1 mole of a gas occupies a specific volume at a particular temperature and pressure.

This is known as the molar volume and given the symbol V_{m}

The units most commonly used for molar volume, V_{m}, are litres per mole, L mol^{-1}

Examples of molar gas volume (V_{m}) for ideal gases^{1}:

Temperature

Pressure

Molar Volume

^{o}C

(K)

kPa

(atm)

(V_{m}) / L mol^{-1}

0^{o}C

(273.15K)

100kPa

(0.987 atm)

22.71

25^{o}C

(298.15 K)

100 kPa

(0.987 atm)

24.79

0^{o}C (273.15K) and 100kPa (0.987 atm) is known as Standard Temperature and Pressure and is often abbreviated to STP ^{2} 25^{o}C (298.15 K) and 100 kPa (0.987 atm) is sometimes referred to as Standard Ambient Temperature and Pressure, SATP, or even as Standard Laboratory Conditions, SLC.^{3}

Calculations involving molar gas volumes:
n(gas) = moles of gas
V(gas) = volume of gas (at some specified temperature and pressure)
V_{m} = molar gas volume (at some specified temperature and pressure)

To calculate moles of gas:

n(gas)

=

V(gas) V_{m}

To calculate volume of gas:

V(gas)

=

n(gas) x V_{m}

Molar Volume of Gas Concept

Molar volume of gas, V_{m}, is defined as the volume of 1 mole of the gas at a specified temperature and pressure.
Molar volume of gas, V_{m}, therefore has the units of volume/mol, or volume ÷ mol
The SI unit for volume is the litre, L, so the molar volume of gas is volume in L ÷ mol

V_{m} in L mol^{-1}

=

volume of gas in litres 1 mole of gas

(at a specified temperature and pressure)

We can use this relationship for molar gas volume (V_{m}) to write an equation for the volume (V in litres) of any amount of gas (n in moles)

V_{m} in L mol^{-1}

=

volume of gas in litres amount of gas in moles

(at a specified temperature and pressure)

V_{m} in L mol^{-1}

=

V (L) n (mol)

(at a specified temperature and pressure)

In order to use this relationship, we will need to know V_{m}.
Below is a list of some of the ways of describing the conditions under which V_{m} is 22.71 L
V_{m} = 22.71 L mol^{-1} at 0^{o}C and 100 kPa
V_{m} = 22.71 L mol^{-1} at 273.15 K and 100 kPa
V_{m} = 22.71 L mol^{-1} at Standard Temperature and Pressure
V_{m} = 22.71 L mol^{-1} at STP

When V_{m} = 22.71 L mol^{-1}, the relationship between volume of gas, V (L), and amount of gas, n (mol), becomes:

V_{m} in L mol^{-1}

=

V (L) n (mol)

(at STP)

22.71 L mol^{-1}

=

V (L) n (mol)

(at STP)

This equation can be rearranged to find the volume of a known amount of gas by multiplying both sides of the equation by the amount of gas in moles, n (mol),

n (mol) x 22.71 (L mol^{-1})

=

V (L) x n (mol) n (mol)

(at STP)

n x 22.71

=

V (L)

(at STP)

This relationship shows us that if we increase the moles of gas, n, by adding more gas while maintaining the same temperature and pressure, the volume of gas, V, will also increase.
Likewise, if we decrease the moles of gas, n, by removing some of the gas while maintaining the same temperature and pressure, the volume of gas, V, will also decrease.

The equation above can be rearranged to find the amount of gas in moles given its volume in litres, by dividing both sides of the equation by the molar volume of gas (22.71 L mol^{-1} at STP),

n (mol) x 22.71 (L mol^{-1}) 22.71 (L mol^{-1})

=

V (L) 22.71 (L mol^{-1})

(at STP)

n (mol)

=

V 22.71

(at STP)

This relationship shows us that the only way to increase the volume of gas, V, while maintaining the same temperature and pressure, is to increase the moles of gas, n, that are present, that is, add more gas.
Likewise, the only way to decrease the volume of gas, V, while maintaining the same temperature and pressure, is to decrease the moles of gas, n, that are present, that is, remove some of the gas.

Example : Calculating Moles of Gas

1. A sample of pure helium gas occupies a volume of 6.8 L at 0^{o}C and 100 kPa.
How many moles of helium gas are persent in the sample?

