{"id":3721,"date":"2018-08-14T19:32:18","date_gmt":"2018-08-15T02:32:18","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=3721"},"modified":"2018-08-14T19:32:18","modified_gmt":"2018-08-15T02:32:18","slug":"the-ideal-gas-law","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/the-ideal-gas-law.htm","title":{"rendered":"The ideal gas law","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

The gas laws of Boyle, Charles law and Gay-Lussac do not apply to all gas conditions, so our analysis becomes more difficult. Therefore, presented the ideal gas model. Ideal gas does not exist in everyday life; the ideal gas is the just perfect form to facilitate analysis. The existence of this ideal gas concept also really helps us in reviewing the relationship between the three laws of gas.<\/span><\/span><\/span><\/p>\n

The relationship among temperature, volume, and gas pressure<\/b><\/span><\/span><\/span><\/p>\n

By referring to the three gas laws above, we can derive a more general relationship between temperature, volume, and gas pressure.<\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

\"The<\/p>\n

P<\/span>1<\/span><\/sub> = initial pressure (Pa or N\/m<\/span>2<\/span><\/sup>), P<\/span>2<\/span><\/sub> = final pressure (Pa or N\/m<\/span>2<\/span><\/sup>), V<\/span>1<\/span><\/sub> = initial volume (m<\/span>3<\/span><\/sup>), V<\/span>2 <\/span><\/sub>= final volume (m<\/span>3<\/span><\/sup>), T<\/span>1 <\/span><\/sub>= initial temperature (K), T<\/span>2 <\/span><\/sub>= end temperature (K)<\/span><\/span><\/span><\/p>\n

(Pa = pascal, N = Newton, m<\/span>2 <\/span><\/sup>= square meter, m<\/span>3<\/span><\/sup> = cubic meter, K = Kelvin)<\/span><\/span><\/span><\/p>\n

The relationship between gas mass (m) and volume (V)<\/b><\/span><\/span><\/span><\/p>\n

The more air that is put into the rubber balloon, the more bloated the balloon is. In other words, the larger the gas mass, the greater the volume of the balloon. We can say that the mass of gas (m) is directly proportional to the volume of gas (V). <\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

Number of moles (n)<\/b><\/span><\/span><\/span><\/p>\n

1 mol = mass of a substance equal to the molecular mass of the substance. Mass and molecular oxygen mass are different.<\/span><\/span><\/span><\/p>\n

Example 1, the oxygen molecular mass (O<\/span>2<\/span><\/sub>) = 16 u + 16 u = 32 u (each oxygen molecule contains 2 Oxygen atoms, where each Oxygen atom has mass 16 u). Thus, 1 mole O<\/span>2<\/span><\/sub> has mass 32 grams. Or the molecular mass of O<\/span>2<\/span><\/sub> = 32 grams \/ mol = 32 kg\/kmol.<\/span><\/span><\/span><\/p>\n

Example 2, the molecular mass of carbon monochide gas (CO) = 12 u + 16 u = 28 u (each molecule of carbon monooksida contains 1 carbon atom (C) and 1 oxygen atom (O) Mass 1 carbon atom = 12 u and mass 1 atom Oxygen = 16 u. 12 u + 16 u = 28 u). Thus, 1 mole of CO has a mass of 28 grams. Or the molecular mass of CO = 28 grams\/mol = 28 kg\/kmol.<\/span><\/p>\n

Example 3, the molecular mass of the carbon dioxide gas (CO2) = [12 u + (2 x 16 u)] = [12 u + 32 u] = 44 u (each carbon dioxide molecule contains 1 carbon atom (C) and 2 oxygen atoms (O). The mass of 1 Carbon atom = 12 u and the mass of 1 oxygen atom = 16 u). Thus, 1 mole of CO<\/span>2 <\/span><\/sub>has mass 44 grams. Or the molecular mass of CO<\/span>2<\/span><\/sub> = 44 grams\/mol = 44 kg\/kmol.<\/span><\/span><\/span><\/p>\n

The number of moles (n) of a substance = the mass ratio of the substance to its molecular mass. Mathematically:<\/span><\/p>\n

