{"id":1632,"date":"2024-07-27T08:33:13","date_gmt":"2024-07-27T08:33:13","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=1632"},"modified":"2024-07-27T08:33:13","modified_gmt":"2024-07-27T08:33:13","slug":"third-law-of-thermodynamics","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/07\/27\/third-law-of-thermodynamics\/","title":{"rendered":"THIRD LAW OF THERMODYNAMICS"},"content":{"rendered":"\n

Third law of thermodynamics is law of entropy. It is a statement about the ability to create an absolute temperature scale, for which absolute zero is the point at which the internal energy of a solid is zero. Third law of thermodynamics states that it is impossible to reduce any system to absolute zero in a finite series of operations.<\/p>\n\n\n\n

1.11\u00a0\u00a0GAS LAWS<\/strong><\/h5>\n\n\n\n

There are some relationships among temperature, volume, pressure, and quantity of a gas that could be described mathematically. This chapter deals with Boyle\u2019s law, Charles\u2019s law, Gay\u2013Lussac\u2019s law, and the combined gas law. These laws have one condition in common, i.e., fixed mass. In addition, some other properties of gases such as internal energy, specific heat capacity, and enthalpy have been introduced. Some of the important non-flow processes such as constant volume process, constant pressure process, isothermal processes, polytropic process, and adiabatic process have been explained with suitable examples. Some laws have been proposed by the various chemists such as Boyle\u2019s law, Charle\u2019s law, Gay\u2013Lussac\u2019s law based on the behaviour of ideal gases. These laws are discussed in the following subsections.<\/p>\n\n\n\n

1.11.1\u00a0\u00a0Boyle\u2019s Law<\/strong><\/h4>\n\n\n\n

Robert Boyle, a British chemist gave the first gas law, now known as Boyle\u2019s law. This law describes the relationship between the pressure and volume of a sample of gas confined in a container. Boyle observed that when the pressure on an ideal gas is increased volume decreases. Similarly, when pressure is released the volume starts to increase. But Boyle\u2019s law is true only when the temperature of the gas remains constant and no additional gas is added to the container or leaks out of the container. On the basis of these observations, the Boyle\u2019s law is stated as: \u2018that the volume and pressure of a sample of gas are inversely proportional to each other at constant temperature\u2019.<\/p>\n\n\n\n

This statement can be expressed as follows.<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

where V<\/em> is volume and P<\/em> is pressure.<\/p>\n\n\n\n

For two different conditions 1 and 2, Boyle\u2019s law can be expressed as<\/p>\n\n\n\n

 <\/p>\n\n\n\n

P<\/em>1<\/sub>V<\/em>1<\/sub> = P<\/em>2<\/sub>V<\/em>2<\/sub><\/p>\n\n\n\n

where P<\/em>1<\/sub> and V<\/em>1<\/sub> are pressure and volume, respectively, at condition 1 and P<\/em>2<\/sub> and V<\/em>2<\/sub> are pressure and volume, respectively, at condition 2.<\/p>\n\n\n\n

Example 1.25:<\/strong> A sample of nitrogen collected in the laboratory occupies a volume of 720 ml at a pressure of 1 atm. What volume will the gas occupy at a pressure of 2 atm, assuming the temperature remains constant?<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: V<\/em>1<\/sub> = 720 ml; P<\/em>1<\/sub> = 1 atm; P<\/em>2<\/sub> = 2 atm; V<\/em>2<\/sub> = ?<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

1.11.2  Charles\u2019s Law<\/h4>\n\n\n\n

Jacques Charles carried out experiments on ideal gas and observed a relationship between the absolute temperature and volume of gases at constant pressure. Volume of the gas increases with increase in temperature and decreases with decrease in temperature. The Charle\u2019s law can be stated as: \u2018that the volume of a sample of gas is directly proportional to the absolute temperature when pressure remains constant\u2019.<\/p>\n\n\n\n

Charles\u2019s law can be expressed as follows.<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

where V<\/em> is volume and T<\/em> is absolute temperature of the gas.<\/p>\n\n\n\n

<\/a>For two different conditions 1 and 2, Boyle\u2019s law can be expressed as<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

where T<\/em>1<\/sub> and V<\/em>1<\/sub> are absolute temperature and volume, respectively, at condition 1 and T<\/em>2<\/sub> and V<\/em>2<\/sub> are absolute temperature and volume, respectively, at condition 2.<\/p>\n\n\n\n

Example 1.26:<\/strong> A container of a gas has a volume of 360 ml at a temperature of 20\u00b0C. What volume will the gas occupy at 60\u00b0C?<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: V<\/em>1<\/sub> = 360 ml; T<\/em>1<\/sub> =273 + 20 = 293 K; T<\/em>2<\/sub> = 273 + 60 = 333 K; V<\/em>2<\/sub> = ?.<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