What is the question asking you to do?
Calculate the moles of helium gas.
n(He_{(g)}) = moles of helium gas = ? mol

What information (data) has been given in the question?
V(He_{(g)}) = volume of helium gas = 6.8 L
conditions: STP (standard temperature and pressure, 0^{o}C and 100 kpa)
So V_{m} = molar volume of gas = 22.71 L mol^{-1} (available on data sheet)

Check for consistency in units, are all the volumes in the same units?
V(He_{(g)}) is given in L
V_{m} is given in L (mol^{-1})
Both volumes are in the same units, L, so no conversion is necessary.

What is the relationship between moles of helium gas and volume of helium gas at a specified temperature and pressure?

n(He_{(g)}) (mol)

=

V(He_{(g)}) V_{m}

Substitute the values into the equation and solve:

n(He_{(g)}) (mol)

=

6.8 22.71

(at STP)

=

0.30 mol

(at STP)

2 A sample of nitrogen gas, N_{2(g)}, has a volume of 956 mL at 273.15 K and 100 kPa.
How many moles of nitrogen gas are present in the sample?

What is the question asking you to do?
Calculate the moles of nitrogen gas.
n(N_{2(g)}) = moles of nitrogen gas = ? mol

What information has been given in the question?
V(N_{2(g)}) = volume of nitrogen gas = 956 mL
Conditions: 273.15 K and 100 kPa (standard temperature and pressure, STP)
So, V_{m} = molar volume of gas = 22.71 L mol^{-1} (available on data sheet)

Check for consistency in units, are all the volumes in the same units?
V(N_{2(g)}) is given in mL
V_{m} is given in L (mol^{-1})
Convert the gas volume, V(N_{2(g)}), from a volume in millilitres, mL, to a volume in litres, L.
V(N_{2(g)}) = 956 mL = 956 mL ÷ 1000 mL L^{-1} = 956 x 10^{-3} L = 0.956 L

What is the relationship between moles of nitrogen gas and volume of nitrogen gas at a specified temperature and pressure?

n(N_{2(g)}) (mol)

=

V(N_{2(g)}) V_{m}

Substitute the values into the equation and solve:

n(N_{2(g)}) (mol)

=

0.956 22.71

(at STP)

=

0. 0421 mol

(at STP)

Example : Calculating Volume of Gas

1. A balloon contains 0.50 moles of pure helium gas at standard temperature and pressure.
What is the volume of the balloon?

What is the question asking you to do?
Calculate the volume of helium gas in the balloon.
V(He_{(g)}) = volume of helium gas = ? L

What information (data) has been given in the question?
n(He_{(g)}) = moles of helium gas = 0.50 mol
conditions: standard temperature and pressure (STP, 0^{o}C and 100 kPa)
So V_{m} = molar volume of gas = 22.71 L mol^{-1}

Are the units consistent?
n(He_{(g)}) is in moles
V_{m} is in moles per litre
Units are therefore consistent and no conversion is required.

What is the relationship between volume of helium gas, V(He_{(g)}), and moles of helium gas, n(He_{(g)}), at a specified temperature and pressure?
V(He_{(g)}) = n(He_{(g)}) x V_{m}

Substitute in the values and solve:
V(He_{(g)}) = n(He_{(g)}) x 22.71 (at STP)
= 0.50 x 22.71
= 11.4 L

2. What is the volume occupied by 3.70 moles of N_{2} gas at STP?

What is the question asking you to do?
Calculate the volume of N_{2} gas.
V(N_{2(g)}) = volume of N_{2} gas = ? L

What information (data) has been given in the question?
n(N_{2(g)}) = moles of N_{2} gas = 3.70 mol
conditions: STP (Standard Temperature and Pressure, 0^{o}C and 100 kPa)
So, V_{m} = molar volume of gas = 22.71 L mol^{-1} (available on data sheet)

Are the units consistent?
amount of N_{2(g)} gas, n(N_{2(g)}), is given in moles
molar gas volume, V_{m} is given in moles per litre
Units are consistent so no conversion is necessary.

