Single Choice

If volume occupied by $$CO_{2}$$ molecules is negligible, then what will be the pressure $$\left (\dfrac {P}{5.277}\right )$$ exerted by one mole of $$CO_{2}$$ gas at $$300\ K? (a = 3.592\ atm\ L^{2} mol^{-2})$$.

A$$7$$
B$$8$$
Correct Answer
C$$9$$
D$$3$$

Solution

$$\left (P + \dfrac {an^{2}}{V^{2}}\right )(V - nb) = nRT$$
$$\left (P + \dfrac {a}{V^{2}}\right )(V - b) = RT (\because n = 1)$$
If $$b$$ is negligible
$$P = \dfrac {RT}{V} - \dfrac {a}{V^{2}}$$
The equation is quadratic in $$V$$, thus
$$V = \dfrac {+RT \pm \sqrt {R^{2}T^{2} - 4aP}}{2P}$$
Since $$V$$ has one value at given $$P$$ and $$T$$, thus numerical value of discriminant $$= 0$$
$$R^{2}T^{2} = 4aP$$
$$P = \dfrac {R^{2}T^{2}}{4a} = \dfrac {(0.0821)^{2}(300)^{2}}{4\times 3.592} \therefore \dfrac {P}{5.277} = 8$$.


SIMILAR QUESTIONS

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States of Matter - Gas and Liquid

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States of Matter - Gas and Liquid

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States of Matter - Gas and Liquid

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States of Matter - Gas and Liquid

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States of Matter - Gas and Liquid

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States of Matter - Gas and Liquid

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