The following data is obtained for one mole o | vedclass

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

Question

(C) For an ideal gas,the compressibility factor $Z = \frac{PV}{RT} = 1$.From the given data table,we observe that for pressure values $P = 1, 2, 	ext

Vedclass · Accepted Answer

$3.1$ to $4.0$

Vedclass · Answer

(C) For an ideal gas,the compressibility factor $Z = \frac{PV}{RT} = 1$.From the given data table,we observe that for pressure values $P = 1, 2, 	ext{ and } 3 	ext{ bar}$,the value of $Z = 1$,which indicates ideal behavior.However,for pressure values ranging from $3.1 	ext{ bar}$ to $4.0 	ext{ bar}$,the value of $Z$ is greater than $1$ $(1.2, 1.4, 1.5)$,which indicates a deviation from ideal behavior.Therefore,the gas deviates from ideal behavior in the pressure range $3.1 	ext{ to } 4.0 	ext{ bar}$.

$P \text{ (bar)}$	$\frac{PV}{RT}$
$1$	$1$
$2$	$1$
$3$	$1$
$3.1$	$1.2$
$3.5$	$1.4$
$4.0$	$1.5$

The following data is obtained for one mole of a gas. The gas deviates from ideal behavior in the pressure (in bar) range:
$P \text{ (bar)}$ $\frac{PV}{RT}$
$1$ $1$
$2$ $1$
$3$ $1$
$3.1$ $1.2$
$3.5$ $1.4$
$4.0$ $1.5$

Similar Questions

At $300 \ K$,one mole of a gas present in a $10 \ L$ flask exerted a pressure of $2.706 \ atm$. What is its compressibility factor $(Z)$? (Given $R=0.082 \ L \ atm \ mol^{-1} \ K^{-1}$).

Positive deviation from ideal behaviour is due to the fact that

Under which one of the following conditions do real gases approach the ideal gas behaviour?

The compressibility factor of a real gas at high pressure is

The correct equation for one mole of a real gas is ($a, b$ are constants)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The following data is obtained for one mole of a gas. The gas deviates from ideal behavior in the pressure (in bar) range:$P \text{ (bar)}$$\frac{PV}{RT}$$1$$1$$2$$1$$3$$1$$3.1$$1.2$$3.5$$1.4$$4.0$$1.5$