What is the relationship between volume of N_{2} gas, V(N_{2(g)}), and moles of N_{2} gas, n(N_{2(g)}), at a specified temperature and pressure?
V(N_{2(g)}) = n(N_{2(g)}) x V_{m}

Substitute in the values and solve:
V(N_{2(g)}) = n(N_{2(g)}) x 22.71 (at STP)
= 3.70 x 22.71
= 84.0 L

Problem Solving Using Molar Volume of Gas

The Problem: Chris the Chemist works in a laboratory in which the temperature is maintained at a constant 25^{o}C and the pressure is always 100 kPa.
Chris needs to analyse some calcium carbonate, CaCO_{3}(s), to determine whether it is pure or has been contaminated.
Chris will analyse the calcium carbonate by taking a small 0.00500 mole sample and adding hydrochloric acid, HCl(aq), to it until all the calcium carbonate has disappeared and no more carbon dioxide gas, CO_{2}(g), is produced.
As the gas is produced it will be collected by a water displacement method.
The balanced chemical equation for this reaction is known to be:

What is the question asking you to do?
Determine the volume of carbon dioxide gas if the calcium carbonate is pure. V(CO_{2(g)}) = volume of carbon dioxide gas = ?

What chemical principle will you need to apply?
Apply stoichoimetry (V_{(g)} = n_{(g)} x V_{m})

n(CaCO_{3(s)}) = amount in moles of calcium carbonate = 0.00500 mol

conditions: 25^{o}C and 100 kPa
So, V_{m} = molar volume of gas = 24.79 L mol^{-1} (from data sheet)

PAUSE!

Plan.

Step 1: Calculate moles of carbon dioxide gas, CO_{2}(g), produced

Assume the CaCO_{3} is 100% pure (no impurities).
Assume that the only source of gas being collected is the reaction given in the problem.

Use the balanced chemical equation to determine moles of CO_{2} produced
CaCO_{3}(s) + 2HCl(aq) → CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l) 1 mole CaCO_{3} produces _______ moles of CO_{2} 0.00500 mol CaCO_{3} produces _____________ moles of CO_{2}

Step 2: Calculate the volume of CO_{2}(g) Assume no loss of CO_{2}(g), that is, all the gas produced is collected. V(CO_{2(g)}) = n(CO_{2(g)}) x V_{m} V(CO_{2(g)}) = n(CO_{2(g)}) x 24.79 L mol^{-1}

GO!

Go with the Plan.

Step 1: Calculate moles of carbon dioxide gas, CO_{2}(g), produced

Assume the CaCO_{3} is 100% pure (no impurities).
Assume that the only source of gas being collected is the reaction given in the problem.

Use the balanced chemical equation to determine moles of CO_{2} produced
CaCO_{3}(s) + 2HCl(aq) → CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l) 1 mole CaCO_{3} produces 1 mole of CO_{2} 0.00500 mol CaCO_{3} produces 0.00500 moles of CO_{2}

Step 2: Calculate the volume of CO_{2}(g) Assume no loss of CO_{2}(g), that is, all the gas produced is collected. V(CO_{2(g)}) = n(CO_{2(g)}) x V_{m} V(CO_{2(g)}) = n(CO_{2(g)}) x 24.79 L mol^{-1} V(CO_{2(g)}) = 0.00500 mol x 24.79 L mol^{-1} = 0.124 L

PAUSE!

Ponder Plausability.

Have you answered the question that was asked?
Yes, we have determined the volume of carbon dioxide that will be collected.

Is your solution to the question reasonable?
At 25^{o}C and 100 kPa, the volume of 1 mole of gas would be 24.79 L (V_{m} from data sheet)
The volume of 0.00500 moles of gas (much less than 1 mole) will be much less than 24.79 L, and our calculated value of 0.124 L is much less than 24.79 L so the answer is reasonable.

Perform a "rough enough" calculation by rounding off the numbers:
that is, let V_{m} ≈ 25 L
and then manipulate the moles of gas so that it is in an easier form for quick "mental" multiplication and division,
for example, V(gas) = 5/1000 x 25 = 10/2000 x 25 = 250/2000 = 125/1000 = 0.125 L
Our "rough enough" answer of 0.125 L is very close to our carefully calculated answer of 0.124 L.

We are confident that our solution to the problem is correct.

STOP!

State the solution.

If the calcium carbonate is pure, 0.124 L of carbon dioxide gas will be collected at 25^{o}C and 100 kPa.

^{1}You can use the ideal gas equation, PV = nRT, to find the volume of 1 mole of ideal gas (molar volume of gas) at 100 kPa and other temperatures.

^{2} Prior to 1982, standard temperature and pressure were defined as 0^{o}C (273.15 K) and 1 atm (101.3 kPa), so 1 mole of gas would occupy 22.41 L

^{3}At 25^{o}C (298.15 K) and 1 atm (101.3 kPa), 1 mole of gas occupies 24.47 L.

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