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Moles and gas volumes

Moles and gas volumes. At the end of this section you should be able to calculate the amount of substance in moles, using gas volume. Avogadro’s hypothesis. Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. At room temperature and pressure (RTP):.

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Moles and gas volumes

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  1. Moles and gas volumes At the end of this section you should be able to calculate the amount of substance in moles, using gas volume.

  2. Avogadro’s hypothesis Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

  3. At room temperature and pressure (RTP): • One mole of gas molecules occupies approximately 24.0dm3 (24000cm3) • The volume per mole of gas molecules is 24.0 dm3 mol-1

  4. To work out amount, in moles, in a solution, use the equations:n = V(in dm3) 24.0n = V(in dm3) 24000Hence, V= n x 24000cm3

  5. Worked examples What amount, in mol, of gas molecules are in the following gas volumes at RTP? (i) 36 dm3 (ii) 250cm3

  6. n = 36 24.0 = 1.5mol(ii)n = 250 24000 = 0.01mol

  7. What is the volume of the following at RTP?(i) 2 mol SO2 (ii) 0.15 mol H2

  8. V = 2 x 24000 cm3= 48000 cm3(ii) V = 0.15 x 24000 cm3= 3600cm3

  9. What is the mass of the following at RTP?(i) 0.6 dm3 N2(ii) 1950 cm3 C2H4NB: mass = mole x molecular mass

  10. Mole = 0.6 24.0= 0.025 molfrom, mass = mole x MM(N2) = 14 x 2 = 28mass = 0.025 x 28 = 0.7g

  11. (ii) Mole = 1950 24000 = 0.0813molmass = mole x M M(C2H4) = (12 x 2) + 1x4 = 28hence, mass = 0.0813 x 28 = 2.275g

  12. What is the volume of the following at RTP?(i) 1.282 g SO2 (ii) 2.5 g CO2NB: For this question, you need to calculate the moles and then follow it up with volume

  13. Mole = mass/molecular massM(SO2) = 32 + 16x2 = 64mole = 1.282 / 64 = 0.02 molfrom, V = n x 24000 cm3 = 0.02 x 24000 = 480.75cm3

  14. Now try question 1 – 3Page 15Question 8 (a) and (b)Page 35

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