
If volume occupied by $C{O_2}$ molecules is negligible, the what will be the pressure $\left( {\dfrac{P}{{5.227}}} \right)$ exerted by one mole of $C{O_2}$ gas at $300K?(a = 3.592atm - {L^2}mo{l^{ - 1}})$.
A.) $7$
B.) $8$
C.) $9$
D.) $3$
Answer
585.9k+ views
Hint: 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 :
$\left( {P + a\dfrac{{{n^2}}}{{{V^2}}}} \right)(V - nb) = nRT$ $ - (1)$
Where, $P = $pressure $n = 1$
$V = $ Volume
$T = $temperature $ = 300K$(given)
$R = 0.0821L - atm{K^{ - 1}}mo{l^{ - 1}}$$ = $real gas constant
$n = 1$ (It is given number of moles)
$a = 3.592atm - {L^2}mo{l^{ - 1}}$(It is the correction factor for the attractive forces between molecules. Its value is given in question)
$b = $ correction factor for volume of molecules.
Since, we are given in question that the volume occupied by carbon dioxide molecules is negligible. Therefore, we will take $b = 0$.
Now, by putting all values in equation $ - (1)$ we get :
$PV + \dfrac{{a(1)}}{V} = RT$
$P = \dfrac{{RT}}{V} - \dfrac{a}{{{V^2}}}$
$
P{V^2} = RTV - a \\
P{V^2} - RTV + a = 0 \\
$
Now, the equation becomes a quadratic equation in $V$. Therefore,
$V = \dfrac{{ + RT \pm \sqrt {{R^2}{T^2} - 4aP} }}{{2P}}$
As, $V$ has only one value at given pressure and temperature, therefore discriminant will become zero.
$
{R^2}{T^2} - 4aP = 0 \\
{R^2}{T^2} = 4aP \\
P = \dfrac{{{R^2}{T^2}}}{{4a}} \\
P = \dfrac{{{{(0.0821)}^2}{{(300)}^2}}}{{4 \times 3.592}} \\
P = 42.22atm \\
\dfrac{P}{{5.227}} = \dfrac{{42.22}}{{5.227}} = 8 \\
$
Hence, the required value of
$\dfrac{P}{{5.227}}$ is $8.$
Hence, B.) is the correct option.
Note:
Remember that the Vander waal’s equation that is $\left( {P + a\dfrac{{{n^2}}}{{{V^2}}}} \right)(V - nb) = nRT$ , is the generalized form used for non- ideal gases. For ideal gases, we can directly write the ideal gas equation that is $PV = nRT$.
Complete step by step answer:
As we know that Vander waals gas equation is a generalized form of ideal gas law hence it can be used for conditions when the gas does not act ideally. The Vander waals equation can be written as :
$\left( {P + a\dfrac{{{n^2}}}{{{V^2}}}} \right)(V - nb) = nRT$ $ - (1)$
Where, $P = $pressure $n = 1$
$V = $ Volume
$T = $temperature $ = 300K$(given)
$R = 0.0821L - atm{K^{ - 1}}mo{l^{ - 1}}$$ = $real gas constant
$n = 1$ (It is given number of moles)
$a = 3.592atm - {L^2}mo{l^{ - 1}}$(It is the correction factor for the attractive forces between molecules. Its value is given in question)
$b = $ 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 :
$PV + \dfrac{{a(1)}}{V} = RT$
$P = \dfrac{{RT}}{V} - \dfrac{a}{{{V^2}}}$
$
P{V^2} = RTV - a \\
P{V^2} - RTV + a = 0 \\
$
Now, the equation becomes a quadratic equation in $V$. Therefore,
$V = \dfrac{{ + RT \pm \sqrt {{R^2}{T^2} - 4aP} }}{{2P}}$
As, $V$ has only one value at given pressure and temperature, therefore discriminant will become zero.
$
{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
$\dfrac{P}{{5.227}}$ is $8.$
Hence, B.) is the correct option.
Note:
Remember that the Vander waal’s equation that is $\left( {P + a\dfrac{{{n^2}}}{{{V^2}}}} \right)(V - nb) = nRT$ , is the generalized form used for non- ideal gases. For ideal gases, we can directly write the ideal gas equation that is $PV = nRT$.
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