Van der Waals equation of state is obeyed by | vedclass

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

Question

(D) The Van der Waals equation for $1$ mole of a real gas is given by $\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$,where $V_m$ is the molar vol

Vedclass · Accepted Answer

$\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$

Vedclass · Answer

(D) The Van der Waals equation for $1$ mole of a real gas is given by $\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$,where $V_m$ is the molar volume $(V_m = V/n)$.Substituting $V_m = V/n$ into the equation:$\left( P + \frac{a}{(V/n)^2} \right) (V/n - b) = RT$$\left( P + \frac{n^2 a}{V^2} \right) \left( \frac{V - nb}{n} \right) = RT$Multiplying both sides by $n$ gives the expression for $n$ moles:$\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$

Van der Waals equation of state is obeyed by real gases. For $n$ moles of a real gas,the expression will be

Similar Questions

In van der Waals equation of state,the constant $b$ is a measure of

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

What is the compressibility factor $(Z)$ for the deviation of real gases from ideal behavior?

Calculate the compressibility factor $Z$ for $CO_2$,if one mole of it occupies $0.4 \ L$ at $300 \ K$ and $40 \ atm$. Also,comment on the result.

At $S.T.P.$,if the compressibility factor $(Z)$ of a gas is less than $1$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module