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Gas laws from KTG

1.00 mol N 2 ...

Question

1.00 mol $N_{2}$ is given and it obeys the van der Waals equation in the form of virial expansion. Assume that the pressure is 10.00 atm throughout.
$V_{c}$ = critical volume
$V_{b}$ = V at the Boyle temperature
$Z_{c}$ = Z at the critical temperature
Take $R = \frac{1}{12} L a t m m o l^{- 1} K^{- 1}$

V_{c} = 0.105 L

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Z_{c} = 0.10

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V_{b} = 3.50 L

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The second virial coefficient is

- v e

T_{c}

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Solution

The correct options are
C $V_{b} = 3.50 L$
D The second virial coefficient is $- v e$ at $T_{c}$
Using the ideal gas equation: $V_{c r i t i c a l} = R T / P = (1 \times 126.3) / (10 \times 12) \times 10^{3} = 1.05 L$
(a) At critical temperature the second virial coefficient;
$B = 0.04 - \frac{(1.40 \times 12)}{(1 \times 126.3)} = - 0.0933 \Rightarrow Z = 1 - \frac{0.0933}{V} p V = Z R T = \frac{V - 0.0933}{V} \times R T \Rightarrow V^{2} = (V - 0.0933) \frac{R T}{p} = (V - 0.0933) \frac{(126.3 \times 1)}{(10 \times 12)} = 1.05 V - 0.098 \Rightarrow V^{2} - 1.05 V + 0.098 = 0 \Rightarrow V = 0.104 L a n d 0.946 L$
Corresponding values of Z are 0.10 and 0.90
A value of 0.09 implies a very large attractive forces, as pressure is reduced to 10% of the ideal gas pressure.
We must choose $V_{c} = 0.946 L$ as our solution since at the given T and P conditions, weak van der Waals forces are applicable.

Boyles temperature:
$T_{B} = \frac{a}{R b} = \frac{(1.4 \times 12)}{(1 \times 0.04)} = 420 K$
At this temperature the gas behaves as an ideal gas.
$V_{B o y l e} = V_{i d e a l}$
at Boyle Temperature, $V_{b} = \frac{R T}{P} = \frac{(1 \times 420)}{(12 \times 10)} = 3.50 L$

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