 # Molar Volume of Gas Calculations (Vm) Chemistry Tutorial

## Key Concepts

• 1 mole of an ideal gas(1) occupies a specific volume at a particular temperature and pressure.
• This is known as the molar volume of gas and given the symbol Vm
• The units most commonly used for molar volume of gas, Vm, are litres per mole, L mol-1
• Examples of molar volume for ideal gases (Vm) are given in the table below:

Temperature Pressure Molar Volume of Gas
°C (K) kPa (atm) (Vm) / L mol-1
0°C (273.15K) 100kPa (0.987 atm) 22.71
25°C (298.15 K) 100 kPa (0.987 atm) 24.79

0°C (273.15K) and 100 kPa (0.987 atm) is known as Standard Temperature and Pressure and is often abbreviated to STP (2)

25°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 ideal gas

V(gas) = volume of ideal gas (at some specified temperature and pressure)

Vm = molar volume of ideal gas (at some specified temperature and pressure)

 To calculate moles of gas: n(gas) = V(gas)   Vm To calculate volume of gas: V(gas) = n(gas) × Vm

No ads = no money for us = no free stuff for you!

## Molar Volume of Gas (Vm) Concept

Molar volume of ideal gas, Vm, is defined as the volume of 1 mole of the ideal gas at a specified temperature and pressure.
Molar volume of gas, Vm, 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

 Vm 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 (Vm) to write an equation for the volume (V in litres) of any amount of gas (n in moles)

 Vm in L mol-1 = volume of gas in litresamount of gas in moles (at a specified temperature and pressure) Vm 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 Vm.

Below is a list of some of the ways of describing the conditions under which Vm is 22.71 L

Vm = 22.71 L mol-1 at 0°C and 100 kPa
Vm = 22.71 L mol-1 at 273.15 K and 100 kPa
Vm = 22.71 L mol-1 at Standard Temperature and Pressure
Vm = 22.71 L mol-1 at STP

When Vm = 22.71 L mol-1, the relationship between volume of gas, V (L), and amount of gas, n (mol), becomes:

 Vm 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.

Do you know this?

Play the game now!

## Worked Examples of Calculating Moles of Gas Using Molar Volume of Gas

Worked Example 1. A sample of pure helium gas occupies a volume of 6.8 L at 0°C and 100 kPa.
How many moles of helium gas are persent in the sample?

1. What is the question asking you to do?

Calculate the moles of helium gas.

n(He(g)) = moles of helium gas = ? mol

2. 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°C and 100 kpa)

So Vm = molar volume of gas = 22.71 L mol-1 (available on data sheet)

3. Check for consistency in units, are all the volumes in the same units?

V(He(g)) is given in L

Vm is given in L (mol-1)

Both volumes are in the same units, L, so no conversion is necessary.

4. 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))     Vm

5. Substitute the values into the equation and solve moles of helium gas:

 n(He(g)) (mol) = 6.8     22.71 (at STP) = 0.30 mol (at STP)

Worked Example 2: A sample of nitrogen gas, N2(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?

1. What is the question asking you to do?

Calculate the moles of nitrogen gas.

n(N2(g)) = moles of nitrogen gas = ? mol

2. What information has been given in the question?

V(N2(g)) = volume of nitrogen gas = 956 mL

Conditions: 273.15 K and 100 kPa (standard temperature and pressure, STP)

So, Vm = molar volume of gas = 22.71 L mol-1 (available on data sheet)

3. Check for consistency in units, are all the volumes in the same units?

V(N2(g)) is given in mL

Vm is given in L (mol-1)

Convert the gas volume, V(N2(g)), from a volume in millilitres, mL, to a volume in litres, L.

V(N2(g)) = 956 mL = 956 mL ÷ 1000 mL L-1 = 956 × 10-3 L = 0.956 L

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

 n(N2(g)) (mol) = V(N2(g))     Vm

5. Substitute the values into the equation and solve for moles of nitrogen gas:

 n(N2(g)) (mol) = 0.956     22.71 (at STP) = 0. 0421 mol (at STP)

Do you understand this?

Take the test now!

