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Standard XII

Chemistry

Compressibility Factor

For a van der...

Question

For a van der Waal's gas, determine Boyle Temperature.
[ given $a = 4.5 a t m L^{2} {m o l}^{- 2}, b = 0.9 L {m o l}^{- 1}$ and $R = 0.082 L a t m K^{- 1} {m o l}^{- 1}$ ].

609.8

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6.09

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273

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60.98

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Solution

The correct option is D $60.98$ K
Boyle Temperature $T_{B} = \frac{a}{b R}$

$T_{B} = \frac{4.5 a t m L^{2} {m o l}^{- 2}}{0.9 L {m o l}^{- 1} \times 0.082 L a t m K^{- 1} {m o l}^{- 1}}$

$T_{B} = 60.98 K$

Hence, the correct option is $D$

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For a van der Waal's gas, determine Boyle Temperature. [ given a=4.5 atm L2mol−2,b=0.9 L mol−1 and R=0.082L atm K−1mol−1].

The correct option is D 60.98KBoyle Temperature TB=abRTB=4.5atm L2 mol−20.9 L mol−1×0.082 L atm K−1 mol−1TB=60.98 KHence, the correct option is D

For a van der Waal's gas, determine Boyle Temperature.
[ given $a = 4.5 a t m L^{2} {m o l}^{- 2}, b = 0.9 L {m o l}^{- 1}$ and $R = 0.082 L a t m K^{- 1} {m o l}^{- 1}$ ].

The correct option is D $60.98$ K
Boyle Temperature $T_{B} = \frac{a}{b R}$

$T_{B} = \frac{4.5 a t m L^{2} {m o l}^{- 2}}{0.9 L {m o l}^{- 1} \times 0.082 L a t m K^{- 1} {m o l}^{- 1}}$

$T_{B} = 60.98 K$

Hence, the correct option is $D$