\"The
\nQuestion 1:<\/span><\/span><\/span><\/p>\n

Calculate the number of moles at 64 g of O<\/span>2<\/span><\/sub><\/span><\/span><\/p>\n

O<\/span>2<\/span><\/sub> mass = 64 grams<\/span><\/span><\/span><\/p>\n

The molecular mass of O<\/span>2 <\/span><\/sub>= 32 grams\/mol<\/span><\/span><\/span><\/p>\n

\"The
\nQuestion 2 :<\/span><\/span><\/span><\/p>\n

Calculate the number of moles at 280 grams of CO<\/span><\/span><\/span><\/p>\n

CO mass = 280 grams<\/span><\/span><\/span><\/p>\n

The molecular mass of CO = 28 grams\/mol<\/span><\/span><\/span><\/p>\n

\"The
\nQuestion 3 :<\/span><\/span><\/span><\/p>\n

Calculate the number of moles at 176 grams of CO<\/span>2<\/span><\/sub><\/span><\/span><\/p>\n

CO<\/span>2<\/span><\/sub> mass = 176 grams<\/span><\/span><\/span><\/p>\n

The molecular mass of CO<\/span>2<\/span><\/sub> = 44 grams\/mol<\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

Universal gas constants (R)<\/b><\/span><\/span><\/span><\/p>\n

Based on research by scientists, it was found that if the number of moles (n) is used to express the size of a substance, then the constant of comparison for each gas has the same value. The corresponding constant is the universal gas constant (R).<\/span><\/span><\/span><\/p>\n

R = 8.315 J \/ mol. K<\/span><\/span><\/span><\/p>\n

= 8315 kJ \/ kmol. K<\/span><\/span><\/span><\/p>\n

= 0.0821 (L.atm) \/ (mol. K)<\/span><\/span><\/span><\/p>\n

= 1.99 cal \/ mol. K<\/span><\/span><\/span><\/p>\n

(J = Joule, K = Kelvin, L = liter, atm = atmosphere, cal = calorie)<\/span><\/span><\/span><\/p>\n

IDEAL GAS LAW (in the number of moles)<\/b><\/span><\/span><\/span><\/p>\n

The above comparison can be converted into equations by entering the number of mol (n) and universal gas constants (R).<\/span><\/span><\/span><\/p>\n

P V = n R T<\/span><\/span><\/span><\/p>\n

This equation is called the ideal gas law or ideal gas state equation.<\/span><\/span><\/span><\/p>\n

P = gas pressure (N\/m<\/span>2<\/span><\/sup>)<\/span><\/span><\/span><\/p>\n

V = volume of gas (m<\/span>3<\/span><\/sup>)<\/span><\/span><\/span><\/p>\n

n = number of moles (mole)<\/span><\/span><\/span><\/p>\n

R = universal gas constants (R = 8.315 J\/mol. K)<\/span><\/span><\/span><\/p>\n

T = absolute temperature of the gas (K)<\/span><\/span><\/span><\/p>\n

In solving the problem, you will find the term STP. STP is the Standard Temperature and Pressure.<\/span><\/span><\/span><\/p>\n

Standard temperature (T) = 0 <\/span>o<\/span><\/sup>C = 273 K<\/span><\/span><\/span><\/p>\n

Standard pressure (P) = 1 atm = 1.013 x 10<\/span>5 <\/span><\/sup>N \/ m<\/span>2 <\/span><\/sup>= 1.013 x 10<\/span>2<\/span><\/sup> kPa = 101 kPa<\/span><\/span><\/span><\/p>\n

In solving the gas law issues, the temperature should be expressed in the Kelvin (K) scale. If the gas pressure is still a measuring pressure, change it first into absolute pressure. Absolute pressure = atmospheric pressure + measuring pressure. If the known is atmospheric pressure (no measuring pressure), solve the problem.<\/span><\/span><\/span><\/p>\n

Question 1 :<\/span><\/span><\/span><\/p>\n

At atmospheric pressure (101 kPa), the temperature of carbon dioxide gas = 20<\/span> o<\/span><\/sup>C and its volume = 2 liters. If the pressure is converted to 201 kPa and the temperature is increased to 40 \u00b0C, calculate the final volume of the carbon dioxide gas.<\/span><\/span><\/span><\/p>\n