## Worked Examples of Calculating Volume of Gas Using Molar Volume of Gas

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

1. 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

2. 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°C and 100 kPa)

So Vm = molar volume of gas = 22.71 L mol-1

3. Are the units consistent?

n(He(g)) is in moles

Vm is in moles per litre

Units are therefore consistent and no conversion is required.

4. 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)) × Vm

5. Substitute in the values and solve for volume of helium gas:

V(He(g)) = n(He(g)) × 22.71         (at STP)

= 0.50 × 22.71

= 11.4 L

Worked Example 2. What is the volume occupied by 3.70 moles of N2 gas at STP?

1. What is the question asking you to do?

Calculate the volume of N2 gas.

V(N2(g)) = volume of N2 gas = ? L

2. What information (data) has been given in the question?

n(N2(g)) = moles of N2 gas = 3.70 mol

conditions: STP (Standard Temperature and Pressure, 0°C and 100 kPa)

So, Vm = molar volume of gas = 22.71 L mol-1 (available on data sheet)

3. Are the units consistent?

amount of N2(g) gas, n(N2(g)), is given in moles

molar gas volume, Vm is given in moles per litre

Units are consistent so no conversion is necessary.

4. What is the relationship between volume of N2 gas, V(N2(g)), and moles of N2 gas, n(N2(g)), at a specified temperature and pressure?

V(N2(g)) = n(N2(g)) × Vm

5. Substitute in the values and solve for volume of nitrogen gas:

V(N2(g)) = n(N2(g)) × 22.71         (at STP)

= 3.70 × 22.71

= 84.0 L

Can you apply this?

Join AUS-e-TUTE!

Take the exam now!

## 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, CaCO3(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, CO2(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:

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(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(CO2(g)) = volume of carbon dioxide gas = ? L PAUSE! Plan. What chemical principle will you need to apply? Apply stoichoimetry (V(g) = n(g) × Vm) What information (data) have you been given? Balanced chemical equation : CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) All the CaCO3 reacts (reaction goes to completion). formula for calcium carbonate: CaCO3(s) formula for hydrochloric acid: HCl(aq) formula for carbon dioxide: CO2(g) n(CaCO3(s)) = amount in moles of calcium carbonate = 0.00500 mol conditions: 25°C and 100 kPa So, Vm = molar volume of gas = 24.79 L mol-1 (from data sheet) What steps do you need to take to solve the problem? Step 1: Calculate moles of carbon dioxide gas, CO2(g), produced Assume the CaCO3 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 CO2 produced CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) 1 mole CaCO3 produces _______ moles of CO2 0.00500 mol CaCO3 produces _____________ moles of CO2 Step 2: Calculate the volume of CO2(g) Assume no loss of CO2(g), that is, all the gas produced is collected. V(CO2(g)) = n(CO2(g)) × Vm V(CO2(g)) = n(CO2(g)) × 24.79 L mol-1 GO! Go with the Plan. Step 1: Calculate moles of carbon dioxide gas, CO2(g), produced Assume the CaCO3 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 CO2 produced CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) 1 mole CaCO3 produces 1 mole of CO2 0.00500 mol CaCO3 produces 0.00500 moles of CO2 Step 2: Calculate the volume of CO2(g) Assume no loss of CO2(g), that is, all the gas produced is collected. V(CO2(g)) = n(CO2(g)) × Vm V(CO2(g)) = n(CO2(g)) × 24.79 L mol-1 V(CO2(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 (Vm 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 Vm ≈ 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(CO2(g)) = 0.124 L at 25°C and 100 kPa.

Footnotes:

(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°C (273.15 K) and 1 atm (101.3 kPa), so 1 mole of gas would occupy a volume of 22.41 L

(3) At 25°C (298.15 K) and 1 atm (101.3 kPa), 1 mole of gas occupies a volume of 24.47 L.

(4) Litre is a metric unit but not an SI base unit. The SI unit for volume would be cubic metres (m3).
Originally 1 L was defined as the volume occupied by 1 kg of water at 3.98°C.
In 1964, 1 L was redefined as one cubic decimetre (dm3), so now it is a derived SI unit ( 1 L = 1 dm3 ).
It is useful to note that this means that 1 mL = 1 cm3