The inversion temperature T_i , (K) of hydrog | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The inversion temperature $(T_i)$ is the temperature below which a gas cools upon Joule-Thomson expansion. It is calculated using the formula: $T_

Vedclass · Accepted Answer

$220$

Vedclass · Answer

(B) The inversion temperature $(T_i)$ is the temperature below which a gas cools upon Joule-Thomson expansion. It is calculated using the formula: $T_i = \frac{2a}{bR}$. Given: $a = 0.244 \, atm \, L^2 \, mol^{-2}$ $b = 0.027 \, L \, mol^{-1}$ $R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$ Substituting these values into the formula: $T_i = \frac{2 	imes 0.244}{0.027 	imes 0.0821} \approx \frac{0.488}{0.0022167} \approx 220.15 \, K$. Rounding to the nearest integer,we get $220 \, K$.

Gas	$a / (kPa \cdot dm^6 \cdot mol^{-2})$	$b / (dm^3 \cdot mol^{-1})$
$A$	$642.32$	$0.05196$
$B$	$155.21$	$0.04136$
$C$	$431.91$	$0.05196$
$D$	$155.21$	$0.4382$

The inversion temperature $T_i \, (K)$ of hydrogen is (given van der Waal's constants $a$ and $b$ are $0.244 \, atm \, L^2 \, mol^{-2}$ and $0.027 \, L \, mol^{-1}$ respectively).

Similar Questions

Consider the following table:

Gas $a / (kPa \cdot dm^6 \cdot mol^{-2})$ $b / (dm^3 \cdot mol^{-1})$

$A$ $642.32$ $0.05196$

$B$ $155.21$ $0.04136$

$C$ $431.91$ $0.05196$

$D$ $155.21$ $0.4382$

$a$ and $b$ are van der Waals constants. The correct statement about the gases is:

For the following gases,the values of $'a'$ (van der Waals constant) are given in parentheses: $CH_4 (2.25)$,$N_2 (1.35)$,$O_2 (1.36)$,and $CO (1.46)$. Which gas will be the most difficult to liquefy?

Under what conditions do real gases behave as ideal gases?

What kind of deviation will $2 \, mol$ of $N_2$ gas show in a $50 \, L$ container at $STP$?

Which of the following formulas is used to calculate the compressibility factor?

Vedclass Test Series

Exam Paper Generator

Online Exam Module