A Level Chemistry

The Ideal Gas Equation – A Level Chemistry Revision

Welcome back to Beyond’s Science Blog! This A Level Chemistry entry explores The Ideal Gas Equation.

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When a substance is in a gaseous state, its particles move around constantly. As they do so, they collide with each other and with their surroundings, creating pressure. There are several variables that can affect the amount of pressure that a gas exerts, including the volume of space that the gas occupies, the temperature of the gas and the number of gaseous particles. The ideal gas equation shows the relationship between these variables:

PV = nRT

P is the pressure pascals (Pa)
V is the volume of gas (m3)
n is the number of moles of gas
R is the gas constant
T is the temperature (K)

You will always be given the value of the gas constant:

The gas constant, R = 8.31 JK-1mol-1

You could be asked to calculate the value of any of the variables within the ideal gas equation so you must be able to rearrange the equation to make P, V, n or T the subject.

In a laboratory setting it is often convenient to measure and record values in units that are different to those stated above. This means that it is usually necessary to convert given values into SI units before carrying out your ideal gas equation calculations.

VariablePressure (P)Volume (V) Temperature (T)
SI Unitpascal, Pacubic metre, m3kelvin, K
Commonly Used Unitkilopascal, kPacubic centimetre, cm3
or
cubic decimetre, dm3		degrees Celsius, °C
ConversionPa = kPa x 1000m3 = \frac{\text{cm}^3}{1000000}
or
m3 = \frac{\text{dm}^3}{1000}		K = °C + 273
Example100kPa x 1000 = 100,000Pa\frac{150\text{cm}^3}{1000000} = 0.00015m3
\frac{0.5\text{dm}^3}{1000} = 0.0005m3		25°C + 273 = 298K

Example Question 1

A stoppered flask has a volume of 1.00dm3. The flask contains chlorine gas. The pressure inside the flask is 100kPa and the temperature is 20.0°C.

Calculate the number of moles of chlorine gas in the flask.

Give your answer to an appropriate number of significant figures.

P = 100kPa = 100 000Pa

V = 1.00dm3 = 0.001m3

R = 8.31JK-1mol-1

T = 20.0°C + 273 = 293K

\text{n} = \frac{\text{PV}}{\text{RT}}=\frac{100000\times 0.001}{8.31 \times 293}=0.410706 = 0.411 \text{ moles (3s.f.)}

Example Question 2

A 250cm3 stoppered flask contains 0.0100 moles of a gas.

The temperature inside the flask is 19.0°C.

Calculate the pressure inside the flask.

Give your answer to an appropriate number of significant figures.

V = 250cm3 = 0.000250m3

n = 0.0100

R = 8.31 JK-1mol-1

T = 19.0°C + 273 = 292K

\text{P}=\frac{\text{nRT}}{\text{V}}=\frac{0.0100 \times 8.31 \times 292}{0.000250}=97060\text{Pa}=97100\text{Pa (3s.f.)}

Example Question 3

A 3.50g sample of nitrogen gas, N2, has a pressure of 101kPa and a volume of 3.00dm3.

Calculate the temperature of the gas.

Give your answer to an appropriate number of significant figures.

P = 101kPa = 101000Pa

V = 3.00dm3 = 0.003m3

n = \frac{\text{mass}}{\text{M}_{\text{r}}}=\frac{3.50}{28.0} = 0.125 moles

R = 8.31 JK-1mol-1

\text{T}=\frac{\text{PV}}{\text{nR}}=\frac{101000 \times 0.003}{0.125\times 8.31}=291.6967509 = 292\text{K (3s.f.)}

Example Question 4

16.0g of liquid methanol, CH3OH, was heated to 67.0°C until all of the methanol had evaporated. The pressure remained constant at 98kPa throughout this process.

Calculate the volume, in cm3, occupied by the methanol vapour.

Give your answer to an appropriate number of significant figures.

P = 98kPa = 98 000Pa

n = \frac{\text{mass}}{\text{M}_{\text{r}}}=\frac{16.0}{32.0} = 0.500 moles

R = 8.31 JK-1mol-1

T = 67.0°C + 273 = 340K

\text{V} = \frac{\text{nRT}}{\text{P}}=\frac{0.500\times 8.31 \times 340}{98000} = 0.1441530612\text{m}^3

0.01441530612m3 × 1 000 000 =14 415.30612cm3 = 14 000cm3 (2s.f.)

Note: The answer is limited to two significant figures because the pressure given (98kPa) was to two significant figures and was the least precise measurement in the calculation.

Practice Questions

1. A stoppered flask, containing oxygen, has a volume of 500.0cm3. The pressure inside the flask is 105kPa and the temperature is 20.0°C.

Calculate the number of moles of oxygen in the flask.

Give your answer to an appropriate number of significant figures.

2. Helium gas is used to inflate a balloon. Inside the balloon, 0.0625 moles of helium gas exert a pressure of 100 000.00Pa at a temperature of 25.0°C.

Calculate the volume, in cm3, of the inflated balloon.

Give your answer to an appropriate number of significant figures.

3. 424mg of gas Z is stored in a sealed 800cm3 flask. The pressure inside the flask is 77kPa and the temperature is 297K.

Calculate the relative molecular mass (Mr) of gas Z.

Give your answer to an appropriate number of significant figures.

Answers

1. A stoppered flask, containing oxygen, has a volume of 500.0cm3. The pressure inside the flask is 105kPa and the temperature is 20.0°C.

Calculate the number of moles of oxygen in the flask.

Give your answer to an appropriate number of significant figures.

P = 105 x 1000 = 105000Pa

V = \boldsymbol{\frac{500}{1000000}} = 0.0005m3

R = 8.31 JK-1mol-1

T = 20 + 273 = 293K

\boldsymbol{\textbf{n}=\frac{\textbf{PV}}{\textbf{RT}}=\frac{105000\times 0.0005}{8.31 \times 293}=0.02156208031 = 0.022 \textbf{ (2s.f.)}}

2. Helium gas is used to inflate a balloon. Inside the balloon, 0.0625 moles of helium gas exert a pressure of 100 000.00Pa at a temperature of 25.0°C.

Calculate the volume, in cm3, of the inflated balloon.

Give your answer to an appropriate number of significant figures.

n = 0.0625mol

R = 8.31 JK-1mol-1

T = 25 + 273 = 298K

P = 100 000Pa

\boldsymbol{\textbf{V}=\frac{\textbf{nRT}}{\textbf{P}}=\frac{0.0625\times 8.31 \times 298}{100000}=0.0015477375\textbf{m}^3}

\boldsymbol{0.0015477375\textbf{m}^3\times 1000000=1547.7375\textbf{cm}^3}

3. 424mg of gas Z is stored in a sealed 800cm3 flask. The pressure inside the flask is 77kPa and the temperature is 297K.

Calculate the relative molecular mass (Mr) of gas Z.

Give your answer to an appropriate number of significant figures.

\boldsymbol{\textbf{n}=\frac{\textbf{PV}}{\textbf{RT}}=\frac{77000\times 0.0008}{8.31 \times 297}=0.02495877345\textbf{mol}}

\boldsymbol{\textbf{M}_{\textbf{r}} = \frac{0.424}{0.02495877345} = 16.988... = 17.0 \textbf{ (3s.f.)}}

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