Solution<\/span><\/span><\/span><\/p>\n

P<\/span>1<\/span><\/sub> = 101 kPa<\/span><\/span><\/span><\/p>\n

P<\/span>2<\/span><\/sub> = 201 kPa<\/span><\/span><\/span><\/p>\n

T<\/span>1 <\/span><\/sub>= 20 <\/span>o<\/span><\/sup>C + 273 K = 293 K<\/span><\/span><\/span><\/p>\n

T<\/span>2<\/span><\/sub> = 40 <\/span>o<\/span><\/sup>C + 273 K = 313 K<\/span><\/span><\/span><\/p>\n

V<\/span>1 <\/span><\/sub>= 2 liters<\/span><\/span><\/span><\/p>\n

\"The
\nQuestion 2 :<\/span><\/span><\/span><\/p>\n

Determine the volume of 2 moles of gas on STP (this gas is the ideal gas)<\/span><\/span><\/span><\/p>\n

Solution<\/span><\/span><\/span><\/p>\n

\"The
\nQuestion 3 :<\/span><\/span><\/span><\/p>\n

Volume of oxygen gas at STP = 20 m<\/span>3<\/span><\/sup>. Calculate the oxygen gas mass.<\/span><\/span><\/span><\/p>\n

Solution<\/span><\/span><\/span><\/p>\n

Volume 1 mol gas at STP = 22.4 liter = 22.4 dm<\/span>3<\/span><\/sup> = 22.4 x 10<\/span>-3 <\/span><\/sup>m<\/span>3 <\/span><\/sup>(22.4 x 10<\/span>-3<\/span><\/sup> m<\/span>3<\/span><\/sup>\/mol)<\/span><\/span><\/span><\/p>\n

Volume of oxygen gas at STP = 20 m<\/span>3<\/span><\/sup><\/span><\/span><\/p>\n

\"The<\/p>\n

The mass of oxygen molecules = 32 grams \/ mol (mass 1 mole oxygen = 32 grams). Thus, the oxygen gas mass is:<\/span><\/span><\/span><\/p>\n

mass (m) = number of mol (n) x molecular mass<\/span><\/span><\/span><\/p>\n

mass = (893 mol) x (32 grams \/ mol) = 28576 grams = 28.576 kg<\/span><\/span><\/span><\/p>\n

Question 4 :<\/span><\/span><\/span><\/p>\n

A tank contains 4 liters of oxygen (O<\/span><\/span><\/span>2<\/span><\/span><\/span><\/sub>). The oxygen temperature is = 20 <\/span><\/span><\/span>o<\/span><\/span><\/span><\/sup>C and measured pressure is = 20 x 10<\/span><\/span><\/span>5<\/span><\/span><\/span><\/sup> N\/m<\/span><\/span><\/span>2<\/span><\/span><\/span><\/sup>. Determine the mass of oxyge<\/span><\/span><\/span>n (<\/span><\/span><\/span>Molecular mass <\/span><\/span><\/em>of <\/span><\/span>oxygen <\/span><\/span><\/em>= 32 kg\/kmol = 32 gram\/mol).<\/span><\/span><\/span><\/p>\n

Solution<\/span><\/span><\/span><\/p>\n

P = P atm + P measured = (1 x 10<\/span>5<\/span><\/sup> N\/m<\/span>2<\/span><\/sup>) <\/span>+ (20 x 10<\/span>5<\/span><\/sup> N\/m<\/span>2<\/span><\/sup>) = 21 x 10<\/span>5<\/span><\/sup> N\/m<\/span>2 <\/span><\/sup><\/span><\/span><\/p>\n

T = 20<\/span>o<\/span><\/sup>C + 273 = 293 K<\/span><\/span><\/span><\/p>\n

V = 4 liters = 4 dm<\/span>3 <\/span><\/sup>= 4 x 10<\/span>-3 <\/span><\/sup>m<\/span>3<\/span><\/sup><\/span><\/span><\/p>\n

R = 8.315 J\/mol. K = 8.315 Nm\/mol. K<\/span><\/span><\/span><\/p>\n

Molecular mass <\/span><\/span><\/span><\/em>of <\/span><\/span><\/span>oxygen <\/span><\/span><\/span><\/em>= 32 kg\/kmol = 32 gram\/mol<\/span><\/span><\/span><\/p>\n

