How Much Volume Does One Mole Of Gas Occupy At NTP? - Vedantu

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Last updated date: 05th Jul 2024•Total views: 404.4k•Views today: 8.04k AnswerStudy MaterialsNCERT Solutions For Class 11Important Questions for Class 11Revision Notes for Class 11NCERT BooksMaths Formula for Class 11Sample PapersClass 11 MathsClass 11 PhysicsClass 11 ChemistryClass 11 Business StudiesClass 11 EconomicsClass 11 AccountancySyllabusMaths SyllabusPhysics SyllabusChemistry SyllabusBiology SyllabusEnglish SyllabusHindi SyllabusAccountancy SyllabusBusiness Studies SyllabusEconomics SyllabusPolitical Science SyllabusPhysical Education SyllabusTextbook SolutionsClass 11 NCERT Exemplar SolutionsClass 11 HC Verma SolutionsClass 11 RD Sharma SolutionsClass 11 RS Aggarwal SolutionsClass 11 DK Goel SolutionsClass 11 TS Grewal SolutionsClass 11 Sandeep Garg SolutionsAnswerVerified404.4k+ viewsHint:NTP stands for normal temperature and pressure. The ideal gas equation which is a relation connecting the pressure, volume, temperature and number of moles present in a sample of an ideal gas can be employed to determine the volume of one mole of gas. Formula used:-The ideal gas equation is given by, $PV = nRT$ where $P$ is the pressure of the ideal gas, $V$ is the volume of the given sample, $n$ is the number of moles present in the given sample, $T$ is the temperature and $R$ is the gas constant.Complete step by step answer.Step 1: Mention the temperature and pressure of a gas at NTP.The temperature at NTP is taken to be $T = 20^\circ {\text{C}}$ .The pressure at NTP is taken to be $P = 1{\text{atm}}$ .We have to determine the volume of one mole of gas. Step 2: Based on the ideal gas equation, obtain the volume of one mole of an ideal gas at NTP. The ideal gas equation can be expressed as $PV = nRT$ ------------- (1)where $P$ is the pressure of the ideal gas, $V$ is the volume of the given sample, $n$ is the number of moles present in the given sample, $T$ is the temperature and $R$ is the gas constant.Rewriting equation (1) in terms of the volume we get, $V = \dfrac{{nRT}}{P}$ ----------- (2)Substituting for $n = 1$ , $T = 293{\text{K}}$ , $P = 101325{\text{Pa}}$ and $R = 8 \cdot 314{\text{Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ in equation (2) we get, $V = \dfrac{{1 \times 8 \cdot 314 \times 293}}{{101325}} = 24 \cdot 12 \times {10^{ - 3}}{{\text{m}}^3}$Thus the volume of one mole of gas at NTP is obtained as $V = 24 \cdot 04 \times {10^{ - 3}}{{\text{m}}^3} = 24 \cdot 04{\text{L}}$ .Note:Here we obtained the volume of one mole of an ideal gas. For a real gas, the value will be slightly different. The S.I unit of volume is cubic-meter and so we converted the temperature and pressure into their respective S.I units as $T = 20^\circ {\text{C}} + 273{\text{K}} = 293{\text{K}}$ and $P = 1{\text{atm}} = 101325{\text{Pa}}$ before substituting in equation (2). STP stands for standard temperature and pressure. The pressure at STP is the same $P = 1{\text{atm}}$ but the temperature at STP will be $T = 0^\circ {\text{C}} = 273{\text{K}}$ .Recently Updated PagesWhy Are Noble Gases NonReactive class 11 chemistry CBSELet X and Y be the sets of all positive divisors of class 11 maths CBSELet x and y be 2 real numbers which satisfy the equations class 11 maths CBSELet x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSELet x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSELet x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSEWhy Are Noble Gases NonReactive class 11 chemistry CBSELet X and Y be the sets of all positive divisors of class 11 maths CBSELet x and y be 2 real numbers which satisfy the equations class 11 maths CBSELet x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSELet x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSELet x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSE

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