Question
Question: If volume occupied by \(C{O_2}\) molecules is negligible, the what will be the pressure \(\left( {\d...
If volume occupied by CO2 molecules is negligible, the what will be the pressure (5.227P) exerted by one mole of CO2 gas at 300K?(a=3.592atm−L2mol−1).
A.) 7
B.) 8
C.) 9
D.) 3
Solution
The pressure in the given question can be calculated with the help of Vander waal’s gas equation because here the value of constant a is given . This equation is used for non-ideal gases.
Complete step by step answer:
As we know that Vander waals gas equation is a generalized form of ideal gas law hence it can be used for conditions when the gas does not act ideally. The Vander waals equation can be written as :
(P+aV2n2)(V−nb)=nRT −(1)
Where, P=pressure n=1
V= Volume
T=temperature =300K(given)
R=0.0821L−atmK−1mol−1 =real gas constant
n=1 (It is given number of moles)
a=3.592atm−L2mol−1(It is the correction factor for the attractive forces between molecules. Its value is given in question)
b= correction factor for volume of molecules.
Since, we are given in question that the volume occupied by carbon dioxide molecules is negligible. Therefore, we will take b=0.
Now, by putting all values in equation −(1) we get :
PV+Va(1)=RT
P=VRT−V2a
PV2=RTV−a PV2−RTV+a=0
Now, the equation becomes a quadratic equation in V. Therefore,
V=2P+RT±R2T2−4aP
As, V has only one value at given pressure and temperature, therefore discriminant will become zero.
R2T2−4aP=0 R2T2=4aP P=4aR2T2 P=4×3.592(0.0821)2(300)2 P=42.22atm 5.227P=5.22742.22=8
Hence, the required value of
5.227P is 8.
Hence, B.) is the correct option.
Note:
Remember that the Vander waal’s equation that is (P+aV2n2)(V−nb)=nRT , is the generalized form used for non- ideal gases. For ideal gases, we can directly write the ideal gas equation that is PV=nRT.