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Grade: 12th pass
        if vol. occupai by co2 molicule is negligiable. then what will be the pressure(P/5.277) exerted by 1 mole of co2 gas at 300k ? ( a=3.592 atm L^2 /mol^-2 )
5 months ago

Answers : (1)

Arun
11598 Points
							
 
van der Waal’s equation for one mole of a gas is
 
[P + a/V2] (V - b) = RT ….(1)
 
Give that volume occupied by CO2 molecules, ‘b’ = 0
 
Hence, (1) becomes [P + a/V2] V = RT or P = RT/V – a/V2
 
Using R = 0.082 , T = 273K, V = 22.4 l for 1 mole of an ideal gas at 1 atm pressure.
 
Case I. 1 * V = 12/m R (t + 273) ….(1)
 
Case I. 1 * V = 12/m R (t + 273) ….(1)
 
∴ P = 0.082 *273/22.4 – 3.592/(22.4)4 = 0.9922 atm
5 months ago
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