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Molar Volume of Gas (V_{m}) Concept
Molar volume of ideal gas, V_{m}, is defined as the volume of 1 mole of the ideal gas at a specified temperature and pressure.
Molar volume of gas, V_{m}, is therefore the gas volume per mole of gas, so the units of molar volume of gas are:
unit volume/mol
or
unit volume ÷ mol
The metric unit^{(4)} 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 volume of ideal gas (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°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 mathematical 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) × 22.71 (L mol^{1}) 
= 
V (L) × n (mol)
n (mol) 
(at STP) 
 n × 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 mathematical 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) × 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.
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°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:
CaCO_{3}(s) + 2HCl(aq) → CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l)
If the sample is pure, what volume of carbon dioxide gas will be collected?
Solving the Problem using the StoPGoPS model for problem solving:
STOP!
 State the question. 
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 = ? L

PAUSE!
 Plan. 
What chemical principle will you need to apply?
Apply stoichoimetry (V_{(g)} = n_{(g)} × V_{m})
What information (data) have you been given?
What steps do you need to take to solve the problem?
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)}) × V_{m}
V(CO_{2(g)}) = n(CO_{2(g)}) × 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)}) × V_{m}
V(CO_{2(g)}) = n(CO_{2(g)}) × 24.79 L mol^{1}
V(CO_{2(g)}) = 0.00500 mol × 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°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,
n(gas) = 0.005 = 5/1000 mol
so, V(gas) = 5/1000 × 25 = (5 × 25)/1000 = 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 reasonably confident that our solution to the problem is correct.

STOP!
 State the solution. 
What volume of carbon dioxide will be collected if the sample is pure calcium carbonate?
V(CO_{2(g)}) = 0.124 L at 25°C and 100 kPa.
