He(g)) = 0.4000 ÷ 4.003 = 0.099925 mol

Rearrange the Ideal Gas equation by dividing both sides by : nR

Substitute in the values and solve for T:

T = PV
nR

= 253.25 × 0.2500
0.099925 × 8.314

= 63.313
0.83078

= 76.21 K

Is your answer plausible?
Round off the numbers to perform a "rough" calculation to check that your answer is in the right "ball park":
Let V = 0.25 = 1/4 ≈ 1/5 L P = 250 kPa R = 10
n = 0.4 g/4 g mol-1 = 0.1 mol
T = PV ÷ R n
T = (250 × 1/5) ÷ 10 × 0.1
T ≈ 50 ÷ 1 = 50 K

Since our "rough" calculation of 50 K is in the same "ball park" as the 76.21 K calculated above, we are reasonably confident that our answer is plausible.