In the following compressibility factor (Z) v | vedclass

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

Question

(A) The compressibility factor is defined as $Z = \frac{p V}{n R T}$.For an ideal gas,$Z = 1$ at all temperatures and pressures.For a real gas,$Z = \f

Vedclass · Accepted Answer

the molar volume of $CH_{4}$ is less than its molar volume in the ideal state

Vedclass · Answer

(A) The compressibility factor is defined as $Z = \frac{p V}{n R T}$.For an ideal gas,$Z = 1$ at all temperatures and pressures.For a real gas,$Z = \frac{p V_{real}}{n R T} \quad (i)$.If the gas shows ideal behaviour,then $V_{ideal} = \frac{n R T}{p}$,which implies $\frac{p}{n R T} = \frac{1}{V_{ideal}} \quad (ii)$.Substituting $(ii)$ into $(i)$,we get $Z = \frac{V_{real}}{V_{ideal}}$.Considering molar volume,$Z = \frac{(V_{m})_{real}}{(V_{m})_{ideal}}$.From the given graph,$Z Therefore,$\frac{(V_{m})_{real}}{(V_{m})_{ideal}} < 1$,which implies $(V_{m})_{CH_{4}, real} < (V_{m})_{CH_{4}, ideal}$.

In the following compressibility factor $(Z)$ versus pressure graph at $300 \, K$,the compressibility of $CH_{4}$ at pressure $< 200 \, bar$ deviates from ideal behaviour because

Similar Questions

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:

$A$ real gas would be most likely to obey the ideal gas laws at

What kind of deviation will $2 \, mol$ of $N_2$ gas show in a $50 \, L$ container at $STP$?

What is Boyle's temperature? Give its relation with the value of $Z$.

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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module