{"id":237452,"date":"2025-06-17T08:53:13","date_gmt":"2025-06-17T08:53:13","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=237452"},"modified":"2025-06-17T08:53:16","modified_gmt":"2025-06-17T08:53:16","slug":"density-of-hydrogen-gas-at-stp-calculate-the-molar-volume-of-this-gas-at-stp-to-be-0-08988-g-l","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/17\/density-of-hydrogen-gas-at-stp-calculate-the-molar-volume-of-this-gas-at-stp-to-be-0-08988-g-l\/","title":{"rendered":"Density of Hydrogen Gas at STP Calculate the molar volume of this gas at STP to be 0.08988 g\/L."},"content":{"rendered":"\n

Density of Hydrogen Gas at STP Calculate the molar volume of this gas at STP to be 0.08988 g\/L. Calculate the percent deviation of this value from the molar volume of an ideal gas, which is 22.414 L. Assuming Charles’ Law is valid, calculate the molar volume of this gas at 0.00 atm. Assuming the molecules of the gas occupy the same equal fractions of the total gas volume, calculate the volume occupied by each molecule. Note that this occupied volume includes both the volume physically occupied by hydrogen and the fraction of the “dead” empty volume associated with it. (The following calculations are optional) Assuming the hydrogen molecule can be approximated by a sphere with a diameter of 130 pm, calculate the physical volume of one hydrogen molecule. Determine the percentage of the total occupied volume per molecule at STP that is actually occupied by the hydrogen molecule. Determine the average center-to-center distance between nearest neighbor molecules in a sample of gas at STP. To do this calculation, it is useful to imagine the sample of gas as a sample of hydrogen cubes, each divided into smaller cubes of equal volume, with one molecule at the center of each cube.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Analyzed<\/p>\n\n\n\n

pythonAlways show detailsCopyfrom math import pi\n\n# Given values\ndensity_H2_STP = 0.08988  # g\/L\nmolar_mass_H2 = 2.016  # g\/mol (approximate)\nideal_molar_volume = 22.414  # L\/mol\ndiameter_H2_pm = 130  # pm\ndiameter_H2_m = diameter_H2_pm * 1e-12  # convert to meters\n\n# Calculate molar volume at STP from density\nmolar_volume_H2 = molar_mass_H2 \/ density_H2_STP  # L\/mol\n\n# Percent deviation from ideal gas molar volume\npercent_deviation = ((ideal_molar_volume - molar_volume_H2) \/ ideal_molar_volume) * 100\n\n# Molar volume at 0.00 atm using Charles' Law (V \u221d T; as P -> 0, interactions disappear, so assume ideal gas behavior)\n# At 0 atm, gas would behave ideally, so the molar volume = ideal molar volume\nmolar_volume_0atm = ideal_molar_volume\n\n# Avogadro's number\nN_A = 6.022e23\n\n# Volume per molecule (includes occupied + dead volume)\nvolume_per_molecule = molar_volume_H2 \/ N_A  # in L\nvolume_per_molecule_m3 = volume_per_molecule * 1e-3  # convert to m^3\n\n# Volume of one H2 molecule (sphere)\nradius_H2_m = diameter_H2_m \/ 2\nvolume_H2_physical = (4 \/ 3) * pi * radius_H2_m**3  # in m^3\n\n# Percentage of total volume actually occupied by molecule\npercent_volume_occupied = (volume_H2_physical \/ volume_per_molecule_m3) * 100\n\n# Average center-to-center distance: assume cubic packing\navg_distance = volume_per_molecule_m3**(1\/3)  # in m\navg_distance_nm = avg_distance * 1e9  # in nm\n\nmolar_volume_H2, percent_deviation, molar_volume_0atm, volume_per_molecule_m3, volume_H2_physical, percent_volume_occupied, avg_distance_nm\n<\/code><\/pre>\n\n\n\n
Result<\/p>\n\n\n\n
(22.429906542056074,\n -0.07096699409329983,\n 22.414,\n 3.724660667893735e-26,\n 1.1503465099894621e-30,\n 0.0030884598962406247,\n 3.3396086014798416)<\/pre>\n\n\n\n
Here are the results and explanation based on your multi-part question:<\/p>\n\n\n\n
1. Molar volume at STP<\/strong><\/h3>\n\n\n\n
