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Question: If volume occupied by \(C{O_2}\) molecules is negligible, the what will be the pressure \(\left( {\d...

If volume occupied by CO2C{O_2} molecules is negligible, the what will be the pressure (P5.227)\left( {\dfrac{P}{{5.227}}} \right) exerted by one mole of CO2C{O_2} gas at 300K?(a=3.592atmL2mol1)300K?(a = 3.592atm - {L^2}mo{l^{ - 1}}).
A.) 77
B.) 88
C.) 99
D.) 33

Explanation

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 aa 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+an2V2)(Vnb)=nRT\left( {P + a\dfrac{{{n^2}}}{{{V^2}}}} \right)(V - nb) = nRT (1) - (1)
Where, P=P = pressure n=1n = 1
V=V = Volume
T=T = temperature =300K = 300K(given)
R=0.0821LatmK1mol1R = 0.0821L - atm{K^{ - 1}}mo{l^{ - 1}} ==real gas constant
n=1n = 1 (It is given number of moles)
a=3.592atmL2mol1a = 3.592atm - {L^2}mo{l^{ - 1}}(It is the correction factor for the attractive forces between molecules. Its value is given in question)
b=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=0b = 0.
Now, by putting all values in equation (1) - (1) we get :
PV+a(1)V=RTPV + \dfrac{{a(1)}}{V} = RT
P=RTVaV2P = \dfrac{{RT}}{V} - \dfrac{a}{{{V^2}}}
PV2=RTVa PV2RTV+a=0  P{V^2} = RTV - a \\\ P{V^2} - RTV + a = 0 \\\
Now, the equation becomes a quadratic equation in VV. Therefore,
V=+RT±R2T24aP2PV = \dfrac{{ + RT \pm \sqrt {{R^2}{T^2} - 4aP} }}{{2P}}
As, VV has only one value at given pressure and temperature, therefore discriminant will become zero.
R2T24aP=0 R2T2=4aP P=R2T24a P=(0.0821)2(300)24×3.592 P=42.22atm P5.227=42.225.227=8  {R^2}{T^2} - 4aP = 0 \\\ {R^2}{T^2} = 4aP \\\ P = \dfrac{{{R^2}{T^2}}}{{4a}} \\\ P = \dfrac{{{{(0.0821)}^2}{{(300)}^2}}}{{4 \times 3.592}} \\\ P = 42.22atm \\\ \dfrac{P}{{5.227}} = \dfrac{{42.22}}{{5.227}} = 8 \\\
Hence, the required value of
P5.227\dfrac{P}{{5.227}} is 8.8.
Hence, B.) is the correct option.

Note:
Remember that the Vander waal’s equation that is (P+an2V2)(Vnb)=nRT\left( {P + a\dfrac{{{n^2}}}{{{V^2}}}} \right)(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=nRTPV = nRT.