1.11.3\u00a0\u00a0Gay\u2013Lussac\u2019s Law<\/strong><\/h4>\n\n\n\n

Pressure of a confined gas increases with increasing temperature. If the temperature of the gas increases enough, the container can explode because of the pressure that builds up inside of it. The relationship between the pressure and temperature of a gas is described by Gay\u2013Lussac\u2019s law. \u2018Gay\u2013Lussac\u2019s law states that the pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant\u2019.<\/p>\n\n\n\n

Gay\u2013Lussac\u2019s law can be expressed as follows.<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

where P<\/em> is the pressure and T<\/em> is the temperature of the gas.<\/p>\n\n\n\n

For two different conditions 1 and 2, Gay\u2013Lussac\u2019s law can be expressed as<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

where T<\/em>1<\/sub> and P<\/em>1<\/sub> are absolute temperature and pressure, respectively, at condition 1 and T<\/em>2<\/sub> and P<\/em>2<\/sub> are absolute temperature and pressure, respectively, at condition 2.<\/p>\n\n\n\n

Example 1.27:<\/strong> A cylinder of a gas has a pressure of 5 atm at 50\u00b0C. At what temperature in \u00b0C will it reach a pressure of 12 atm?<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: P<\/em>1<\/sub> = 5 atm; T<\/em>1<\/sub> =273 + 50 = 323 K; P<\/em>2<\/sub> = 12 atm; T<\/em>2<\/sub> = ?.<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

<\/a>1.11.4  The Combined Gas Law<\/h4>\n\n\n\n

We have three different relationships among temperature, volume, and pressure of a gas; these are as follows:<\/p>\n\n\n\n

Boyle\u2019s Law:<\/em><\/strong> PV<\/em> = k<\/em> at constant temperature.<\/p>\n\n\n\n

Charle\u2019s Law:<\/em><\/strong> \"equation\" at constant pressure.<\/p>\n\n\n\n

Gay\u2013Lussac\u2019s Law:<\/em><\/strong> \"equation\" at constant volume.<\/p>\n\n\n\n

These three gas laws can be combined in one combined gas law. This law can be expressed as<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

Example 1.28:<\/strong> A sample of a gas has a volume of 80.0 ml at a pressure of 1 atm and a temperature of 20\u00b0C. What volume will the gas occupy at 1.5 atm and 45\u00b0C?<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: V<\/em>1<\/sub> = 80 ml; P<\/em>1<\/sub> = 1 atm; T<\/em>1<\/sub> = 273 + 20 = 293 K; P<\/em>2<\/sub> = 1.5 atm; T<\/em>2<\/sub> = 273 + 45 = 318 K; V<\/em>2<\/sub> = ?<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

1.11.5\u00a0\u00a0Gas Constant<\/strong><\/h4>\n\n\n\n

Since 1 mole of a gas occupies 22.4 l at standard temperature (273 K) and pressure (1 atm), it is possible to arrive at a mathematical expression to relate moles, pressure, temperature, and volume. This expression is called the ideal gas law. This law contains an additional term \u2018R<\/em>\u2019 which is called the universal gas constant. In this expression \u2018N<\/em>\u2019 equals the number of moles of a gas, the volume \u2018V<\/em>\u2019 must be expressed in litres, the pressure \u2018P<\/em>\u2019 must be expressed in atmospheres and the temperature must be expressed in degrees Kelvin.<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

This constant can be calculated by using the above values in this law.<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

When the values of 22.4 l and 273 degrees Kelvin are applied, the value of R<\/em> is found to be<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

<\/a>If we use CGS units, P<\/em> will be expressed in dynes per square cm, V<\/em> is the volume of a mole (i.e., the volume occupied by 6.0221 \u00d7 1023<\/sup> molecules) and the value of the universal gas constant is 8.3145 \u00d7 107<\/sup> erg\/mole K. If we use SI units, P<\/em> will be expressed in Pascal (N\/m2<\/sup>), V<\/em> will be the volume of a kilomole (i.e., the volume occupied by 6.0221 \u00d7 1026<\/sup> molecules) and the value of the universal gas constant is 8.3145 \u00d7 103<\/sup> J\/kilomole K.<\/p>\n","protected":false},"excerpt":{"rendered":"

Third law of thermodynamics is law of entropy. It is a statement about the ability to create an absolute temperature scale, for which absolute zero is the point at which the internal energy of a solid is zero. Third law of thermodynamics states that it is impossible to reduce any system to absolute zero in […]<\/p>\n","protected":false},"author":1,"featured_media":1601,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-1632","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-blog"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/07\/thermodynamics-1-3.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/1632","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=1632"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/1632\/revisions"}],"predecessor-version":[{"id":1633,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/1632\/revisions\/1633"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/1601"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=1632"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=1632"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=1632"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}