A real gas obeying Van der Waals equation wil | vedclass

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

Question

(A) The Van der Waals equation for $n$ moles of a real gas is given by: $(P + \frac{an^2}{V^2})(V - nb) = nRT$. For a real gas to behave like an ideal

Vedclass · Accepted Answer

constants $a$ and $b$ are small

Vedclass · Answer

(A) The Van der Waals equation for $n$ moles of a real gas is given by: $(P + \frac{an^2}{V^2})(V - nb) = nRT$. For a real gas to behave like an ideal gas,the deviations from ideal behavior must be minimized. $1$. The term $\frac{an^2}{V^2}$ accounts for intermolecular forces of attraction. If $a$ is small,these forces are negligible. $2$. The term $nb$ accounts for the finite volume occupied by gas molecules. If $b$ is small,the volume of the molecules is negligible compared to the total volume of the container. Therefore,when both Van der Waals constants $a$ and $b$ are small,the real gas equation reduces to the ideal gas equation $PV = nRT$.

$A$ real gas obeying Van der Waals equation will resemble an ideal gas if the:

Similar Questions

$CH_4$ gas is behaving non-ideally. The compressibility factor for the gas is $1.5$ at $2 \ atm$ and $400 \ K$. Calculate the molar volume for the gas in $litre$. [Given $R = 0.08 \ L \ atm \ K^{-1} \ mol^{-1}$]

The compressibility factor $(Z)$ of a gas is less than unity at $STP$. Therefore:

At very low pressure,if the molar volume of $CO_2$ is constant,its compressibility factor is ..........

At high pressure,the van der Waals equation for a real gas reduces to:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module