For NH_3 gas,the Z vs pressure (P) curve at d | vedclass

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(B) The compressibility factor $Z$ is given by $Z = \frac{PV_m}{RT}$.For real gases,at low pressure,the attractive forces dominate,and $Z < 1$. As tem

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$T_1 < T_2 < T_3 < T_B$

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(B) The compressibility factor $Z$ is given by $Z = \frac{PV_m}{RT}$.For real gases,at low pressure,the attractive forces dominate,and $Z From the graph,the depth of the dip represents the deviation from ideal behavior. The curve with the deepest dip corresponds to the lowest temperature,and the curve with the shallowest dip corresponds to the highest temperature.Thus,the order of temperatures is $T_1 Since all these curves show a dip below $Z = 1.0$ at low pressures,all these temperatures are below the Boyle temperature $(T_B)$.Therefore,the correct order is $T_1 < T_2 < T_3 < T_B$.

Gas	$a$ value
$1. C_6H_{6(g)}$	$a. 0.217$
$2. C_6H_5CH_{3(g)}$	$b. 5.464$
$3. Ne_{(g)}$	$c. 18.000$
$4. H_2O_{(g)}$	$d. 24.060$

For $NH_3$ gas,the $Z$ vs pressure $(P)$ curve at different temperatures is shown. What is the correct order of temperatures? (Given $T_B$ = Boyle temperature)

Similar Questions

Under which conditions do real gases deviate most from ideal gas behavior?

The pressure exerted by $5 \, mol$ of a real gas in a $1 \, L$ container at $47 \, ^oC$ is ............. $atm$. $\left( a = 3.592 \, atm \, L^2 \, mol^{-2}, b = 0.0427 \, L \, mol^{-1} \right)$

Assertion : Greater the value of van der Waal's constant $a$,greater is the liquefaction of gas.
Reason : $a$ indirectly measures the magnitude of attractive forces between the molecules.

The van der Waals equation of state for a real gas is given. For $n$ moles of a real gas,the equation can be expressed as:

Match the Van der Waals constant $a$ (in $L^2 \cdot bar \cdot mol^{-2}$) for the following gases:

Gas $a$ value

$1. C_6H_{6(g)}$ $a. 0.217$

$2. C_6H_5CH_{3(g)}$ $b. 5.464$

$3. Ne_{(g)}$ $c. 18.000$

$4. H_2O_{(g)}$ $d. 24.060$

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