Exercise 1.15 - One mole of an ideal gas at standard temperature - Exercise Questions

part 2 - Exercise 1.15 - Exercise Questions - Chapter 1 - Units and Measurement - Physics Class 11

part 3 - Exercise 1.15 - Exercise Questions - Chapter 1 - Units and Measurement - Physics Class 11

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Exercise 1.15 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 ร…). Why is this ratio so large ? Given: ๐‘‰_(gas)=22.4 L=22.4ร—10^(โˆ’3) m^(3) Take the stated molecular size as an effective radius: ๐‘Ÿโ‰ˆ1โ„ซ=10^(โˆ’10) m Step 1: Calculate the volume of one molecule Approximating the molecule as a sphere: Volume=v =(4)/(3)๐œ‹๐‘Ÿ^(3) =(4)/(3)๐œ‹(10^(โˆ’10))^(3) โ‰ˆ๐Ÿ’.๐Ÿ๐Ÿ— ร— ๐Ÿ๐ŸŽ^(โˆ’๐Ÿ‘๐ŸŽ) ๐’Ž^(๐Ÿ‘) [; ] Step 2: Calculate the summed volume of one mole of molecules One mole contains approximately 6.022 ร— 10^(23) molecules: ๐‘‰_(molecules )=๐‘_(๐ด)๐‘ฃ =(6.022ร—10^(23))(4.19ร—10^(โˆ’30)) โ‰ˆ๐Ÿ.๐Ÿ“๐Ÿ ร— ๐Ÿ๐ŸŽ^(โˆ’๐Ÿ”) ๐’Ž^(๐Ÿ‘) Step 3: Calculate the ratio ๐‘…=(๐‘ฃ_(๐‘”๐‘Ž๐‘ ))/(๐‘‰_(๐‘š๐‘œ๐‘™๐‘’๐‘๐‘ข๐‘™๐‘’๐‘ )) [; =(22.4ร—10^(โˆ’3))/(2.52ร—10^(โˆ’6))] โ‰ˆ๐Ÿ–.๐Ÿ—ร—๐Ÿ๐ŸŽ^(๐Ÿ‘) This is approximately 10^(4) as an order-of-magnitude estimate: ๐‘นโ‰ƒ๐Ÿ๐ŸŽ^(๐Ÿ’) Why is the ratio so large? The gas volume includes both the molecules and the spaces between them. In a gas, the intermolecular separation is much larger than the size of a molecule, so most of the volume is space between molecules. =(4)/(3)๐œ‹(10^(โˆ’10))^(3) โ‰ˆ๐Ÿ’.๐Ÿ๐Ÿ— ร— ๐Ÿ๐ŸŽ^(โˆ’๐Ÿ‘๐ŸŽ) ๐’Ž^(๐Ÿ‘) [; ] Step 2: Calculate the summed volume of one mole of molecules One mole contains approximately 6.022 ร— 10^(23) molecules: ๐‘‰_(molecules )=๐‘_(๐ด)๐‘ฃ =(6.022ร—10^(23))(4.19ร—10^(โˆ’30)) โ‰ˆ๐Ÿ.๐Ÿ“๐Ÿ ร— ๐Ÿ๐ŸŽ^(โˆ’๐Ÿ”) ๐’Ž^(๐Ÿ‘) Step 3: Calculate the ratio ๐‘…=(๐‘ฃ_(๐‘”๐‘Ž๐‘ ))/(๐‘‰_(๐‘š๐‘œ๐‘™๐‘’๐‘๐‘ข๐‘™๐‘’๐‘ )) [; =(22.4ร—10^(โˆ’3))/(2.52ร—10^(โˆ’6))] โ‰ˆ๐Ÿ–.๐Ÿ—ร—๐Ÿ๐ŸŽ^(๐Ÿ‘) This is approximately 10^(4) as an order-of-magnitude estimate: ๐‘นโ‰ƒ๐Ÿ๐ŸŽ^(๐Ÿ’) Why is the ratio so large? The gas volume includes both the molecules and the spaces between them. In a gas, the intermolecular separation is much larger than the size of a molecule, so most of the volume is space between molecules. This is approximately 10^(4) as an order-of-magnitude estimate: ๐‘นโ‰ƒ๐Ÿ๐ŸŽ^(๐Ÿ’) Why is the ratio so large? The gas volume includes both the molecules and the spaces between them. In a gas, the intermolecular separation is much larger than the size of a molecule, so most of the volume is space between molecules.

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