The inversion temperature $T_{i} (K)$ of hydrogen is:

[Given: Van der Waal's constants a and b are $0.244 a t m L^{2} m o l^{- 2}$ and $0.027 L m o l^{- 1}$ respectively]

440

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220

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110

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330

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Solution

The correct option is B $220$
Gases become cooler during Joule Thomson's expansion only if they are below a certain temperature known as inversion temperature $(T_{i})$ .

The inversion temperature is characteristic of each gas and is given by $T_{i} = \frac{2 a}{b R}$
where $R$ is gas constant

Given:
$a = 0.244 a t m L^{2} m o l^{- 2}$
$b = 0.027 L m o l^{- 1}$
$R = 0.0821 L a t m K^{- 1} m o l^{- 1}$

$∴ T_{i} = \frac{2 \times 0.244}{0.027 \times 0.0821} = 220 K$

Hence, the correct answer is option $B$ .

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The inversion temperature Ti(K) of hydrogen is:[Given: Van der Waal's constants a and b are 0.244 atm L2mol−2 and 0.027 L mol−1 respectively]

The inversion temperature $T_{i} (K)$ of hydrogen is:

[Given: Van der Waal's constants a and b are $0.244 a t m L^{2} m o l^{- 2}$ and $0.027 L m o l^{- 1}$ respectively]