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CBSE Class 11-science Answered

28.32 li!tres of chlorine w€re liberated at nornal conditioDs of temperature and pressure. Calc.lllate the \olume ofthe gas at l2oCaod 780 mm pressure.
Asked by ashutoshkumarchoubey845 | 21 Jul, 2021, 08:56: AM
answered-by-expert Expert Answer
No. of moles at NTP, n= PV/RT
R= 0.082055 L atm mol-1 K-1
V= 28.32 L
T = 293 K
P = 1 atm
​
straight n space equals space fraction numerator 1 space atm cross times 28.32 space straight L over denominator 0.082055 Latm space mol to the power of negative 1 end exponent straight K to the power of negative 1 end exponent cross times 293 straight K end fraction space equals space 1.179 space mol Volume space of space the space gas space at space 12 to the power of straight o straight C space and space 780 space mm space pressure straight T space equals space 285 space straight K straight P space equals space 1.026 space atm straight V space equals space nRT over straight P space equals space fraction numerator 1.179 cross times 0.082055 cross times 285 over denominator 1.026 end fraction space space equals space 26.873 space straight L 
 
 
 
Answered by Ramandeep | 21 Jul, 2021, 12:25: PM

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