Calculate The Pressure-volume Work Done By The System ... - Vedantu

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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsCalculate the pressure-volume work done by the system when the gas expands from 1-litre to 2-litre against a constant external pressure of 10 atmosphere. Express the answer in calorie and joule.AnswerVerified584.7k+ viewsHint: Pressure -Volume work done occurs when the volume of the gas or system changes. Work done by a system is the energy transferred from it to its surroundings. Express that work in joules and calories, -${{P}_{external}}$$\times $$\vartriangle V$.Complete step by step answer:If the gas is heated, the gas molecules are energized. The increase in average kinetic energy of the gas molecules can be observed by measuring how the temperature of the gas particles increased. The gas molecules collide with the piston more frequently, by the fast movement of the gas particles. These frequent collisions transfer energy to the piston and allow it to move against an external pressure thus, increasing the net volume of the gas. The work done by the gas is specifically the expansion and compression of the gas. Work done by gases is sometimes called pressure-volume or PV work.To calculate how much work is done (or has done to it) by the gas against a constant external pressure, we have to use an equation:WORK= (w) = -${{P}_{external}}$$\times $$\vartriangle V$ where external pressure is denoted by ${{P}_{external}}$. The external pressure opposes the pressure of the gas in the system. ∆V is a change in the volume of the gas, which is calculated as ${{V}_{final}}-{{V}_{\operatorname{in}itial}}$.Sign convention of work :When a system does work on the surroundings, negative work is said to have occurred.When the gas does work, the volume of the gas particles decreases ($\vartriangle V$<0).The work done is positive.When the gas does work, the volume of the gas particles increases ($\vartriangle V$>0).The work done is negative.Units conversion:Work has its units in JOULES. Pressure-volume work has its unit in L-atm which is to be converted to Joules using the conversion factor 101.325J/1L-atmTo convert joules to calories the conversion factor is 0.239cal/1J.Calculate the pressure-volume work done by the gas:Given: ${{P}_{external}}$= 10 atmospheres ${{V}_{initial}}$= 1 litre ${{V}_{final}}$ = 2 litresUse the formula, to calculate the work done by the gas:WORK= (w) = -${{P}_{external}}$$\times $$\vartriangle V$ Substitute the values as ${{V}_{final}}-{{V}_{\operatorname{in}itial}}$The net formula is = -${{P}_{external}}$$\times $(${{V}_{final}}-{{V}_{\operatorname{in}itial}}$)Substitute the values of ${{P}_{external}}$,${{V}_{initial}}$, ${{V}_{final}}$ as 10 atmospheres, 1 litre and 2 litres.The work done is = - (10 atm)$\times $(2 litres – 1 litre)= - (10 atm)$\times $(1 litre)= - (10 litre-atm)Convert litre atm to joules using the conversion factor:1 L-atm has 101.325J, so -10 L-atm has (-10)$\times $(101.325J) is equal to (-1013.25 Joules. )To convert joules into calories multiply the joule by 0.239 factor, (0.239) $\times $(-1013.25) is equal to - 242.16 Calories.The answer of the question is -1013.25 Joules and -242.16 calories.Note:(i) Take care of the sign convention while solving the numerical of the work done. As the sign convention by the gas and on the gas is different.(ii) Do not forget to convert to use conversion factor for the required units as mentioned in the question.Recently Updated PagesWhy are manures considered better than fertilizers class 11 biology CBSEFind the coordinates of the midpoint of the line segment class 11 maths CBSEDistinguish between static friction limiting friction class 11 physics CBSEThe Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSEThe first National Commission on Labour NCL submitted class 11 social science CBSENumber of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSEWhy are manures considered better than fertilizers class 11 biology CBSEFind the coordinates of the midpoint of the line segment class 11 maths CBSEDistinguish between static friction limiting friction class 11 physics CBSEThe Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSEThe first National Commission on Labour NCL submitted class 11 social science CBSENumber of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

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