2.1.3 (g) Ideal Gas Equation

Syllabus

(f) the ideal gas equation: pV = nRT

What does this mean?

This is one of the additions to the syllabus.

Enjoy!

You don't really need to know what is meant by an Ideal Gas - that's Physics.

No gas can be perfectly ideal but some approach it and they mostly conform to the equation:

PV = nRT

Where:

P = Pressure in Pascals (Pa) - which is equivalent to N/m²

V = Volume in m³ - to match the m² term above

n = the number of moles

T = Temperature in Kelvin (K)

R = The Gas Constant - which is given to you on the data sheet as 8.314 Jmol^-1K^-1

Converting Units

^Temperature

^{The Kelvin scale is based on Celsius.}

^{But has zero in a different place.}

0 K = - 273 ^oC

This is Absolute Zero - the lowest possible temperature.

So to convert Celsius to Kelvin we add 273.

And to convert Kelvin to Celsius we subtract 273.

eg.

400 ^oC in Kelvin = 400 + 273 = 673K

300 K in Celsius = 300 - 273 = 27 ^oC

Volume

Uniquely (for Chemistry) the volume must be in m³.

Blame Physics!

But Chemistry tends to use dm³ (litres)

or cm³(mililitres)

a) cm³

There are 100 cm in 1m

And 1m³ = 1m x 1m x 1m

So 1m³ = 100cm x 100cm x 100cm

Therefore, 1m³ = 1x 10⁶ cm³

So to convert cm³ to m³ we divide by a million

And to convert m³ back to cm³ we multiply by a million

a) dm³

1 dm³ = 1000 cm³

We've seen that 1m³ = 1,000,000 cm³

So 1m³ = 1000 dm³

So to convert dm³ to m³ we divide by a thousand

And to convert m³ back to dm³ we multiply by a thousand

Using the Equation

Providing that you remember to use the correct units it is just a matter of re-arranging the equation and plugging in the numbers.

a) How many moles of gas occupy 2000 cm³ at 37^oC and 1100 Pa?

2000 cm³ = 0.002 m³ 37^oC = 310K

n = PV

RT

n= 1100 x 0.002 = 2.2 = 0.00085

8.314 x 310 2577.34

b) What pressure is exerted by 2 moles of gas compressed to 4 dm³ at 25^oC?

4 dm³ = 0.004 m³ 25^oC = 298 K

P = nRT / V

P = (2 x 8.314 x 298) / 0.004

P = 1,238,786 Pa

c) What volume is occupied by 5 moles of gas at 20,000 Pa at 300K?

V = nRT / P

V = (5 x 8.314 x 300) / 20,000

V = 12471 / 20,000

V = 0.62355 m³

V = 623.55 dm³

d) What temperature is a gas at if 5 moles exerts a pressure of 10,000 Pa when it occupies 0.025m³?

T = PV / nR

T = (10,000 x 0.025) / (5 x 8.314)

T = 250 / 41.57

T = 6.01 K

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Exam style Questions

(1) A gas cylinder, of volume 5.00 × 10–3 m3, contains 325 g of argon gas.

(i) Give the ideal gas equation.

...........………………….............................................................................................................................……………………………………………………….……………………………………………………….......................................................................

(ii) Use the ideal gas equation to calculate the pressure of the argon gas in the cylinder at a temperature of 298 K.

(The gas constant R = 8.31 J K–1 mol–1)

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32 (a) A sample of ethanol vapour, C2H5OH (Mr = 46.0), was maintained at a pressure of

100 kPa and at a temperature of 366K.

(i) State the ideal gas equation.

...........………………….............................................................................................................................……………………………………………………….……………………………………………………….........

(ii) Use the ideal gas equation to calculate the volume, in cm3, that 1.36 g of ethanol vapour would occupy under these conditions.

(The gas constant R = 8.31 J K–1 mol–1)

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...........…………………......................................................................................................................................

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(5)

3.

(a) Nitromethane, CH3NO2, burns in oxygen forming three gases.

2CH₃NO_2(l)+ 1.5 O_2(g)--> 2CO_2(g) + 3H₂O_(g) + N_2(g)

(i) A 100g sample of nitromethane was completely burned in oxygen. Calculate the number of moles of nitromethane that were burned and also calculate the total volume of gaseous products at 400K and l00kPa.

Moles of nitromethane. .................................................................................................................................................................

.......................................................................................................................................................................

Total volume of gaseous products. ..............................................................................................................................................

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(ii) The combustion reaction is very exothermic and heats the products to a temperature of 1000 K. Calculate the total volume of gaseous products at this temperature and 100 kPa.

.......................................................................................................................................................................

....................................................................................................................................................................(7)

(b) A 153 kg sample of ammonia gas, NH₃, was compressed at 800 K into a cylinder of volume 3.00 m³.

(i) Calculate the pressure in the cylinder assuming that the ammonia remained

as a gas.

.......................................................................................................................................................................

(ii) Calculate the pressure in the cylinder when the temperature is raised to 1000K.

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(iii) Calculate the molarity of the solution formed by dissolving this mass of ammonia in water to make 1.0 m³ of solution.

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.....................................................................................................................................................................(7)

(Total 14 marks)

Answers

(1) (i) PV = nRT 1

(ii) Moles Ar = 325/39.9 = 8.15 1

(accept Mr = 40)

P = nRT/V = (8.15 × 8.31 × 298)/5.00 × 10–3 1

= 4.03 × 106 Pa or = 4.03 × 103 kPa 1

Range = 4.02 × 106 Pa to 4.04 × 106 Pa

(If equation incorrectly rearranged, M3 & M4 = 0 If n =325, lose M2)

(Allow M1 if gas law in (ii) if not given in (i))

2. (a) (i) pV = nRT (1)

(ii) Moles ethanol = n = 1.36/46 (=0.0296 mol) (1)

V = nrT/p = (0.0296 x 8.31 x 366)/100,000 (1)

if V = p/nRT lose M3 and M4

= 8.996 × 10–4 (m3) (1)

= 899 (900) cm3 (1)

range = 895 – 905

If final answer = 0.899 award (2 + M1); if = 0.899 dm3 or if = 912 award (3 + M1)

Note: If 1.36 or 46 or 46/1.36 used as number of moles (n) then M2 and M4 not available

Note: If pressure = 100 then, unless answer = 0.899 dm3, deduct M3 and mark consequentially

3.

(a) (i) Mass/Mr = 100/61 (1) = 1.64 (1)

allow 1.63 to 1.64

PV = nRT (1)

V = nRT/P = (3 x 1.64 x 8.31 x 400)/ 100,000 (1) if no × 3 CE

allow use of p = 100 if answer in dm3

0.164 m³(1)

allow 0.162 to 0.166

allow conseq on moles CH2NO2

(ii) V = V1 × T₂/T₁= 0.164 x 1000/400 = 0.41o m³

allow 0.4 to 0.42

or V= nRT/P = (3 x 1.64 x 8.31 x 1000)/100,000 (1) or (for 1st mark)

Thus if ans = ans to (i) × 10/4 allow 2 marks ignore units

allow conseq on moles CH3NO2 and on omission of 3

b) (i) P = nRT/V (or pV = nRT) (1)

n= mass/Mr = 153 x 10³/17 =9000 (1)

P= (9000 x 8.31 x 800)/3 (1)

= 1.99 × 107 Pa (1) (allow 1.99 to 2.00)

(ii) 1.99 × 107 × 1000/800 = 2.49 × 107 Pa (1)

(allow conseq)

(iii) 9000/1000 = 9.0 mol dm–3 (or M) (1) (1)

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