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Calculate the pressure exerted by one mole of CO 2 gas at 273 K if the van der Waal’s constant a = 3.592 dm 6 atm mol -2 . Assume that the volume occupied by CO 2 molecules in negligible.

Calculate the pressure exerted by one mole of CO2 gas at 273 K if the van der Waal’s constant a = 3.592 dm6 atm mol-2. Assume that the volume occupied by CO2 molecules in negligible.

Grade:upto college level

3 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago
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
Chandan Kumar
15 Points
5 years ago
Vanderwal eq for one mole of gas 
[p+a/v2](v-b)=RT... (1)
Give that volume occupied by CO2molecules 
, b'=0
Hence (1)becomes [p+a/v2]v=RT
Or p=RT/V-a/V2   
Pv2+a-RTV=0
D=0
(RT)-4ap=0
R=0.0821, T=273, a=3.59  
Then the currect answer is 
=34.98 atm
Then the currect answer is 
 
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
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 = 22.4L for 1 mole of an ideal gas at 1 atm pressure.
∴ P = 0.082 *273/22.4 – 3.592/(22.4)2 = 0.9922 atm
 
Thanks and regards,
Kushagra

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