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CBSE Class 11-science Answered

12th?
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Asked by smanishkumar2002 | 04 Aug, 2018, 05:39: AM
answered-by-expert Expert Answer
Given:
 
T= 273 K
 
a= 3.592 atm lit2 mol-2 
 
For 1 mole
 
open square brackets straight P plus straight a over straight V squared close square brackets open square brackets straight V minus straight b close square brackets space equals space RT

If space straight b space is space negligible comma space then

open square brackets straight P plus straight a over straight V squared close square brackets open parentheses straight V close parentheses space equals space RT

straight P space equals space RT over straight V minus straight a over straight V squared

Multiply space by space straight V squared space

PV squared equals RTV minus straight a

PV squared minus RTV plus straight a equals 0

or

straight V equals fraction numerator plus RT plus-or-minus square root of open square brackets open parentheses negative RT close parentheses squared minus 4 Pa close square brackets end root over denominator 2 straight P end fraction

open parentheses RT close parentheses squared space minus 4 Pa space equals 0

or

straight P space equals space fraction numerator straight R squared straight T squared over denominator 4 straight a end fraction

space space space equals fraction numerator open parentheses 0.082 close parentheses squared open parentheses 273 close parentheses squared over denominator 4 cross times 3.592 end fraction

space straight P equals 34.96 space atm
Pressure exrted by gas is 34.96 atm.
 
 
Answered by Varsha | 05 Aug, 2018, 07:52: PM
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