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

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

Question

(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

Vedclass · Accepted Answer

$T_1 < T_2 < T_3 < T_B$

Vedclass · Answer

(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 $T_1 < T_2 < T_3 < T_B$.

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

The compressibility factor $(Z)$ of a gas is greater than unity at $1 \, atm$ and $273 \, K$. Therefore:

For $1 \ mol$ of gas,the plot of $pV$ vs $p$ is shown below. $p$ is the pressure and $V$ is the volume of the gas. What is the value of compressibility factor at point $A$?

Why do real gases deviate from ideal behavior in the $pV$ vs $p$ and $p$ vs $V$ graphs at specific temperatures?

Which statement is correct regarding a real gas?

Calculate the compressibility factor for $SO_2$,if $1 \ mol$ of it occupies $0.35 \ L$ at $300 \ K$ and $50 \ atm$ pressure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module