Using the given density of hydrogen gas at STP (0.08988 g\/L) and its molar mass (2.016 g\/mol):Molar volume=2.016 g\/mol0.08988 g\/L\u224822.43 L\/mol\\text{Molar volume} = \\frac{2.016\\ \\text{g\/mol}}{0.08988\\ \\text{g\/L}} \\approx 22.43\\ \\text{L\/mol}Molar volume=0.08988 g\/L2.016 g\/mol\u200b\u224822.43 L\/mol<\/p>\n\n\n\n
2. Percent deviation from ideal gas molar volume<\/strong><\/h3>\n\n\n\n
The ideal molar volume at STP is 22.414 L\/mol. The percent deviation is calculated as:Percent deviation=22.414\u221222.4322.414\u00d7100\u2248\u22120.071%\\text{Percent deviation} = \\frac{22.414 – 22.43}{22.414} \\times 100 \\approx -0.071\\%Percent deviation=22.41422.414\u221222.43\u200b\u00d7100\u2248\u22120.071%<\/p>\n\n\n\n
This small negative deviation suggests the measured molar volume is very close to ideal behavior.<\/p>\n\n\n\n
3. Molar volume at 0.00 atm<\/strong><\/h3>\n\n\n\n
Assuming Charles\u2019 Law and extrapolating to 0.00 atm, gas behavior approaches ideal. Thus:Molar volume at 0.00 atm=22.414 L\/mol\\text{Molar volume at 0.00 atm} = 22.414\\ \\text{L\/mol}Molar volume at 0.00 atm=22.414 L\/mol<\/p>\n\n\n\n
4. Volume occupied by each molecule (including empty space)<\/strong><\/h3>\n\n\n\n
Using Avogadro\u2019s number 6.022\u00d710236.022 \\times 10^{23}6.022\u00d71023, the volume per molecule is:22.43 L\/mol6.022\u00d71023\u22483.72\u00d710\u221226 m3\\frac{22.43\\ \\text{L\/mol}}{6.022 \\times 10^{23}} \\approx 3.72 \\times 10^{-26}\\ \\text{m}^36.022\u00d7102322.43 L\/mol\u200b\u22483.72\u00d710\u221226 m3<\/p>\n\n\n\n
5. Physical volume of one hydrogen molecule<\/strong><\/h3>\n\n\n\n
Assuming the hydrogen molecule is spherical with diameter 130 pm (radius = 65 pm = 6.5\u00d710\u2212116.5 \\times 10^{-11}6.5\u00d710\u221211 m):V=43\u03c0r3\u22481.15\u00d710\u221230 m3V = \\frac{4}{3} \\pi r^3 \\approx 1.15 \\times 10^{-30}\\ \\text{m}^3V=34\u200b\u03c0r3\u22481.15\u00d710\u221230 m3<\/p>\n\n\n\n
6. Percentage of space truly occupied by the molecule<\/strong><\/h3>\n\n\n\n
1.15\u00d710\u2212303.72\u00d710\u221226\u00d7100\u22480.0031%\\frac{1.15 \\times 10^{-30}}{3.72 \\times 10^{-26}} \\times 100 \\approx 0.0031\\%3.72\u00d710\u2212261.15\u00d710\u221230\u200b\u00d7100\u22480.0031%<\/p>\n\n\n\n
This means most of the gas volume is “empty” space.<\/p>\n\n\n\n
7. Average center-to-center molecular distance<\/strong><\/h3>\n\n\n\n
Assuming a simple cube model, the average spacing is the cube root of the volume per molecule:(3.72\u00d710\u221226)1\/3\u22483.34 nm(3.72 \\times 10^{-26})^{1\/3} \\approx 3.34\\ \\text{nm}(3.72\u00d710\u221226)1\/3\u22483.34 nm<\/p>\n\n\n\n
Summary of Key Results:<\/h3>\n\n\n\n
    \n
  • Molar volume: 22.43 L\/mol<\/strong><\/li>\n\n\n\n
  • Percent deviation: \u22120.071%<\/strong><\/li>\n\n\n\n
  • Volume per molecule: 3.72 \u00d7 10\u207b\u00b2\u2076 m\u00b3<\/strong><\/li>\n\n\n\n
  • Physical volume of H\u2082 molecule: 1.15 \u00d7 10\u207b\u00b3\u2070 m\u00b3<\/strong><\/li>\n\n\n\n
  • Occupied volume percentage: 0.0031%<\/strong><\/li>\n\n\n\n
  • Average molecular spacing: 3.34 nm<\/strong><\/li>\n<\/ul>\n\n\n\n
    These values show how sparsely gas molecules are distributed in space even under standard conditions.<\/p>\n\n\n\n
    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
    Density of Hydrogen Gas at STP Calculate the molar volume of this gas at STP to be 0.08988 g\/L. Calculate the percent deviation of this value from the molar volume of an ideal gas, which is 22.414 L. Assuming Charles’ Law is valid, calculate the molar volume of this gas at 0.00 atm. Assuming the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-237452","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237452","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=237452"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237452\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=237452"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=237452"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=237452"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}