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 )

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Arun · Answer

van der Waal’s equation for one mole of a gas is [P + a/V 2 ] (V - b) = RT ….(1) Give that volume occupied by CO 2 molecules, ‘b’ = 0 Hence, (1) becomes [P + a/V 2 ] V = RT or P = RT/V – a/V 2 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

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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 )

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 )

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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 )

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 )

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