O<\/span>2<\/span><\/sub> mass ?<\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

\"The<\/p>\n

IDEAL GAS LAW (In the number of molecules)<\/span><\/span><\/span><\/strong><\/p>\n


\nIf we state the size of the substance not in the form of mass (m), but in the number of moles (n), then the universal gas constant (R) applies to all gases. It was first discovered by Amadeo Avogadro (1776-1856), an Italian scientist.<\/span><\/span><\/span><\/p>\n

Avogadro says that when the volume, pressure, and temperature of each gas are equal, each gas has the same number of molecules.<\/span><\/span><\/span><\/p>\n

The Avogadro hypothesis corresponds to the fact that the constant R is the same for all gases. Here are some proofs:<\/span><\/span><\/span><\/p>\n

First, if we solve the problem using the ideal gas law equation (PV = nRT), we will find that when the number of moles (n) is equal, <\/span><\/span><\/span><\/p>\n

the pressure and temperature are also similar, then the volume of all gases will be the same if we use universal gas constants R = 8.315 J \/ mol. K). In STP, any gas having the same number of moles (n) will have the same volume. Volume 1 mol of gas at STP = 22.4 liters. Volume 2 mol of gas = 44.8 liters. Volume 3 mol of gas = 67.2 liters. And so on. This applies to all gas.<\/span><\/span><\/span><\/p>\n

Second, the number of molecules in 1 mol is the same for all gases. The number of molecules in 1 mol = number of molecules per mole = Avogadro number (NA). So, the Avogadro number is the same for all gas.<\/span><\/span><\/span><\/p>\n

Avogadro’s number obtained through measurement:<\/span><\/span><\/span><\/p>\n

NA = 6.02 x 10<\/span>23<\/span><\/sup> molecule \/ mol<\/span><\/span><\/span><\/p>\n

To get the total number of molecules (N), multiply the number of molecules per mole (N<\/span>A<\/span><\/sub>) by the number of moles (n).<\/span><\/span><\/span><\/p>\n

Total number of molecules (N) = number of molecules per mole (N<\/b><\/span>A<\/b><\/span><\/sub>) x number of moles (n)<\/b><\/span><\/span><\/span><\/p>\n

\"The
\nP = Pressure, V = Volume, N = Total number of molecules, k = Boltzmann constant (k = 1.38 x 10-23 J \/ K), T = Temperature<\/span><\/span><\/span><\/p>\n

\"The<\/p>\n

Volume Units<\/b><\/span><\/span><\/span><\/p>\n

1 liter (L) = 1000 milliliters (mL) = 1000 cubic centimeters (cm<\/span>3<\/span><\/sup>)<\/span><\/span><\/span><\/p>\n

1 liter (L) = 1 cubic decimeter (dm<\/span>3<\/span><\/sup>) = 1 x 10<\/span>-3 <\/span><\/sup>m<\/span>3<\/span><\/sup><\/span><\/span><\/p>\n

Pressure Units<\/b><\/span><\/span><\/span><\/p>\n

1 N\/m<\/span>2<\/span><\/sup> = 1 Pa<\/span><\/span><\/span><\/p>\n

1 atm = 1.013 x 10<\/span>5 <\/span><\/sup>N\/m<\/span>2<\/span><\/sup> = 1.013 x 10<\/span>5 <\/span><\/sup>Pa = 1.013 x 10<\/span>2<\/span><\/sup> kPa = 101.3 kPa (usually used 101 kPa)<\/span><\/span><\/span><\/p>\n

Pa = pascal<\/span><\/span><\/span><\/p>\n

atm = atmosphere<\/span><\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

The gas laws of Boyle, Charles law and Gay-Lussac do not apply to all gas conditions, so our analysis becomes more difficult. Therefore, presented the ideal gas model. Ideal gas does not exist in everyday life; the ideal gas is the just perfect form to facilitate analysis. The existence of this ideal gas concept also … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"The ideal gas law","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[2],"tags":[],"class_list":["post-3721","post","type-post","status-publish","format-standard","hentry","category-basic-physics-tutorials"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3721","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=3721"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3721\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=3721"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=3721"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=3721"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}